Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
@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? –  FranckStudiesCommEng Jun 25 '12 at 5:36

1 Answer 1

up vote 3 down vote accepted

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.