Show that there exists a bounded linear functional $\Lambda$ on a Hilbert space $H$ such that $\Lambda(y_{0})=\|\Lambda\|_{H^{*}} \|y_{0}\|$

Question

Let $H$ be a Hilbert space.

I want to show that for $y_{0} \in H$ that there exists a bounded linear functional $\Lambda \in H^{*}$ such that

$$\Lambda(y_{0})=\|\Lambda\|_{H^{*}} \|y_{0}\|$$

I wanted to show this as an application of the Hahn-Banach theorem applied to the norm on $H$, that is to the sublinear functional $p: y \mapsto \|y\|$.

Answer 1

So you really want to start by picking $\lambda>0$ and then constructing the functional. This constant will end up being the norm.

Okay, so you want to define $\Lambda$ on $\operatorname{span}\{y_0\}$ so that $\Lambda (t y_0) = t\Lambda(y_0) = t\lambda \|y_0\|$. This defines a linear functional on the span of $y_0$. Now define a sublinear function $p(x) = \lambda\|x\|$. Then on the span of $y_0$, $$p(ty_0) = \lambda\|ty_0\| = |t|\lambda\|y_0\|\geq t\lambda\|y_0\| = \Lambda(ty_0).$$ This gives that $\Lambda$ is dominated by the sublinear function $p$ on the subspace $\operatorname{span}\{y_0\}$, and Hahn-Banach gives the result while preserving the bound, i.e. for any $x\in H$, we have $$\Lambda(x) \leq p(x) = \lambda\|x\|.$$ But since $\Lambda(y_0)= \lambda \|y_0\|$, this gives that $\|\Lambda\|_{H^*} = \lambda$.