# Inverse mapping

(inverse operator) of a single-valued onto mapping (operator) $f: M \to f[M]$

A single-valued mapping $g$ such that \begin{align} g \circ f & = \mathsf{Id}_{X} \quad \text{on} ~ M, \qquad (1) \\ f \circ g & = \mathsf{Id}_{Y} \quad \text{on} ~ f[M], \qquad (2) \end{align} where $M \subseteq X$, $f[M] \subseteq Y$, and $X$ and $Y$ are any sets.

If $g$ satisfies only Condition (1), then it is called a left-inverse mapping of $f$, and if it satisfies only Condition (2), then it is called a right-inverse mapping of $f$. The inverse mapping $f^{-1}$ exists if and only if for each $y \in f[M]$, the inverse image ${f^{\leftarrow}}[\{ y \}]$ consists of just a single element $x \in M$. If $f$ has an inverse mapping $f^{-1}$, then the equation $$f(x) = y \qquad (3)$$ has a unique solution for each $y \in f[M]$. If only a right inverse $f_{R}^{-1}$ exists, then a solution of (3) exists, but its uniqueness is an open question. If only a left inverse $f_{L}^{-1}$ exists, then any solution is unique, assuming that it exists.

If $X$ and $Y$ are vector spaces, and if $A$ is a linear operator from $X$ into $Y$, then $A^{-1}$ is also linear, if it exists. In general, if $X$ and $Y$ are endowed with some kind of structure, it may happen that certain properties of $A$ are also inherited by $A^{-1}$, assuming that it exists. For example:

• If $X$ and $Y$ are Banach spaces, and if $A: X \to Y$ is a closed operator, then $A^{-1}$ is also closed.
• If $\mathcal{H}$ is a Hilbert space and $A: \mathcal{H} \to \mathcal{H}$ is self-adjoint, then $A^{-1}$ is also self-adjoint.
• If $f: \mathbf{R} \to \mathbf{R}$ is an odd function, then $f^{-1}$ is also odd.

The continuity of $A$ does not always imply the continuity of $A^{-1}$ for many important classes of linear operators, e.g., completely-continuous operators. The following are important tests for the continuity of the inverse of a linear operator.

Let $X$ be a finite-dimensional vector space, with a certain basis, and let $A: X \to X$ be given by the matrix $[a_{ij}]$ with respect to this basis. Then $A^{-1}$ exists if and only if $\det([a_{ij}]) \neq 0$ (in this case, $A$ and $A^{-1}$ are automatically continuous).

Let $X$ and $Y$ be Banach spaces, and let $A$ be a continuous linear operator from $X$ into $Y$.

1. If $\| A(x) \| \geq m \| x \|$, where $m > 0$, then $A^{-1}$ exists and is continuous.
2. If $X = Y$ and $\| A \| < 1$, then $(\mathsf{Id} - A)^{-1}$ exists, is continuous, and $$(\mathsf{Id} - A)^{-1} = \sum_{n = 0}^{\infty} A^{n},$$ where the series on the right-hand side converges in the norm of the space $\mathcal{L}(X)$.
3. The operator $A^{-1}$ exists and is continuous on all of $Y$ if and only if the conjugate $A^{*}$ has an inverse that is defined and continuous on $X^{*}$. Here, $(A^{-1})^{*} = (A^{*})^{-1}$.
4. If $A^{-1}$ exists and is continuous, and if $\| A - B \| < \| A^{-1} \|^{-1}$, then $B^{-1}$ also exists, is continuous, and $$B^{-1} = A^{-1} \sum_{n = 0}^{\infty} [(A - B) A^{-1}]^{n}.$$ Therefore, the set of invertible operators is open in $\mathcal{L}(X,Y)$ in the uniform topology of this space.
5. Banach’s Open-Mapping Theorem. If $A$ is a one-to-one mapping of $X$ onto $Y$, then the inverse mapping, which exists, is continuous. This theorem has the following generalization: A one-to-one continuous linear mapping of a fully-complete space $X$ onto a separated barrelled space $Y$ is a topological isomorphism.

The spectral theory of linear operators on a Hilbert space contains a number of results on the existence and continuity of the inverse of a continuous linear operator. For example, if $A$ is self-adjoint and $\lambda$ is not real, then $(A - \lambda \cdot \mathsf{Id})^{-1}$ exists and is continuous.

#### References

 [1] N. Dunford, J.T. Schwartz, “Linear operators. General theory”, 1, Interscience (1958). [2] L.V. Kantorovich, G.P. Akilov, “Functional analysis”, Pergamon (1982). (Translated from Russian) [3] W. Rudin, “Functional analysis”, McGraw-Hill (1979). [4] A.P. Robertson, W.S. Robertson, “Topological vector spaces”, Cambridge Univ. Press (1964).
How to Cite This Entry:
Inverse mapping. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Inverse_mapping&oldid=40159
This article was adapted from an original article by V.I. Sobolev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article