the functional to the double dual is isometric

[[concept]]

Topics

const fieldName = "theme"; // Your field with links
const oldPrefix = "Thoughts/01 Themes/";
const newPrefix = "Digital Garden/Topics/";
 
const relatedLinks = dv.current()[fieldName];
 
if (Array.isArray(relatedLinks)) {
    // Map over the links, replace the path, and output only clickable links
    dv.el("span",
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                    let newPath = link.path.startsWith(oldPrefix)
                        ? link.path.replace(oldPrefix, newPrefix)
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            .filter(Boolean).join(", ") 
	// Remove any undefined/null items
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} else {
    dv.el(dv.current().theme);
}
 
 

Theorem

Let $v\in V$ and define $T_v:V^{\prime }\to \mathbb{C}$ via $T_v(v^{\prime })=v^{\prime }(v)$ Then the map $T:V\to V^{\prime \prime }$ where $v\mapsto T_v$ is isometric

Proof

Define $T_v:V^{\prime }\to \mathbb{C}$ by $T_v(v^{\prime })=v^{\prime }v,v^{\prime }\in V^{\prime }$ We’ve shown already that $v\mapsto T_v\in \mathcal{B}(V,V^{\prime \prime })$ (see this example)

1. Clearly $T_v$ is linear
2. $T_v$ is bounded since $|T_v(v^{\prime })|=|v^{\prime }(v)|\le ||v^{\prime }||\times \underset{\text{constant}}{\underbrace{||v||}}⟹T_v\in (V^{\prime }{)}^{\prime }=V^{\prime \prime }$ Thus we can see $||T_v||\le ||v||$

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

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

Let $v\in V\setminus \{0\}$. Then by the corollary of Hahn-Banach, there exists $f\in V^{\prime }$ such that $||f||=1$ and $f(v)=||v||$. Then $||v||=|f(v)|\le ||T_v||\times ||f||=||T_v||$ Thus $||T_v||=||v||$. Thus the map is indeed isometric.

\tag*{$\blacksquare$}

Note

isometric bounded linear operators are one-to-one because the only thing that can be sent to the 0 vector is the 0 vector itself.

References

References

See Also

Mentions

Mentions

const modules = await cJS()
 
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