Past exam of the analysis course of the University of Oxford / 2011 / 4

The graph $G(T)=\{(x,Tx):x\in X\}$ is closed if it is a closed subset of $X\times Y$. Equivalently, $x_n\to x$ and $T{x}_n\to y$ imply $Tx=y$.

The closed graph theorem states that an everywhere-defined linear map between Banach spaces is bounded if and only if its graph is closed. Completeness of both spaces is part of the theorem. A bounded map has a closed graph by continuity.

Conversely, if the graph is closed, it is a Banach space in the product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$. The coordinate projection $P:G(T)\to X$, $P(x,Tx)=x$, is a bounded linear bijection. The open mapping theorem makes $P^{-1}$ bounded. The other coordinate projection $Q:G(T)\to Y$ is bounded, and $T=Q{P}^{-1}$ is therefore bounded. This proves the theorem from the permitted result.

For scalars $a,b$ and $f,h\in X$, linearity of the Fourier coefficients givesThe assumed uniqueness of the function with these coefficients implies $T(af+bh)=aTf+bTh$. Thus $T$ is linear.

For each fixed $n$, the coefficient functional is bounded sinceSuppose $f_j\to f$ and $T{f}_j\to g$ uniformly. Taking coefficients and passing to the limit givesfor every fixed $n$. No uniform bound on ${\alpha }_n$ is needed for this passage. Uniqueness gives $g=Tf$, so $T$ has a closed graph. The space $X$ is Banach, being the closed subspace defined by $f(-\pi )=f(\pi )$ inside the continuous functions with the uniform norm. The closed graph theorem therefore proves $T$ is bounded.

The normalization givesThus ${\alpha }_n{e}_n$ has exactly the coefficients prescribed for $T{e}_n$. Uniqueness yieldsThese are eigenvectors, including the zero-eigenvalue cases.

Since every $e_n$ is nonzero, boundedness givesCancel its norm to obtain $\boxed{\underset{n\in \mathbb{Z}}{\sup }|{\alpha }_n|\le \Vert T\Vert }$. Thus the multiplier sequence is bounded.

Evaluation at zero is a bounded linear functional for the uniform norm. DefineIt is linear and satisfies $|\varphi (f)|\le \sqrt{2\pi }\Vert T\Vert \Vert f{\Vert }_{\infty }$. Since $e_n(0)=(2\pi {)}^{-1/2}$,The factor $\sqrt{2\pi }$ is required by the source's basis normalization.