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Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r1300701.png" /> be a [[Hilbert space|Hilbert space]] of functions defined on an abstract set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r1300702.png" />.
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Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r1300703.png" /> denote the [[Inner product|inner product]] and let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r1300704.png" /> be the norm in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r1300705.png" />. The space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r1300706.png" /> is called a reproducing-kernel Hilbert space if there exists a function <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r1300707.png" />, the [[Reproducing kernel|reproducing kernel]], on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r1300708.png" /> such that:
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1) <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r1300709.png" /> for any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007010.png" />;
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Let $H$ be a [[Hilbert space|Hilbert space]] of functions defined on an abstract set $E$.
  
2) <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007011.png" /> for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007012.png" /> (the reproducing property). From this definition it follows that the value <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007013.png" /> at a point <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007014.png" /> is a continuous [[Linear functional|linear functional]] in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007015.png" />:
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Let $( f , g )$ denote the [[Inner product|inner product]] and let $\| f \| = ( f , f ) ^ { 1 / 2 }$ be the norm in $H$. The space $H$ is called a reproducing-kernel Hilbert space if there exists a function $K ( x , y )$, the [[Reproducing kernel|reproducing kernel]], on $E \times E$ such that:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007016.png" /></td> </tr></table>
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1) $K ( x , y ) \in H$ for any $y \in E$;
  
The converse is also true. The following theorem holds: A Hilbert space of functions on a set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007017.png" /> is a reproducing-kernel Hilbert space if and only if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007018.png" /> for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007019.png" />.
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2) $( f ( . ) , K (. , y ) ) = f ( y )$ for all $f \in H$ (the reproducing property). From this definition it follows that the value $f ( y )$ at a point $y \in E$ is a continuous [[Linear functional|linear functional]] in $H$:
  
By the [[Riesz theorem(2)|Riesz theorem]], the above assumption implies the existence of a linear functional <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007020.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007021.png" />. By the construction, the kernel <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007022.png" /> is the reproducing kernel for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007023.png" />.
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\begin{equation*} | f ( y ) | \leq c ( y ) \| f \| , c ( y ) : = \| K (\, .\, , y ) \|. \end{equation*}
  
An example of a construction of a reproducing-kernel Hilbert space is the rigged triple of Hilbert spaces <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007024.png" />, which is defined as follows [[#References|[a5]]] (cf. also [[Rigged Hilbert space|Rigged Hilbert space]]). Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007025.png" /> be a Hilbert space of functions, let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007026.png" /> be a linear densely defined [[Self-adjoint operator|self-adjoint operator]] on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007027.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007028.png" /> (the eigenvalues <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007029.png" /> are counted according to their multiplicities) and assume that
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The converse is also true. The following theorem holds: A Hilbert space of functions on a set $E$ is a reproducing-kernel Hilbert space if and only if $| f ( y ) | \leq c ( y ) \| f \|$ for all $y \in E$.
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007030.png" /></td> </tr></table>
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By the [[Riesz theorem(2)|Riesz theorem]], the above assumption implies the existence of a linear functional $K (\ \cdot \ , y )$ such that $f ( y ) = ( f , K (\, .\, , y ) )$. By the construction, the kernel $K ( x , y )$ is the reproducing kernel for $H$.
  
Define <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007031.png" /> to be the Hilbert space with inner product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007032.png" />. <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007033.png" /> is the completion of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007034.png" /> in the norm <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007035.png" />. Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007036.png" /> be the dual space to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007037.png" /> with respect to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007038.png" />. Then the inner product in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007039.png" /> is defined by the formula <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007040.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007041.png" />, equipped with the inner product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007042.png" />, is a Hilbert space.
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An example of a construction of a reproducing-kernel Hilbert space is the rigged triple of Hilbert spaces $H _ { + } \subset H _ { 0 } \subset H _ { - }$, which is defined as follows [[#References|[a5]]] (cf. also [[Rigged Hilbert space|Rigged Hilbert space]]). Let $H _ { 0 }$ be a Hilbert space of functions, let $A > 0$ be a linear densely defined [[Self-adjoint operator|self-adjoint operator]] on $H _ { 0 }$, $A \varphi _ { j } = \lambda _ { j } \varphi _ { j }$ (the eigenvalues $\lambda_j > 0$ are counted according to their multiplicities) and assume that
  
Define <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007043.png" />, where the overline stands for complex conjugation. For any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007044.png" />, one has <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007045.png" />. Indeed,
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\begin{equation*} \Lambda ^ { 2 } : = \sum _ { j = 1 } ^ { \infty } \lambda _ { j } < \infty , | \varphi _ { j } ( x ) | < c , \forall j , x. \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007046.png" /></td> </tr></table>
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Define $H _ { - } \supset H _ { 0 }$ to be the Hilbert space with inner product $( u , v ) _ { - } = ( A ^ { 1 / 2 } u , A ^ { 1 / 2 } v ) _ { 0 }$. $H_-$ is the completion of $H _ { 0 }$ in the norm $\| u \| : = ( u , u ) ^ { 1 / 2 }_ { - }$. Let $H _ { + } \subset H _ { 0 }$ be the dual space to $H_-$ with respect to $H _ { 0 }$. Then the inner product in $H _ { + }$ is defined by the formula $( u , v ) _ { + } : = ( A ^ { - 1 / 2 } u , A ^ { - 1 / 2 } v ) _ { 0 },$ and $H _ { + } = R ( A ^ { 1 / 2 } )$, equipped with the inner product $( u , v )_+$, is a Hilbert space.
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007047.png" /></td> </tr></table>
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Define $B ( x , y ) = \sum _ { j = 1 } ^ { \infty } \lambda _ { j } \overline { \varphi _ { j } ( x ) } \varphi _ { j } ( y )$, where the overline stands for complex conjugation. For any $y$, one has $B ( x , y ) \in H _ { + }$. Indeed,
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\begin{equation*} \| B ( x , y ) \| _ { + } \leq c \sum _ { j = 1 } ^ { \infty } \| \lambda _j \varphi_j ( x ) \| _ { + } = \end{equation*}
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 +
\begin{equation*} = c \sum _ { j = 1 } ^ { \infty } ( A \varphi _ { j } , \varphi _ { j } ) _ { 0 } = c \Lambda ^ { 2 } < \infty. \end{equation*}
  
 
Furthermore,
 
Furthermore,
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007048.png" /></td> </tr></table>
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\begin{equation*} ( u , B ( x , y ) ) _ { + } = ( u , A ^ { - 1 } B ) = u ( y ), \end{equation*}
  
so that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007049.png" /> is the [[Reproducing kernel|reproducing kernel]] in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007050.png" />. Moreover <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007051.png" />, where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007052.png" /> is a constant independent of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007053.png" />. Indeed, if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007054.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007055.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007056.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007057.png" /> <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007058.png" />, and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007059.png" />.
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so that $B ( x , y )$ is the [[Reproducing kernel|reproducing kernel]] in $H _ { + }$. Moreover $| u ( y ) | \leq c ( y ) \| u \|_+$, where $c ( y ) > 0$ is a constant independent of $u \in H _ { + }$. Indeed, if $u _ { j } : = ( u , \varphi _ { j } ) _ { 0 }$ and $u \in H _ { + }$, then $u = A ^ { 1 / 2 } v$, $v \in H _ { 0 }$ $v _ { j } \lambda _ { j } ^ { 1 / 2 } = u _ { j }$, and $| u ( y ) | \leq \sum _ { j = 1 } ^ { \infty } | u _ { j } \varphi _ { j } ( y ) | \leq c \Lambda \| v \| = c \Lambda \| u \| _ { + }$.
  
Thus <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007060.png" /> is a reproducing kernel Hilbert space with the reproducing kernel <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007061.png" /> defined above. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007062.png" /> is a function on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007063.png" /> such that
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Thus $H _ { + }$ is a reproducing kernel Hilbert space with the reproducing kernel $B ( x , y )$ defined above. If $K ( x , y )$ is a function on $E \times E$ such that
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007064.png" /></td> <td valign="top" style="width:5%;text-align:right;">(a1)</td></tr></table>
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\begin{equation} \tag{a1} \sum _ { i ,\, j = 1 } ^ { n } K ( x _ { i } , x _ { j } ) t _ { j } \overline { t } _ { i } \geq 0 ,\, \forall t \in \mathbf{C} ^ { n } ,\, \forall x _ { i } \in E, \end{equation}
  
then one can define a [[Pre-Hilbert space|pre-Hilbert space]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007065.png" /> of functions of the form
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then one can define a [[Pre-Hilbert space|pre-Hilbert space]] $H ^ { 0 }$ of functions of the form
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007066.png" /></td> </tr></table>
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\begin{equation*} f ( x ) : = \sum _ { j = 1 } ^ { J } K ( x , y_j ) c_j , c_j =\text{const.} \end{equation*}
  
The inner product of two functions from <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007067.png" /> is defined by
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The inner product of two functions from $H ^ { 0 }$ is defined by
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007068.png" /></td> </tr></table>
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\begin{equation*} ( f , g ) : = \left( \sum _ { j = 1 } ^ { J } K ( x , y _ { j } ) c _ { j } , \sum _ { m = 1 } ^ { M } K ( x , z _ { m } ) \beta _ { m } \right) = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007069.png" /></td> </tr></table>
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\begin{equation*} = \sum _ { j , m } K ( z _ { m } , y _ { j } ) c _ { j } \overline { \beta _ { m } }. \end{equation*}
  
This definition makes sense because of (a1) and because of reproducing property 2). In particular, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007070.png" />, as follows from (a1), and if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007071.png" /> then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007072.png" />, as follows from property 2).
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This definition makes sense because of (a1) and because of reproducing property 2). In particular, $( f , f ) \geq 0$, as follows from (a1), and if $( f , f ) = 0$ then $f = 0$, as follows from property 2).
  
 
Indeed,
 
Indeed,
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007073.png" /></td> </tr></table>
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\begin{equation*} ( f ( x ) , K ( x , y ) ) = \left( \sum _ { j = 1 } ^ { J } K ( x , y _ { j } ) c _ { j } , K ( x , y ) \right) = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007074.png" /></td> </tr></table>
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\begin{equation*} = \sum _ { j = 1 } ^ { J } K ( y , y _ { j } ) c _ { j } = f ( y ) , \forall y \in E. \end{equation*}
  
Thus, if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007075.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007076.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007077.png" />, so <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007078.png" /> as claimed.
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Thus, if $( f , f ) = 0$, then $\| f \| = 0$ and $| f ( y ) | \leq \| f \| \| K ( x , y ) \| = 0$, so $f ( y ) = 0$ as claimed.
  
Denote by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007079.png" /> the completion of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007080.png" /> in the norm <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007081.png" />. Then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007082.png" /> is a reproducing-kernel Hilbert space and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007083.png" /> is its reproducing kernel.
+
Denote by $H$ the completion of $H ^ { 0 }$ in the norm $\| f \|$. Then $H$ is a reproducing-kernel Hilbert space and $K ( x , y )$ is its reproducing kernel.
  
A reproducing-kernel Hilbert space is uniquely defined by its reproducing kernel. Indeed, if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007084.png" /> is another reproducing-kernel Hilbert space with the same reproducing kernel <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007085.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007086.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007087.png" /> is dense in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007088.png" />: If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007089.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007090.png" /> for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007091.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007092.png" />. Using this and the equality <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007093.png" /> for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007094.png" />, one can check that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007095.png" /> and vice versa, so <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007096.png" />, that is, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007097.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007098.png" /> consist of the same set of elements. Moreover, the norms in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r13007099.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070100.png" /> are equal. Indeed, take an arbitrary <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070101.png" /> and a sequence <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070102.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070103.png" />. Then
+
A reproducing-kernel Hilbert space is uniquely defined by its reproducing kernel. Indeed, if $H _ { 1 }$ is another reproducing-kernel Hilbert space with the same reproducing kernel $K ( x , y )$, then $H ^ { 0 } \subset H _ { 1 }$ and $H ^ { 0 }$ is dense in $H _ { 1 }$: If $f \in H _ { 1 }$, $0 = ( f , K ( x , y ) ) _ { H _ { 1 } } = f ( y )$ for all $y \in E$, then $f \equiv 0$. Using this and the equality $( f , g ) _ { H _ { 1 } } = ( f , g ) _ { H }$ for all $f , g \in H ^ { 0 }$, one can check that $H \subset H _ { 1 }$ and vice versa, so $H = H _ { 1 }$, that is, $H$ and $H _ { 1 }$ consist of the same set of elements. Moreover, the norms in $H$ and $H _ { 1 }$ are equal. Indeed, take an arbitrary $f \in H _ { 1 }$ and a sequence $f _ { n } \in H ^ { 0 }$, $\| f _ { n } - f \| _ { 1 } \rightarrow 0$. Then
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070104.png" /></td> </tr></table>
+
\begin{equation*} \| f \| _ { 1 } ^ { 2 } = \operatorname { lim } _ { n \rightarrow \infty } \| f _ { n } \| _ { 1 } ^ { 2 } = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070105.png" /></td> </tr></table>
+
\begin{equation*} = \operatorname { lim } _ { n \rightarrow 0 } \left( \sum _ { j_n = 1 } ^ { J _ { n } } K ( x , y _ { j _n } ) c _ { j _n } , \sum _ { m_n = 1 } ^ { J _ { n } } K ( x , y _ { m_n } ) c _ { m_n } \right) _ { 1 } = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070106.png" /></td> </tr></table>
+
\begin{equation*} = \sum _ { j _ { n } ,\, m _ { n } } ^ { J _ { n } } K ( y _ { m _ { n } } , y _ { j _ { n } } ) c _ { j _ { n } } \overline { c_{m _ { n }}} = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070107.png" /></td> </tr></table>
+
\begin{equation*} = \operatorname { lim } _ { n \rightarrow \infty } ( f _ { n } , f _ { n } ) = \| f \| ^ { 2 }. \end{equation*}
  
Thus, the norms in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070108.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070109.png" /> are equal, as claimed, and so are the inner products (by the [[polarization identity]]).
+
Thus, the norms in $H _ { 1 }$ and $H$ are equal, as claimed, and so are the inner products (by the [[polarization identity]]).
  
Define a [[Linear operator|linear operator]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070110.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070111.png" />, where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070112.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070113.png" /> is the range <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070114.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070115.png" />, which will be equipped with the structure of a Hilbert space below:
+
Define a [[Linear operator|linear operator]] $L : {\cal H} \rightarrow H$, $D ( L ) = \mathcal{H}$, where $\mathcal{H} = L ^ { 2 } ( T , d m )$ and $H$ is the range $R ( L )$ of $L$, which will be equipped with the structure of a Hilbert space below:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070116.png" /></td> <td valign="top" style="width:5%;text-align:right;">(a2)</td></tr></table>
+
\begin{equation} \tag{a2} f ( x ) = L F : = \int _ { T } F ( t ) \overline { h ( t , x ) } d m ( t ). \end{equation}
  
Here, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070117.png" /> is a domain in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070118.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070119.png" /> is a positive [[Measure|measure]] on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070120.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070121.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070122.png" /> for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070123.png" />, and it is assumed that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070124.png" /> is injective, that is, the system <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070125.png" /> is total in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070126.png" /> (cf. also [[Total set|Total set]]).
+
Here, $T$ is a domain in ${\bf R} ^ { n }$ and $m$ is a positive [[Measure|measure]] on $T$, $m ( T ) < \infty$, $h ( t , x ) \in \mathcal{H}$ for all $x \in E$, and it is assumed that $L$ is injective, that is, the system $\{ h ( t , x ) \}_{\forall x \in E}$ is total in $\mathcal{H}$ (cf. also [[Total set|Total set]]).
  
 
Define
 
Define
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070127.png" /></td> <td valign="top" style="width:5%;text-align:right;">(a3)</td></tr></table>
+
\begin{equation} \tag{a3} K ( x , y ) : = \int _ { T } h ( t , y ) \overline { h ( t , x ) } d m ( t ) = \end{equation}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070128.png" /></td> </tr></table>
+
\begin{equation*} = ( h (\, . \,  , y ) , h (\, . \, , x ) ) _ { \mathcal{H} }. \end{equation*}
  
This kernel clearly satisfies condition (a1) and therefore is a reproducing kernel for the reproducing-kernel Hilbert space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070129.png" /> which it generates. Clearly <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070130.png" /> for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070131.png" />. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070132.png" />, that is, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070133.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070134.png" />, then
+
This kernel clearly satisfies condition (a1) and therefore is a reproducing kernel for the reproducing-kernel Hilbert space $H _ { K }$ which it generates. Clearly $K ( x , y ) \in H$ for all $y \in E$. If $f \in H$, that is, $f = L F$, $f \in \mathcal{H}$, then
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070135.png" /></td> </tr></table>
+
\begin{equation*} ( f ( \cdot ) , K ( \cdot , y ) ) _ { H } = ( L F , K ( \cdot , y ) ) _ { H } = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070136.png" /></td> </tr></table>
+
\begin{equation*} = ( ( F ( \cdot ) , h ( \cdot , x ) ) _ { \mathcal{H} } , ( h ( \text{..} , y ) , h ( \text{..} , x ) ) _ { \mathcal{H} } ) _ { H } = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070137.png" /></td> </tr></table>
+
\begin{equation*} = ( F ( . ) , ( h ( .. , y ) , ( h (. , x ) , h ( .. , x ) ) _ { H } ) _ {\cal H } ) _ {\cal H } = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070138.png" /></td> </tr></table>
+
\begin{equation*} = ( F ( \cdot ) , h ( \cdot , y ) ) _ { \mathcal{H} } = f ( y ), \end{equation*}
  
if one equips <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070139.png" /> with the inner product such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070140.png" />. This requirement is formally equivalent to the following one: <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070141.png" />, where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070142.png" />, so that the distributional kernel <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070143.png" /> is not the usual [[Delta-function|delta-function]], but the one which acts by the rule
+
if one equips $H$ with the inner product such that $( f , g ) _ { H } = ( F , G ) _ { \mathcal{H} }$. This requirement is formally equivalent to the following one: $( h ( s , x ) , h ( t , x ) ) _ { H } = \delta _ { m } ( t - s )$, where $( h ( s , y ) , \delta _ { m } ( t - s ) ) _ { \mathcal{H} } = h ( t , y )$, so that the distributional kernel $\delta _ { m } ( t - s )$ is not the usual [[Delta-function|delta-function]], but the one which acts by the rule
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070144.png" /></td> </tr></table>
+
\begin{equation*} \int _ { T } d m ( t ) F ( t ) \int _ { T } d m ( s ) G ( s ) \delta _ { m } ( t - s ) = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070145.png" /></td> </tr></table>
+
\begin{equation*} = \int _ { T } d m ( t ) F ( t ) G ( t ), \end{equation*}
  
and formally one has <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070146.png" />.
+
and formally one has $\int _ { T } d m ( s ) G ( s ) \delta _ { m } ( t - s ) = G ( t )$.
  
With the inner product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070147.png" />, the linear set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070148.png" /> becomes a Hilbert space:
+
With the inner product $( f , g ) _ { H }$, the linear set $R ( L )$ becomes a Hilbert space:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070149.png" /></td> <td valign="top" style="width:5%;text-align:right;">(a4)</td></tr></table>
+
\begin{equation} \tag{a4} ( f , g ) _ { H } = ( L F , L G ) _ { H } = \end{equation}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070150.png" /></td> </tr></table>
+
\begin{equation*} = \int_{T} \int _ { T } d m ( t ) d m ( s ) F ( t ) \overline { G ( s ) } ( h ( s , x ) , h ( t , x ) ) _ { H } = \end{equation*}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070151.png" /></td> </tr></table>
+
\begin{equation*} = \int _ { T } d m ( t ) F ( t ) \overline { G ( t ) } = ( F , G ) _ { \mathcal{H} }. \end{equation*}
  
Thus, this inner product makes <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070152.png" /> an [[Isometric operator|isometric operator]] defined on all of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070153.png" /> and makes <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070154.png" /> a (complete) Hilbert space, namely <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070155.png" />, a reproducing-kernel Hilbert space. Since <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070156.png" /> is assumed injective, it follows that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070157.png" /> is defined on all of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070158.png" /> and, since <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070159.png" /> is complete in the norm <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070160.png" />, one concludes that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070161.png" /> is continuous (by the Banach theorem). Consequently, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070162.png" /> is a co-isometry, that is, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070163.png" />, where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070164.png" /> is the [[Adjoint operator|adjoint operator]] to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070165.png" />. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070166.png" />, then one can write an inversion formula for the linear transform <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070167.png" /> similar to the well-known inversion formula for the [[Fourier transform|Fourier transform]]. Formally one has:
+
Thus, this inner product makes $L$ an [[Isometric operator|isometric operator]] defined on all of $\mathcal{H}$ and makes $H = R ( L )$ a (complete) Hilbert space, namely $H = H _ { K }$, a reproducing-kernel Hilbert space. Since $L$ is assumed injective, it follows that $L ^ { - 1 }$ is defined on all of $R ( L ) = H$ and, since $H$ is complete in the norm $\|\, f \| = ( f , f ) ^ { 1 / 2 } _ { H }$, one concludes that $L ^ { - 1 }$ is continuous (by the Banach theorem). Consequently, $L$ is a co-isometry, that is, $L ^ { * } = L ^ { - 1 }$, where $L ^ { * }$ is the [[Adjoint operator|adjoint operator]] to $L$. If $L ^ { * } = L ^ { - 1 }$, then one can write an inversion formula for the linear transform $L$ similar to the well-known inversion formula for the [[Fourier transform|Fourier transform]]. Formally one has:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070168.png" /></td> </tr></table>
+
\begin{equation*} f ( x ) = ( F ( t ) , h ( t , x ) ) _ { \cal H } , ( f ( x ) , h ( s , x ) ) _ { H } = F ( s ). \end{equation*}
  
The space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070169.png" /> is the reproducing-kernel Hilbert space generated by kernel (a3) which is the reproducing kernel for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070170.png" />. The above formal inversion formulas may be of practical interest if the norm in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070171.png" /> is a standard one. In this case the second formula should be suitably interpreted, since <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070172.png" /> is defined at <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070173.png" />-almost all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070174.png" />.
+
The space $H = H _ { K }$ is the reproducing-kernel Hilbert space generated by kernel (a3) which is the reproducing kernel for $H$. The above formal inversion formulas may be of practical interest if the norm in $H$ is a standard one. In this case the second formula should be suitably interpreted, since $F ( s )$ is defined at $m$-almost all $s$.
  
In [[#References|[a6]]] it is claimed that the characterization of the range of the linear operator <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070175.png" />, defined in (a3), can be given as follows: <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070176.png" />, where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070177.png" /> is the reproducing-kernel Hilbert space generated by kernel (a3).
+
In [[#References|[a6]]] it is claimed that the characterization of the range of the linear operator $L$, defined in (a3), can be given as follows: $R ( L ) = H _ { K }$, where $H _ { K }$ is the reproducing-kernel Hilbert space generated by kernel (a3).
  
However, in fact such a characterization does not give, in general, practically useful necessary and sufficient conditions for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070178.png" /> because the norm in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070179.png" /> is not defined in terms of standard norms such as Sobolev or Hölder ones (see [[#References|[a3]]], [[#References|[a4]]], [[#References|[a5]]]). However, when the norm in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r130/r130070/r130070180.png" /> is equivalent to a standard norm, the above characterization becomes efficient (see [[#References|[a3]]], [[#References|[a4]]], [[#References|[a5]]], and also [[#References|[a6]]]).
+
However, in fact such a characterization does not give, in general, practically useful necessary and sufficient conditions for $f ( x ) \in R ( L )$ because the norm in $H _ { K }$ is not defined in terms of standard norms such as Sobolev or Hölder ones (see [[#References|[a3]]], [[#References|[a4]]], [[#References|[a5]]]). However, when the norm in $H _ { K }$ is equivalent to a standard norm, the above characterization becomes efficient (see [[#References|[a3]]], [[#References|[a4]]], [[#References|[a5]]], and also [[#References|[a6]]]).
  
 
Many concrete examples of reproducing-kernel Hilbert spaces can be found in [[#References|[a1]]], [[#References|[a2]]] and [[#References|[a6]]].
 
Many concrete examples of reproducing-kernel Hilbert spaces can be found in [[#References|[a1]]], [[#References|[a2]]] and [[#References|[a6]]].
Line 122: Line 130:
  
 
====References====
 
====References====
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  N. Aronszajn,  "Theory of reproducing kernels"  ''Trans. Amer. Math. Soc.'' , '''68'''  (1950)  pp. 337–404</TD></TR><TR><TD valign="top">[a2]</TD> <TD valign="top">  S. Bergman,  "The kernel function and conformal mapping" , Amer. Math. Soc.  (1950)</TD></TR><TR><TD valign="top">[a3]</TD> <TD valign="top">  A.G. Ramm,  "On the theory of reproducing kernel Hilbert spaces"  ''J. Inverse Ill-Posed Probl.'' , '''6''' :  5  (1998)  pp. 515–520</TD></TR><TR><TD valign="top">[a4]</TD> <TD valign="top">  A.G. Ramm,  "On Saitoh's characterization of the range of linear transforms"  A.G. Ramm (ed.) , ''Inverse Problems, Tomography and Image Processing'' , Plenum  (1998)  pp. 125–128</TD></TR><TR><TD valign="top">[a5]</TD> <TD valign="top">  A.G. Ramm,  "Random fields estimation theory" , Longman/Wiley  (1990)</TD></TR><TR><TD valign="top">[a6]</TD> <TD valign="top">  S. Saitoh,  "Integral transforms, reproducing kernels and their applications" , ''Pitman Res. Notes'' , Longman  (1997)</TD></TR><TR><TD valign="top">[a7]</TD> <TD valign="top">  L. Schwartz,  "Sous-espaces hilbertiens d'espaces vectoriels topologiques et noyaux associès (noyaux reproduisants)"  ''J. Anal. Math.'' , '''13'''  (1964)  pp. 115–256</TD></TR></table>
+
<table><tr><td valign="top">[a1]</td> <td valign="top">  N. Aronszajn,  "Theory of reproducing kernels"  ''Trans. Amer. Math. Soc.'' , '''68'''  (1950)  pp. 337–404</td></tr><tr><td valign="top">[a2]</td> <td valign="top">  S. Bergman,  "The kernel function and conformal mapping" , Amer. Math. Soc.  (1950)</td></tr><tr><td valign="top">[a3]</td> <td valign="top">  A.G. Ramm,  "On the theory of reproducing kernel Hilbert spaces"  ''J. Inverse Ill-Posed Probl.'' , '''6''' :  5  (1998)  pp. 515–520</td></tr><tr><td valign="top">[a4]</td> <td valign="top">  A.G. Ramm,  "On Saitoh's characterization of the range of linear transforms"  A.G. Ramm (ed.) , ''Inverse Problems, Tomography and Image Processing'' , Plenum  (1998)  pp. 125–128</td></tr><tr><td valign="top">[a5]</td> <td valign="top">  A.G. Ramm,  "Random fields estimation theory" , Longman/Wiley  (1990)</td></tr><tr><td valign="top">[a6]</td> <td valign="top">  S. Saitoh,  "Integral transforms, reproducing kernels and their applications" , ''Pitman Res. Notes'' , Longman  (1997)</td></tr><tr><td valign="top">[a7]</td> <td valign="top">  L. Schwartz,  "Sous-espaces hilbertiens d'espaces vectoriels topologiques et noyaux associès (noyaux reproduisants)"  ''J. Anal. Math.'' , '''13'''  (1964)  pp. 115–256</td></tr></table>

Latest revision as of 16:39, 2 February 2024

Let $H$ be a Hilbert space of functions defined on an abstract set $E$.

Let $( f , g )$ denote the inner product and let $\| f \| = ( f , f ) ^ { 1 / 2 }$ be the norm in $H$. The space $H$ is called a reproducing-kernel Hilbert space if there exists a function $K ( x , y )$, the reproducing kernel, on $E \times E$ such that:

1) $K ( x , y ) \in H$ for any $y \in E$;

2) $( f ( . ) , K (. , y ) ) = f ( y )$ for all $f \in H$ (the reproducing property). From this definition it follows that the value $f ( y )$ at a point $y \in E$ is a continuous linear functional in $H$:

\begin{equation*} | f ( y ) | \leq c ( y ) \| f \| , c ( y ) : = \| K (\, .\, , y ) \|. \end{equation*}

The converse is also true. The following theorem holds: A Hilbert space of functions on a set $E$ is a reproducing-kernel Hilbert space if and only if $| f ( y ) | \leq c ( y ) \| f \|$ for all $y \in E$.

By the Riesz theorem, the above assumption implies the existence of a linear functional $K (\ \cdot \ , y )$ such that $f ( y ) = ( f , K (\, .\, , y ) )$. By the construction, the kernel $K ( x , y )$ is the reproducing kernel for $H$.

An example of a construction of a reproducing-kernel Hilbert space is the rigged triple of Hilbert spaces $H _ { + } \subset H _ { 0 } \subset H _ { - }$, which is defined as follows [a5] (cf. also Rigged Hilbert space). Let $H _ { 0 }$ be a Hilbert space of functions, let $A > 0$ be a linear densely defined self-adjoint operator on $H _ { 0 }$, $A \varphi _ { j } = \lambda _ { j } \varphi _ { j }$ (the eigenvalues $\lambda_j > 0$ are counted according to their multiplicities) and assume that

\begin{equation*} \Lambda ^ { 2 } : = \sum _ { j = 1 } ^ { \infty } \lambda _ { j } < \infty , | \varphi _ { j } ( x ) | < c , \forall j , x. \end{equation*}

Define $H _ { - } \supset H _ { 0 }$ to be the Hilbert space with inner product $( u , v ) _ { - } = ( A ^ { 1 / 2 } u , A ^ { 1 / 2 } v ) _ { 0 }$. $H_-$ is the completion of $H _ { 0 }$ in the norm $\| u \| : = ( u , u ) ^ { 1 / 2 }_ { - }$. Let $H _ { + } \subset H _ { 0 }$ be the dual space to $H_-$ with respect to $H _ { 0 }$. Then the inner product in $H _ { + }$ is defined by the formula $( u , v ) _ { + } : = ( A ^ { - 1 / 2 } u , A ^ { - 1 / 2 } v ) _ { 0 },$ and $H _ { + } = R ( A ^ { 1 / 2 } )$, equipped with the inner product $( u , v )_+$, is a Hilbert space.

Define $B ( x , y ) = \sum _ { j = 1 } ^ { \infty } \lambda _ { j } \overline { \varphi _ { j } ( x ) } \varphi _ { j } ( y )$, where the overline stands for complex conjugation. For any $y$, one has $B ( x , y ) \in H _ { + }$. Indeed,

\begin{equation*} \| B ( x , y ) \| _ { + } \leq c \sum _ { j = 1 } ^ { \infty } \| \lambda _j \varphi_j ( x ) \| _ { + } = \end{equation*}

\begin{equation*} = c \sum _ { j = 1 } ^ { \infty } ( A \varphi _ { j } , \varphi _ { j } ) _ { 0 } = c \Lambda ^ { 2 } < \infty. \end{equation*}

Furthermore,

\begin{equation*} ( u , B ( x , y ) ) _ { + } = ( u , A ^ { - 1 } B ) = u ( y ), \end{equation*}

so that $B ( x , y )$ is the reproducing kernel in $H _ { + }$. Moreover $| u ( y ) | \leq c ( y ) \| u \|_+$, where $c ( y ) > 0$ is a constant independent of $u \in H _ { + }$. Indeed, if $u _ { j } : = ( u , \varphi _ { j } ) _ { 0 }$ and $u \in H _ { + }$, then $u = A ^ { 1 / 2 } v$, $v \in H _ { 0 }$ $v _ { j } \lambda _ { j } ^ { 1 / 2 } = u _ { j }$, and $| u ( y ) | \leq \sum _ { j = 1 } ^ { \infty } | u _ { j } \varphi _ { j } ( y ) | \leq c \Lambda \| v \| = c \Lambda \| u \| _ { + }$.

Thus $H _ { + }$ is a reproducing kernel Hilbert space with the reproducing kernel $B ( x , y )$ defined above. If $K ( x , y )$ is a function on $E \times E$ such that

\begin{equation} \tag{a1} \sum _ { i ,\, j = 1 } ^ { n } K ( x _ { i } , x _ { j } ) t _ { j } \overline { t } _ { i } \geq 0 ,\, \forall t \in \mathbf{C} ^ { n } ,\, \forall x _ { i } \in E, \end{equation}

then one can define a pre-Hilbert space $H ^ { 0 }$ of functions of the form

\begin{equation*} f ( x ) : = \sum _ { j = 1 } ^ { J } K ( x , y_j ) c_j , c_j =\text{const.} \end{equation*}

The inner product of two functions from $H ^ { 0 }$ is defined by

\begin{equation*} ( f , g ) : = \left( \sum _ { j = 1 } ^ { J } K ( x , y _ { j } ) c _ { j } , \sum _ { m = 1 } ^ { M } K ( x , z _ { m } ) \beta _ { m } \right) = \end{equation*}

\begin{equation*} = \sum _ { j , m } K ( z _ { m } , y _ { j } ) c _ { j } \overline { \beta _ { m } }. \end{equation*}

This definition makes sense because of (a1) and because of reproducing property 2). In particular, $( f , f ) \geq 0$, as follows from (a1), and if $( f , f ) = 0$ then $f = 0$, as follows from property 2).

Indeed,

\begin{equation*} ( f ( x ) , K ( x , y ) ) = \left( \sum _ { j = 1 } ^ { J } K ( x , y _ { j } ) c _ { j } , K ( x , y ) \right) = \end{equation*}

\begin{equation*} = \sum _ { j = 1 } ^ { J } K ( y , y _ { j } ) c _ { j } = f ( y ) , \forall y \in E. \end{equation*}

Thus, if $( f , f ) = 0$, then $\| f \| = 0$ and $| f ( y ) | \leq \| f \| \| K ( x , y ) \| = 0$, so $f ( y ) = 0$ as claimed.

Denote by $H$ the completion of $H ^ { 0 }$ in the norm $\| f \|$. Then $H$ is a reproducing-kernel Hilbert space and $K ( x , y )$ is its reproducing kernel.

A reproducing-kernel Hilbert space is uniquely defined by its reproducing kernel. Indeed, if $H _ { 1 }$ is another reproducing-kernel Hilbert space with the same reproducing kernel $K ( x , y )$, then $H ^ { 0 } \subset H _ { 1 }$ and $H ^ { 0 }$ is dense in $H _ { 1 }$: If $f \in H _ { 1 }$, $0 = ( f , K ( x , y ) ) _ { H _ { 1 } } = f ( y )$ for all $y \in E$, then $f \equiv 0$. Using this and the equality $( f , g ) _ { H _ { 1 } } = ( f , g ) _ { H }$ for all $f , g \in H ^ { 0 }$, one can check that $H \subset H _ { 1 }$ and vice versa, so $H = H _ { 1 }$, that is, $H$ and $H _ { 1 }$ consist of the same set of elements. Moreover, the norms in $H$ and $H _ { 1 }$ are equal. Indeed, take an arbitrary $f \in H _ { 1 }$ and a sequence $f _ { n } \in H ^ { 0 }$, $\| f _ { n } - f \| _ { 1 } \rightarrow 0$. Then

\begin{equation*} \| f \| _ { 1 } ^ { 2 } = \operatorname { lim } _ { n \rightarrow \infty } \| f _ { n } \| _ { 1 } ^ { 2 } = \end{equation*}

\begin{equation*} = \operatorname { lim } _ { n \rightarrow 0 } \left( \sum _ { j_n = 1 } ^ { J _ { n } } K ( x , y _ { j _n } ) c _ { j _n } , \sum _ { m_n = 1 } ^ { J _ { n } } K ( x , y _ { m_n } ) c _ { m_n } \right) _ { 1 } = \end{equation*}

\begin{equation*} = \sum _ { j _ { n } ,\, m _ { n } } ^ { J _ { n } } K ( y _ { m _ { n } } , y _ { j _ { n } } ) c _ { j _ { n } } \overline { c_{m _ { n }}} = \end{equation*}

\begin{equation*} = \operatorname { lim } _ { n \rightarrow \infty } ( f _ { n } , f _ { n } ) = \| f \| ^ { 2 }. \end{equation*}

Thus, the norms in $H _ { 1 }$ and $H$ are equal, as claimed, and so are the inner products (by the polarization identity).

Define a linear operator $L : {\cal H} \rightarrow H$, $D ( L ) = \mathcal{H}$, where $\mathcal{H} = L ^ { 2 } ( T , d m )$ and $H$ is the range $R ( L )$ of $L$, which will be equipped with the structure of a Hilbert space below:

\begin{equation} \tag{a2} f ( x ) = L F : = \int _ { T } F ( t ) \overline { h ( t , x ) } d m ( t ). \end{equation}

Here, $T$ is a domain in ${\bf R} ^ { n }$ and $m$ is a positive measure on $T$, $m ( T ) < \infty$, $h ( t , x ) \in \mathcal{H}$ for all $x \in E$, and it is assumed that $L$ is injective, that is, the system $\{ h ( t , x ) \}_{\forall x \in E}$ is total in $\mathcal{H}$ (cf. also Total set).

Define

\begin{equation} \tag{a3} K ( x , y ) : = \int _ { T } h ( t , y ) \overline { h ( t , x ) } d m ( t ) = \end{equation}

\begin{equation*} = ( h (\, . \, , y ) , h (\, . \, , x ) ) _ { \mathcal{H} }. \end{equation*}

This kernel clearly satisfies condition (a1) and therefore is a reproducing kernel for the reproducing-kernel Hilbert space $H _ { K }$ which it generates. Clearly $K ( x , y ) \in H$ for all $y \in E$. If $f \in H$, that is, $f = L F$, $f \in \mathcal{H}$, then

\begin{equation*} ( f ( \cdot ) , K ( \cdot , y ) ) _ { H } = ( L F , K ( \cdot , y ) ) _ { H } = \end{equation*}

\begin{equation*} = ( ( F ( \cdot ) , h ( \cdot , x ) ) _ { \mathcal{H} } , ( h ( \text{..} , y ) , h ( \text{..} , x ) ) _ { \mathcal{H} } ) _ { H } = \end{equation*}

\begin{equation*} = ( F ( . ) , ( h ( .. , y ) , ( h (. , x ) , h ( .. , x ) ) _ { H } ) _ {\cal H } ) _ {\cal H } = \end{equation*}

\begin{equation*} = ( F ( \cdot ) , h ( \cdot , y ) ) _ { \mathcal{H} } = f ( y ), \end{equation*}

if one equips $H$ with the inner product such that $( f , g ) _ { H } = ( F , G ) _ { \mathcal{H} }$. This requirement is formally equivalent to the following one: $( h ( s , x ) , h ( t , x ) ) _ { H } = \delta _ { m } ( t - s )$, where $( h ( s , y ) , \delta _ { m } ( t - s ) ) _ { \mathcal{H} } = h ( t , y )$, so that the distributional kernel $\delta _ { m } ( t - s )$ is not the usual delta-function, but the one which acts by the rule

\begin{equation*} \int _ { T } d m ( t ) F ( t ) \int _ { T } d m ( s ) G ( s ) \delta _ { m } ( t - s ) = \end{equation*}

\begin{equation*} = \int _ { T } d m ( t ) F ( t ) G ( t ), \end{equation*}

and formally one has $\int _ { T } d m ( s ) G ( s ) \delta _ { m } ( t - s ) = G ( t )$.

With the inner product $( f , g ) _ { H }$, the linear set $R ( L )$ becomes a Hilbert space:

\begin{equation} \tag{a4} ( f , g ) _ { H } = ( L F , L G ) _ { H } = \end{equation}

\begin{equation*} = \int_{T} \int _ { T } d m ( t ) d m ( s ) F ( t ) \overline { G ( s ) } ( h ( s , x ) , h ( t , x ) ) _ { H } = \end{equation*}

\begin{equation*} = \int _ { T } d m ( t ) F ( t ) \overline { G ( t ) } = ( F , G ) _ { \mathcal{H} }. \end{equation*}

Thus, this inner product makes $L$ an isometric operator defined on all of $\mathcal{H}$ and makes $H = R ( L )$ a (complete) Hilbert space, namely $H = H _ { K }$, a reproducing-kernel Hilbert space. Since $L$ is assumed injective, it follows that $L ^ { - 1 }$ is defined on all of $R ( L ) = H$ and, since $H$ is complete in the norm $\|\, f \| = ( f , f ) ^ { 1 / 2 } _ { H }$, one concludes that $L ^ { - 1 }$ is continuous (by the Banach theorem). Consequently, $L$ is a co-isometry, that is, $L ^ { * } = L ^ { - 1 }$, where $L ^ { * }$ is the adjoint operator to $L$. If $L ^ { * } = L ^ { - 1 }$, then one can write an inversion formula for the linear transform $L$ similar to the well-known inversion formula for the Fourier transform. Formally one has:

\begin{equation*} f ( x ) = ( F ( t ) , h ( t , x ) ) _ { \cal H } , ( f ( x ) , h ( s , x ) ) _ { H } = F ( s ). \end{equation*}

The space $H = H _ { K }$ is the reproducing-kernel Hilbert space generated by kernel (a3) which is the reproducing kernel for $H$. The above formal inversion formulas may be of practical interest if the norm in $H$ is a standard one. In this case the second formula should be suitably interpreted, since $F ( s )$ is defined at $m$-almost all $s$.

In [a6] it is claimed that the characterization of the range of the linear operator $L$, defined in (a3), can be given as follows: $R ( L ) = H _ { K }$, where $H _ { K }$ is the reproducing-kernel Hilbert space generated by kernel (a3).

However, in fact such a characterization does not give, in general, practically useful necessary and sufficient conditions for $f ( x ) \in R ( L )$ because the norm in $H _ { K }$ is not defined in terms of standard norms such as Sobolev or Hölder ones (see [a3], [a4], [a5]). However, when the norm in $H _ { K }$ is equivalent to a standard norm, the above characterization becomes efficient (see [a3], [a4], [a5], and also [a6]).

Many concrete examples of reproducing-kernel Hilbert spaces can be found in [a1], [a2] and [a6].

The papers [a1] and [a7] are important in this area, the book [a6] contains many references, while [a2] is an earlier book important for the development of the theory of reproducing-kernel Hilbert spaces.

References

[a1] N. Aronszajn, "Theory of reproducing kernels" Trans. Amer. Math. Soc. , 68 (1950) pp. 337–404
[a2] S. Bergman, "The kernel function and conformal mapping" , Amer. Math. Soc. (1950)
[a3] A.G. Ramm, "On the theory of reproducing kernel Hilbert spaces" J. Inverse Ill-Posed Probl. , 6 : 5 (1998) pp. 515–520
[a4] A.G. Ramm, "On Saitoh's characterization of the range of linear transforms" A.G. Ramm (ed.) , Inverse Problems, Tomography and Image Processing , Plenum (1998) pp. 125–128
[a5] A.G. Ramm, "Random fields estimation theory" , Longman/Wiley (1990)
[a6] S. Saitoh, "Integral transforms, reproducing kernels and their applications" , Pitman Res. Notes , Longman (1997)
[a7] L. Schwartz, "Sous-espaces hilbertiens d'espaces vectoriels topologiques et noyaux associès (noyaux reproduisants)" J. Anal. Math. , 13 (1964) pp. 115–256
How to Cite This Entry:
Reproducing-kernel Hilbert space. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Reproducing-kernel_Hilbert_space&oldid=37590
This article was adapted from an original article by A.G. Ramm (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article