If $B$ is continuous, then there exists $T \in BL(\mathcal{H})$ so that $B(x,y)=\langle Tx,y\rangle $

Question

Let $B:H \times H \to \mathbb C$ continuous and sesquilinear where $H$ is a Hilbert space. Show that there exists a $T \in BL(\mathcal{H})$ so that $B(x,y)=\langle Tx,y\rangle $ for all $x,y \in H$.

My ideas on $T$:

since $B$ is continuous, we know that $x \mapsto B(x,y)$ is continuous on $H$. I am shooting in the dark here, I do not see how we can define $T$ appropriately without knowing more about $B$. If $y$ were fixed, it would be simple to define $Tx=B(x,y)\frac{y}{\vert \vert y \vert \vert}$ and then we get $B(x,y)=\langle Tx,y\rangle $ for all $x,y \in H$. But, as stated, this only works for a fixed $y$, but in our case $y$ is variable.

Answer 1

Hint: it is well known that when $H$ is a Hilbert space, the map

$$ v \in H \mapsto (-,v) \in H^* $$

is an isomorphism. That is, for any linear continuous functional $\varphi : H \to \mathbb{k}$ (= $\mathbb{R}$ or $\mathbb{C}$) there exists a unique $v \in H$ such that $\varphi(x) = (x,v)$ for all $x \in H$.

Now, following your idea, fix $x \in H$ and consider the function $f_x(y) := \overline{B(x,y)}$. Does it have any special properties you could use?

Note that $f_x(y)$ is linear and continuous, since $B$ is continuous and sesquilinear. Hence there has to exist a unique vector which I will define as $Tx$ for which we have $f_x(y) = (y,Tx)$ for all $y \in H$. You are left to show that $T$ is linear and continuous: this follows from the uniqueness of each $Tx$ and the continuity of $B$.