The proof of $||T|| = \sup\{|(T(f),f)| \;|\; ||f|| = 1 \}$ when $T = T^*$

Question

I'm studying a property of symmetric linear operator in Hilbert space in Stein's Real analysis chapter 4.

When $T = T^*$, then $\Vert T\Vert = \sup\{|(T(f),f)| \;|\; \Vert f\Vert = 1 \}$

The following picture is the proof of this.

I can follow most of the proof, except for the red line. If $g$ is replaced with $e^{i\theta}g$, then $Re(T(f),e^{i\theta}g) = Re(T(f), g\cos\theta + ig\sin\theta) = Re(T(f),g\cos\theta) + Re(T(f),ig\sin\theta)$.

$cos\theta Re(T(f),g) - isin\theta Re(T(f),g)$ has no special meaning because its absolute value returns to |Re($T(f), g$)|.

Then, I tried as follows;

$Re(T(f),e^{i\theta}g) = \cos\theta Re(T(f),g) + \sin\theta Re(-i(T(f),g))$ $ = \cos\theta Re(T(f),g) + \sin\theta Im(T(f),g)$.

However, this doesn't make any useful results too, and this is where I'm stuck on.

Any help will be appreciated. Thank you.

For reference, let me write the lemma 5.1 of this book.

Lemma 5.1 $\Vert T\Vert = \sup\{|(T(f), g) \;|\;||f|| \le 1,\; \Vert g\Vert \le 1,\; f \in H_1, g \in H_2 \}$

Answer 1

Let $\langle Tf, g \rangle =|\langle Tf, g \rangle|e^{i\theta}$ with $\theta \in \mathbb R$. Repalce $g$ by $ge^{i\theta}$ in $|\Re( \langle Tf, g \rangle)|\leq M$ to get $|\langle Tf, g \rangle| \leq M$.

[$\Re( \langle Tf, g \rangle)=\Re e^{-i\theta} \langle Tf, g \rangle=\Re |\langle Tf, g \rangle|=|\langle Tf, g \rangle|$].