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The open mapping theorem states that a surjective bounded linear operator $T:X\to Y$ between Banach spaces is an open map: $T(U)$ is an open set in $Y$ whenever $U$ is an open set in $X$. Equivalently, there is $\delta >0$ such thatwhere both displayed balls are open. In particular, a bijective bounded linear operator between Banach spaces has a bounded inverse, the bounded inverse theorem.

Put $C=R/(1-k)$. The hypothesis scales to every residual $r\in F$: if $r\ne 0$, apply it to $r/\Vert r\Vert$ and multiply the resulting vector by $\Vert r\Vert$ to obtain an approximate preimage of norm at most $R\Vert r\Vert$ with error at most $k\Vert r\Vert$. For $r=0$, choose zero.

Start with $r_0=y$. Inductively choose $u_j\in E$ so thatThen $\Vert r_j\Vert \le k^j\Vert y\Vert$ and ${\sum }_j\Vert u_j\Vert \le R\Vert y\Vert /(1-k)$. Completeness of the Banach space $E$ gives $x={\sum }_j{u}_j$. Boundedness of $T$ givesThusThis geometric correction for approximate surjectivity uses no completeness assumption on $F$: every infinite sum was formed in $E$.

To establish the completeness forced by uniformly bounded lifting, let $(y_n)$ be a Cauchy sequence in $F$. Choose a subsequence $(y_{n_j})$ with $\Vert y_{n_{j+1}}-y_{n_j}\Vert \le 2^{-j}$. Lift $y_{n_1}$ to $v_0\in E$ and each difference to $v_j\in E$ with $T{v}_j=y_{n_{j+1}}-y_{n_j}$ and $\Vert v_j\Vert \le C{2}^{-j}$. The series $v_0+{\sum }_j{v}_j$ converges in $E$, and its image under $T$ is the limit of the subsequence. The original Cauchy sequence has that same limit, by the triangle inequality. Hence $F$ is complete.

The open mapping theorem states that a bounded surjective linear map between Banach spaces is open. In particular, a bounded linear bijection has a bounded inverse. This is the theorem the question permits us to state without proof.

For the closed graph theorem, suppose $T:E\to F$ is everywhere-defined and linear with closed graph $G(T)$. The product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$ makes $E\times F$ a Banach space, so its closed subspace $G(T)$ is complete. Projection $\pi :G(T)\to E$, $(x,Tx)\mapsto x$, is a bounded linear bijection. Its inverse is bounded by the open mapping theorem; thereforeIt follows that $T$ is bounded, proving the closed graph theorem rather than assuming it.

For the final clause, $S$ is the separating space of a linear map. It contains zero, and if $x_n\to 0$, $T{x}_n\to y$ and $z_n\to 0$, $T{z}_n\to w$, then $\alpha x_n+\beta z_n\to 0$ and its image tends to $\alpha y+\beta w$. Thus $S$ is a vector subspace. If $y_j\in S$ and $y_j\to y$, choose one vector $w_j$ from a witnessing sequence for $y_j$ withThen $w_j\to 0$ and $T{w}_j\to y$, so $y\in S$: it is closed. If $T$ is continuous, every defining image limit is zero, hence $S=\{0\}$. Conversely, if $S=\{0\}$ and $x_n\to x$, $T{x}_n\to y$, then $x_n-x\to 0$ and $T(x_n-x)\to y-Tx\in S$, so $y=Tx$. The graph is sequentially closed and therefore closed in the normed, metrizable product. The proved closed graph theorem now gives continuity. ConsequentlyThe original PDF places $S$ in $F$; the TeX's first reference to a subset of $E$ is a transcription error.

The open mapping lemma says: if a bounded linear map between Banach spaces satisfies $\overline{T(B_X)}\supset r{B}_Y$, then $T(B_X)$ contains a (possibly smaller) ball. Given $y$, repeatedly choose corrections $x_n$ whose residuals shrink geometrically; completeness makes $\sum x_n$ converge and maps it to $y$. Baire category applied to $Y={\bigcup }_{n}T(n{B}_X)$ supplies the closure hypothesis for a surjective map. This proves the Open mapping theorem.

If $\Vert Tx\Vert \ge \delta \Vert x\Vert$ and $T{x}_n\to y$, then $x_n$ is Cauchy, so $x_n\to x$ and $y=Tx$; the range is closed. Conversely a closed range is Banach, and $T:X\to T(X)$ has bounded inverse by the Open mapping theorem, giving the lower bound. This is the closed-range criterion for an injection.

Surjectivity and Open mapping theorem give $r{B}_Y\subset T(B_X)$. If $\Vert S-T\Vert <r$, iteratively solve with $T$ and correct the error $(S-T)x$; the errors shrink geometrically. Completeness yields an exact preimage under $S$.