the functional to the double dual is isometric

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Proof

Define $T_{v} : V^{'} \to C$ by

T_{v} (v^{'}) = v^{'} v, v^{'} \in V^{'}

We've shown already that $v \mapsto T_{v} \in B (V, V^{″})$ (see this example)

Clearly $T_{v}$ is linear
$T_{v}$ is bounded since

| T_{v} (v^{'}) | = | v^{'} (v) | \leq | | v^{'} | | \times \underset{constant}{\underset{⏟}{| | v | |}} ⟹ T_{v} \in (V^{'})^{'} = V^{″}

Thus we can see $| | T_{v} | | \leq | | v | |$

What is left is to show is that it is isometric, ie that $| | T_{v} v | | = | | v | |$

Since $| | T_{v} | | \leq | | v | |$ , we know that $| | T | | \leq 1$ . Now we show for all $v \in V$ that $| | T_{v} v | | = | | v | |$

Let $v \in V ∖ {0}$ . Then by the corollary of Hahn-Banach, there exists $f \in V^{'}$ such that $| | f | | = 1$ and $f (v) = | | v | |$ . Then

| | v | | = | f (v) | \leq | | T_{v} | | \times | | f | | = | | T_{v} | |

Thus $| | T_{v} | | = | | v | |$ . Thus the map is indeed isometric.

\begin{matrix} ◼ \end{matrix}

File	Last Modified
Functional Analysis Lecture 6	2025-07-08

File

Last Modified

Functional Analysis Lecture 6

2025-07-08

the functional to the double dual is isometric

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