My CVFunctional Analysis 11.13
13. Use the Uniform Boundedness Principle to prove the Hellinger–Toeplitz Theorem.
Construct a functional $f_y$ such that
\[f_y(x)=(Ax,y)=(x,Ay).\]In fact, they satisfy the norm relation $f_y = \Vert Ay \Vert$. We first prove this relation:
\[\Vert f_y\Vert=\sup_{\Vert x \Vert =1}|(x,Ay)|\leq \Vert x\Vert\cdot\Vert Ay \Vert=\Vert Ay \Vert.\]Moreover, we also have
\[f_y(Ay)=(Ay,Ay)=\Vert Ay\Vert^2,\]which implies that $\Vert f_y\Vert\geq \Vert Ay \Vert$. Thus, in fact, $\Vert f_y\Vert=\Vert Ay \Vert$. Now we prove the theorem. Since it is defined everywhere, we obtain
\[\sup_{\Vert y\Vert= 1}\Vert f_y(x)\Vert=\sup_{\Vert y\Vert= 1}|(Ax,y)|\leq\Vert Ax\Vert<\infty.\]We regard ${f_y}$ as a family of bounded functionals indexed by $y$. By the Uniform Boundedness Principle together with the definition of the operator norm in an inner product space,
\[\Vert A\Vert=\sup_{||y||=1}\Vert Ay\Vert=\sup_{\Vert y\Vert= 1}\Vert f_y\Vert<\infty.\]This shows that $A$ is bounded.
14. Suppose $A,B$ are linear operators defined everywhere on a Hilbert space $H$, and satisfy
\((Ax,y)=(x,By),\quad \forall x,y\in H.\)
Prove: $A,B$ are bounded and $B=A^*$.
The proof idea is similar to the previous problem.
\[\Vert f_y\Vert=\sup_{\Vert x \Vert =1}|(x,By)|\leq \Vert x\Vert\cdot\Vert By \Vert=\Vert By \Vert,\]and
\[f_y(By)=(By,By)=\Vert By\Vert^2.\]Hence we also have $\Vert f_y\Vert=\Vert By \Vert$. Moreover,
\[\sup_{\Vert y\Vert= 1}\Vert f_y(x)\Vert=\sup_{\Vert y\Vert= 1}|(Ax,y)|\leq\Vert Ax\Vert<\infty.\]Exactly as in Problem 13, by the Uniform Boundedness Principle, $B$ is bounded.
Since
\[(Ax,y)=(x,By),\]it follows that $A$ is also bounded. By the definition of the adjoint operator, we have $B=A^*$.
15. Suppose ${x_n}_{n=1}^\infty$ is a sequence in a Banach space $X$, and for every $f\in X’$,
\[\sum^{\infty}_{n=1}|f(x_n)|<\infty,\]then there exists a constant $\mu>0$ such that for all $f\in X’$,
\[\sum^\infty_{n=1}|f(x_n)|\leq \mu\ \|f\|.\]First, let us clarify that $X’$ denotes the dual space, i.e. the space of all bounded linear functionals on $X$.
Now, construct a sequence of operators $T_n\in \mathcal{L}(X’,l^1)$, which are bounded linear operators from the Banach space $X’$ to the normed space $l^1$, defined by
\[T_k(f)=\sum^k_{n=1}|f(x_n)|,\qquad k\in \mathbb{N^*}.\]By assumption,
\[\sup_{k\in \mathbb{N^*}}|T_k(f)|=\sum^{\infty}_{n=1}|f(x_n)|<\infty.\]By the Uniform Boundedness Principle,
\[\sup_n \|T_n\|\leq \mu <\infty.\]Therefore, we have
\[\sum^\infty_{n=1}|f(x_n)|=\lim_{n\to\infty}T_n(f)\leq \|T_n\|\ \|f\|\leq \mu \|f\|.\]Hence the claim is proved.