corollary of Hahn-Banach

[[concept]]

Topics

const fieldName = "theme"; // Your field with links
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Matrix Analysis

Corollary of Hahn-Banach

Let $V,||\cdot ||$ be a NLS over $K$. If $x\in V$, then there exists $f\in V^{\ast }$ such that $||f|{|}_{V^{\ast }}=1$ and $f(x)=||x|{|}_V$

(without proof)

Functional Analysis

Theorem

if $V$ is a normed space, then for all $v\in V\setminus \{0\}$, there exists $f\in V^{\prime }$ such that

\lvert \lvert f \rvert \rvert &=1 \\ f(v) &= \lvert \lvert v \rvert \rvert \end{align}$$

Proof

Define $u:\mathbb{C}v\to \mathbb{C}$ as $u(\lambda v)=\lambda ||v||$. Then $|u(t)|\le ||t||$ for all $t\in \mathbb{C}v$ and also $u(v)=||v||$. Thus, by the Hahn-Banach theorem, there exists some $f\in V^{\prime }$ that extends $u$, ie $f(v)=u(v)=||v||$ . But then

Then $||f(t)||\le ||t||$ for all $t\in V$, ie $1=f\left(\frac{v}{||v||}\right)\le ||f||⟹||f||=1$

Thus we have found a linear functional as desired.

References

References

See Also

• Hahn-Banach theorem

Mentions

Mentions

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