Proving the existence and uniqueness of an operator over a Hilbert space

Question

Let $H$ be a Hilbert space and let $f \colon H \times H \to \mathbb C$ be a sesquilinear map such that $$ M = \sup \{ |f(x, y)| \mid x, y \in H, \|x\| = \|y\| = 1 \} < \infty$$

Prove that there exists a unique operator $S \in B(H)$, such that $$ f(x, y) = \langle Sx, y \rangle,\qquad\forall x, y \in H.$$ Finally, prove that $\|S\| = M$.

My incomplete attempt
Fix $x \in H$ and consider the functional \begin{align*} \Phi_x \colon H &\to \mathbb C\\ y &\mapsto \overline{f(x, y)} \end{align*}

1. $\Phi_x \in H^\star$ is well-defined by construction, since by hypothesis $f(x, y) \in \mathbb C$.
2. $\Phi_x$ is linear. Let $y, z \in H$ and $\alpha, \beta \in \mathbb C$. Then $$\Phi_x(\alpha y + \beta z) = \overline{f(x, \alpha y + \beta z)} = \alpha \overline{f(x, y)} + \beta \overline{f(x, z)} = \alpha \Phi_x(y) + \beta \Phi_x(z).$$
3. $\Phi_x$ is bounded. Let $y \in H$. We have to prove that there exists a constant $c \in \mathbb R$ such that $|\Phi_x(y)| \leq c\|y\|$. If $y = 0$, the inequality is trivially true. Therefore, suppose $y \neq 0$. Using the bilinearity of $f$ we have that $$|\Phi_x(y)| = |\overline{f(x, y)}| = |f(x, y)| = \|x\|\cdot\|y\|\left|f\left(\frac{x}{\|x\|}, \frac{y}{\|y\|}\right)\right| \leq M\|x\| \cdot \|y\|$$ Therefore we can take $c = M\|x\| \in \mathbb R$, since $x$ is fixed. We also observe that $\|\Phi_x\| \leq M\|x\|$.

By Riesz's representation theorem, there exists a unique element in $H$, which we call $Sx \in H$, such that $\Phi_x(y) = \overline{f(x, y)} = \langle y, Sx \rangle$.

We will now prove that $S \in B(H)$.

1. $S$ is linear. Let $x, z \in H$ and $\alpha, \beta \in \mathbb C$. For all $y \in H$, we have that $$\langle y, S(\alpha x + \beta z) \rangle = \overline{f(\alpha x + \beta z, y)} = \overline{\alpha f(x, y) + \beta f(z, y)} = \bar\alpha \langle y, Sx \rangle + \bar\beta \langle y, Sz \rangle = \langle y, \alpha Sx + \beta Sz \rangle$$ From the arbitrariness of $y$, it follows that $S(\alpha x + \beta z) = \alpha Sx + \beta Sz$.
2. $S$ is bounded. From Riesz's theorem, we have that $\|Sx\| = \|\Phi_x\|$, and at point $3.$ above we proved that $\|\Phi_x\| \leq M\|x\|$. So $\|S\| \leq M \in \mathbb R$ and $S$ is bounded.

Finally, we use the definition of operator's norm: $$\|S\| = \sup \{ \|Sx\| \mid x \in H, \|x\| = 1 \} = {\large\textbf{?}}$$

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Questions
1. Is the proof correct up to the last point? I'm particularly interested in the second part, where I applied Riesz's theorem.
2. How should I conclude it? I know it must be related to the definition of $M$, probably I'm just missing something simple.

Answer 1

Your proof looks fine. What you have to use is $$ \|S\|=\sup\{|\langle Sx,y\rangle|:\ \|x\|=\|y\|=1\}. $$