# Sesquilinear forms on Hilbert spaces

### 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!

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@FrackStudiesCommEng I would proceed as follows: (i) implies (ii) is obvious. For (ii) implies (iii) you can proceed (essentially) as in your proof of (i) implies (iii). Now, (iii) implies (i) is straightforward just write down the definition of bounded. – azarel Jun 24 '12 at 23:58
@azarel Thank you for your answer. For the last implication, I showed that $\left|B\left(x+x_{0},y\right)-B\left(x,y\right)\right|\leq c\left\Vert x_{0}\right\Vert \left\Vert y\right\Vert \rightarrow0$ as $x_{0}\rightarrow0$, and that $\left|B\left(x,y+y_{0}\right)-B\left(x,y\right)\right|=\left|B\left(x,y+y_{0}-y‌​\right)\right|\leq c\left\Vert x\right\Vert \left\Vert y_{0}\right\Vert \rightarrow0$ as $y_{0}\rightarrow0$, which proves that $B$ is continuous. But I still do not see how to prove the "obvious" implication. Could you be more precise? – franckysnow Jun 25 '12 at 5:36

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 $$|B(x_n,y_n)-B(x,y)|=|B(x_n,y_n)-B(x,y_n)+B(x,y_n)-B(x,y)|\leq$$ $$|B(x_n,y_n)-B(x,y_n)|+|B(x,y_n)-B(x,y)|=|B(x_n-x,y_n)|+|B(x,y_n-y)|\leq$$ $$c\Vert x_n-x\Vert\Vert y\Vert +c\Vert x\Vert\Vert y_n-y\Vert$$