Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2007 / A / 3 / Solution

A linear operator has closed graph when $\{(x,Tx):x\in X\}$ is closed in $X\times Y$. The closed graph theorem states that an everywhere-defined linear operator between Banach spaces is bounded if and only if it has closed graph. Boundedness gives the closed-graph property by continuity. Conversely a closed graph is Banach in the product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$. Projection of the graph onto $X$ is a bounded linear bijection, so the open mapping theorem makes its inverse bounded. Composing that inverse with the second-coordinate projection proves $T$ is bounded.

For the composition assertion, suppose $x_n\to x$ and $T{x}_n\to y$. Boundedness of $S$ gives $ST{x}_n\to Sy$, and boundedness of $ST$ gives $ST{x}_n\to STx$. Injectivity of $S$ forces $y=Tx$. Hence $T$ has closed graph and is bounded by the theorem.

For the derivative bound take $Y=C^1[0,1]$ with norm $\Vert v{\Vert }_{C^1}=\Vert v{\Vert }_{\infty }+\Vert v^{\prime }{\Vert }_{\infty }$, and $Z=C[0,1]$ with its supremum norm; both are Banach. Regard $T(x)=Ux$ as a map into $Y$, and let $S:Y\to Z$ be inclusion. This inclusion is bounded and injective, and $ST=U$ is bounded by hypothesis. The preceding result gives a constant $c$ with $\Vert Ux{\Vert }_{C^1}\le c\Vert x\Vert$. In particular $|(Ux{)}^{\prime }(t)|\le c\Vert x\Vert$ uniformly in $t$.