Open mapping theorem (functional analysis)

A bounded surjective linear map between Banach spaces is open. Baire category first puts a ball in the closure of an image; iterative correction removes the closure.

A bounded linear operator $T:X\to Y$ from a Banach space to a normed space either maps onto $Y$ or has meagre set image in $Y$. To prove this, write $T(X)={\bigcup }_{n\ge 1}T(n{\overline{B}}_X)$. If every $\overline{T(n{\overline{B}}_X)}$ has empty interior, each term is a nowhere dense set. Otherwise scaling and geometric correction for approximate surjectivity place a ball about zero inside $T({\overline{B}}_X)$; the vector subspace $T(X)$ then equals $Y$. In particular a nonclosed operator image is a meagre set. A proper closed operator image is also a meagre set, so the closed/meagre alternatives need not be exclusive.

Let $T:E\to F$ be a bounded linear operator, $E$ a Banach space, and $F$ a normed vector space. Suppose each residual $r$ has an approximate preimage of norm at most $R\Vert r\Vert$ leaving error at most $k\Vert r\Vert$, where $0<k<1$. Correct successive residuals to get vectors $x_j$ with $\Vert x_j\Vert \le R{k}^j\Vert y\Vert$. Their absolutely convergent sum in $E$ maps exactly to $y$, and its norm is at most $R\Vert y\Vert /(1-k)$. Completeness of $F$ is not needed for this argument.

If a bounded linear operator from a Banach space onto a normed vector space admits the displayed lift for every $y$, then its target is complete. Given a Cauchy sequence, choose a subsequence with summable successive differences. Lift those differences to a summable sequence in the source, and use its convergent sum to produce a limit of the target subsequence. The original Cauchy sequence has the same limit.

A continuous surjective linear map between Fréchet spaces is an open map. The Baire category theorem first shows that the closure of the image of every convex balanced zero-neighbourhood is a zero-neighbourhood: the target is the countable union of scalar multiples of this image, and an interior point of its closure can be translated to zero using convexity and symmetry.

To remove the closure, fix a domain zero-neighbourhood $U$ and choose convex balanced zero-neighbourhoods $U_n$ so that every series ${\sum }_n{u}_n$, $u_n\in U_n$, converges to a point of $U$. With an increasing defining sequence of seminorms $p_j$, one can ensure $p_j(u_n)<2^{-n}$ for $j\le n$ and make the finitely many seminorms defining $U$ summable within its bounds. Completeness gives convergence. Choose target zero-neighbourhoods $V_n\subset \overline{A(U_n)}$ shrinking to zero. For $y\in V_1$, choose $u_n\in U_n$ successively so that $y-A{\sum }_{j\le n}{u}_j\in V_{n+1}$. Density permits each correction. Continuity then gives $y=A{\sum }_n{u}_n\in A(U)$, proving openness.

Let $T:X\to X$ be a bounded linear operator on a Banach space, and let $B$ be the open unit ball. If $B\subseteq \overline{T(B)}$, then $B\subseteq T(B)$. Scale the density statement and correct a target by a geometrically decreasing sequence of approximate preimages. Completeness makes their sum converge inside $B$, and continuity of $T$ makes its image equal the target.

A bounded bijective linear map between Banach spaces has a bounded inverse. This follows because an open bijection has continuous inverse.