I'm trying to solve below exercise in Brezis' Functional Analysis
Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $G$ be a vector subspace of $H$. Let $(F, [\cdot])$ be a Banach space. Let $S:G \to F$ be a bounded linear operator. Prove that there is a bounded linear operator $T:H \to F$ that extends $S$ and that $$ \| T\|_{\mathcal L (H, F)} = \| S\|_{\mathcal L (G, F)}. $$
Could you verify my below attempt?
Let $\overline S :\overline G \to F$ be the bounded linear operator that extends $S$. Here $\overline G$ is the closure of $G$. Such $\overline S$ is well-defined and unique [A proof can be found here or here]. We have $$ \| \overline S\|_{\mathcal L (\overline G, F)} = \| S\|_{\mathcal L (G, F)}. $$
Let $\pi_{\overline G} : H \to \overline G$ be the orthogonal projection onto $\overline G$. Then $\pi_{\overline G}$ is a bounded linear operator such that $\| \pi_{\overline G} \|_{\mathcal L (H, \overline G)} \le 1$. Let $T := \overline S \circ \pi_{\overline G}$. Then $T : H \to F$ is a bounded linear operator that extends $S$. Also, $$ \| T\|_{\mathcal L (H, F)} \le \| \overline S\|_{\mathcal L (\overline G, F)} \| \pi_{\overline G} \|_{\mathcal L (H, \overline G)} \le \| \overline S\|_{\mathcal L (\overline G, F)} = \| S\|_{\mathcal L (G, F)}. $$
Let's prove the reverse inequality. Indeed, $$ \| T\|_{\mathcal L (H, F)} = \sup_{x \in H} \frac{[T(x)]}{|x|} \ge \sup_{x \in G} \frac{[T(x)]}{|x|} = \sup_{x \in G} \frac{[S(x)]}{|x|} = \| S\|_{\mathcal L (G, F)}. $$
This completes the proof.
