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!