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

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.