Definition of the problem
Let $\mathcal{H}$ be a Hilbert space, and let $B:\mathcal{H}\times\mathcal{H}\rightarrow\mathbb{K}$
be a sesquilinear form. Prove that TFAE:
$(i)$ $B$ is continuous.
$(ii)$ For each $x\in\mathcal{H}$ the mapping $y\mapsto B\left(x,y\right)$
is continuous, and for each $y\in\mathcal{H}$ the mapping $x\mapsto B\left(x,y\right)$
is continuous.
$(iii)$ $B$ is bounded.
My efforts
$(i)\Rightarrow(iii)$: $B$ is continous sesquilinear on $\mathcal{H}$ a Hilbert
space, then by the Riesz Representation Theorem:
$$
\exists!A\in L\left(\mathcal{H}\right),\,\forall(u,v)\in\mathcal{H}\times\mathcal{H},\, B\left(u,v\right)=\left\langle Au,v\right\rangle .
$$
Then we have that $\left|B\left(u,v\right)\right|=\left|\left\langle Au,v\right\rangle \right|\leq\left\Vert Au\right\Vert \left\Vert v\right\Vert \leq\left\Vert A\right\Vert \left\Vert u\right\Vert \left\Vert v\right\Vert =c\left\Vert u\right\Vert \left\Vert v\right\Vert $,
since $A$ bounded linear operator, and by Cauchy-Schwarz inequality.
Then $B$ is bounded.
My question
I feel that the step $(i)\Rightarrow(ii)\Rightarrow(iii)$ is kind of contained within my $(i)\Rightarrow(iii)$ step, but how could I show these two implications? By Riesz Representation Theorem again? How could I properly express that relationship using continuity?
Idea
For the step $(iii)\Rightarrow(i)$, could you hint me a theorem? I was actually thinking of the Closed Graph Theorem, by showing that $B$ is a closed operator, I could conclue that $B$ is continuous. But how could I show that $B$ is a closed operator, by having that $B$ is bounded?
Thank you a lot, Franck!
Best Answer
Implication $(i)\Longrightarrow(ii)$ follows from the fact that inclusion maps $$ i_y:\mathcal{H}\to\mathcal{H}\times\mathcal{H}:x\to(x,y) $$ $$ i_x:\mathcal{H}\to\mathcal{H}\times\mathcal{H}:y\to(x,y) $$ are continuous. The maps you are considering are $B \circ i_x$, $B\circ i_y$. They are continuous as compositions of continuous maps.
Implication $(ii)\Longrightarrow(iii)$ follows from uniform boundness principle. Apply it to the family of continuous linear operators $\{T_y:y\in \mathcal{H}\}$ where $$ T_y:\mathcal{H}\to \mathbb{K}:x\mapsto B(x,y) $$ Implication $(iii)\Longrightarrow(i)$ follows from inequalities $$ |B(x_n,y_n)-B(x,y)|=|B(x_n,y_n)-B(x,y_n)+B(x,y_n)-B(x,y)|\leq $$ $$ |B(x_n,y_n)-B(x,y_n)|+|B(x,y_n)-B(x,y)|=|B(x_n-x,y_n)|+|B(x,y_n-y)|\leq $$ $$ c\Vert x_n-x\Vert\Vert y\Vert +c\Vert x\Vert\Vert y_n-y\Vert $$