Prove that $\{T_v: \|v\|\leq 1\}$ is a family of pointwise uniformly bounded functionals

Question

Let $V,W$ be Hilbert spaces and $A:V\rightarrow W$, $B:W^*\rightarrow V^*$ linear operators such that $l(Av)=(Bl)(v)$ for all $v\in V$ and for all $l\in W^*$. For $v\in V$ define $T_v(l):=l(A(v))$ for all $l\in W^*$. Then I need to show that for all $l\in W^*$ $$\sup_{\|v\|\leq 1}|T_v(l)|<\infty$$

I wanted to do it as follows: $$\begin{align}\sup_{\|v\|\leq 1}|T_v(l)|&=\sup_{\|v\|\leq 1}|l(A(v))|\\&\leq\sup_{\|v\|\leq 1}\|l\|~\|A\|~\|v\|\\&\leq \|l\|~\|A\|\end{align}$$Now since $l\in W^*$ I know that $\|l\|<\infty$. But I don't know anything about $A$. I only know that $\|A\|=\|B\|$,i.e. if $A$ is bounded then $B$ is bounded but I don't see how this helps me.

Can maybe someone help me how to conclude?

Answer 1

$$|T_v(l)|=|l(A(v))|=|(B(l))(v)|\le\|B(l)\|\|v\|.$$

Answer 2

What you are computing is the norm of $lA = l \circ A \in V^*$. Indeed $$ \sup_{\|v \| \leq 1} | T_v(l) | = \sup_{\|v \| \leq 1} | l (A(v)) |\sup_{\|v \| \leq 1} | (l \circ A)v | = \| l \circ A \|_{V^*}. $$ Now the last number is finite for any $l \in W^*$ if and only if $A : V \rightarrow W$ is bounded. Indeed if $A$ is bounded then $l \circ A$ is also bounded. Conversly if all the $(l \circ A)_{l \in W^*}$ are bounded, you then know that $A$ is continuous from $V$ endowed of the norm topology to $W$ endowed of its weak topology. This is enough to say that $A$ is continuous, indeed then the graph of $A$ is strongly closed and by the closed graph theorem $A$ is strongly continuous.

So I think your problem miss the hypothesis "$A$ bounded".

Added : After the explanations of Anne Beauval I realized that to define the adjoint (what you call $B$) as a map $A^* : W^* \rightarrow V^*$ you need to require $A$ continuous. Indeed you trapped yourself saying that $V^*$ and $W^*$ stand for the topological dual spaces, hence continuous mappings. Then using what I showed you above you get $$ \begin{align} A^* \text{ well defined} &\Longrightarrow \left[ \forall w^* \in W^*,\quad A^*w^* \in V^* \right]\\ &\Longleftrightarrow \left[ \forall w^* \in W^*,\quad w^* \circ A \in V^* \right]\\ &\Longleftrightarrow A \text{ continuous} \end{align} $$

Answer 3

Both $A$ and $B$ are bounded (it seems this is what you need to finish your argument).

Firstly, let us see the boundedness of $A$. Since $V$ and $W$ are Banach spaces and $A$ is already linear, by the Closed Graph Theorem, it suffices to prove that $$v_n \to v, \ Av_n \to w \Rightarrow Av = w.$$ In fact, assuming the left hand side above, we have for $l \in W^*$: $$l(w) = l\left(\lim_{n\to\infty} Av_n\right) = \lim_{n\to\infty}l(Av_n)=\lim_{n\to\infty} (Bl)(v_n) = (Bl)(v) = l(Av).$$ Since this is true for all $l \in W^*$, the Hahn-Banach Theorem implies $Av=w$ as desired. This proves the boundedness of $A$.

The boundedness of $B$ now follows trivially from $$\vert Bl(v) \vert = \vert l(Av)\vert \leq \Vert y'\Vert\Vert A\Vert\Vert x \Vert.$$

Does this solve your problem?