What does it mean by "there is a unique linear norm preserving extension $f$ of $g$ on $H$"?

Question

I'm trying to do this exercise:

I'm unable to understand the meaning of the requirement

"there is a unique linear norm preserving extension $f$ of $g$ on $H$"

What do "linear norm" and "preserving extension" mean? Could you please elaborate on this issue? Thank you so much!

Answer 1

I think you cluttered the words exactly in the wrong way. The statement is, that you have a map, that is linear and defined on the whole Hilbert space $H$ (not just $G$). But its norm stays the same or you can say is preserved,e.g. $\|g\| = \|f\|$ and $f\bigg|_G = g$.

Answer 2

It means that there is a unique linear extension $f: H \to \mathbb R$ of $g$ which is norm preserving, i.e. satisfies $\lVert f \rVert = \lVert g \rVert$.

Answer 3

I post my attempt here. It would be great if someone helps me verify it ^o^

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It follows from $G$ is a closed linear subspace of the Hilbert space $H$ that $G$ is also a Hilbert space. By Riesz theorem, there is a unique $x_G \in G$ such that $g(x) = \langle x_G,x \rangle$ for all $x \in G$ and that $\|g\| = \|x_G\|$. We define $f:H \to \mathbb R$ by $f(x) = \langle x_G,x \rangle$. Clearly, $f$ is a linear functional.

By Cauchy-Schwarz inequality, $|f(x)| \le \|x_G\| \cdot \|x\|$. As such, $\|f\| \le \|x_G\| = \|g\|$. Hence $f$ is bounded and thus continuous. As such, $f \in H^*$. Since $f \restriction G = g$, $\|g\| \le \|f\|$. As such, $\|f\| = \|g\|$.

Assume that there is another $f' \in H^*$ that satisfies the requirement. Then by Riesz theorem, there is a unique $x_H \in H$ such that $f'(x) = \langle x_H,x \rangle$ for all $x \in H$. Because $f \restriction G = g = f' \restriction G$, $f'(a) = f(a)$ for all $a \in G$. As such, $\langle x_G,a \rangle = \langle x_H,a \rangle$ or $\langle x_G-x_H,a \rangle = 0$ for all $a \in G$. This means $(x_G-x_H) \in G^\perp = \{0\}$. Hence $x_G= x_H$ and thus $f'=f$.

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Here is my fix for the uniqueness part ^^:

Assume that there is another $f' \in H^*$ that satisfies the requirement. Then by Riesz theorem, there is a unique $x_H \in H$ such that $f'(x) = \langle x_H,x \rangle$ for all $x \in H$.

Because $f'(x_G) = f(x_G)$, $\langle x_H,x_G \rangle = \langle x_G,x_G \rangle$ or equivalently $\langle x_H,x_G \rangle = \|x_G\|^2$ or equivalently $\langle x_H - x_G,x_G \rangle =0$.

Because $\|f\| =\|f'\|= \|g\|$ and $\|f\| = \|x_G\|$ and $\|f'\| = \|x_H\|$, we have $\|x_G\| = \|x_H\|$ or equivalently $\|x_G\|^2 = \|x_H\|^2$. As such, $\langle x_H,x_G \rangle = \|x_H\|^2$ or equivalently $\langle x_H,x_G \rangle = \langle x_H,x_H \rangle$. This implies $\langle x_H - x_G,x_H \rangle =0$.

Finally, we have $\langle x_H - x_G,x_G \rangle =0$ and $\langle x_H - x_G,x_H \rangle =0$. This implies $\langle x_H - x_G,x_H -x_G \rangle =0$. As such, $x_G-x_H =0$ and thus $x_G=x_H$.