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For any vector $h \in H$, where $H$ is a Hilbert space, show that $\exists$ a bounded linear functional $\psi \in H^{*}$ such that:

$$\|\psi\| = 1 \ \text{and} \ \psi(h) = \|h\|$$

Can anyone propose what this $\psi$ should be? I don't really have a clue as to how to start this. Unfortunately, we are not allowed to use Hahn-Banach.

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Hint: Try a linear functional of the form $\psi(f) = (f,g)_H$ for some $g \in H$. (In fact, by the Riesz representation theorem, every bounded linear functional has this form.) What should $g$ be to get $\psi$ as desired by the problem?

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Could you please explain the notation $(f,g)_H$? What does it denote? – user44069 Mar 11 '13 at 21:37
@Stefan: $(f,g)_H$ denotes the inner product of $f,g$ in the Hilbert space $H$. – Nate Eldredge Mar 11 '13 at 22:22

I think you can prove this result in the more general setting of a Banach space. On the subspace $Y$ generated by $h$, let $\psi(\alpha h) = \alpha ||h||$.

Now your entire job is to ensure that this $\psi$, defined as a linear functional on $Y$, extends to the whole space with norm preservation. This can be realized as a (nice) exercise from the Hahn-Banach theorem.

(for details, see for example the real analysis book by Aliprantis and Burkinshaw pp 237)

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Unfortunately, we are not allowed to use Hahn-Banach for this one. – user44069 Mar 11 '13 at 22:01

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