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Let E be a normed linear space. Let $G \subset E$ a linear subspace. Show that if $g : G \to \mathbb{R}$ is a continuous linear operator, then $\exists f \in E^*$ such that $f_{/G} = g$ and $\|f\|_{E^*} = \|g\|_{G^*}$.

Let $p : E \to \mathbb{R}, p(x) := \|g\|_{G^*}\|x\|, \forall x \in E$ . Then $$g(x) <p(x), \forall x \in G.$$ From Hahn-Banach Theorem we have that $\exists f : E \to \mathbb{R}$ linear such that $$g(x) = f(x), \forall x \in G$$ and $$f(x) \leq p(x), \forall x \in E.$$ I can't show that $\|f\|_{E^*} = \|g\|_{G^*}$.

Thank you!

g.pomegranate
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1 Answers1

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Since $f(x) \leq p(x)$, $\| f \| \leq \| g \|$ by the definition of $p$. On the other hand, $\| f \| \geq \| g \|$ is essentially trivial because $\sup \{ f(x) : x \in E, \| x \| = 1 \} \geq \sup \{ f(x) : x \in G,\| x \| = 1 \}$. In other words an extension can only increase the norm or keep it the same.

Ian
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  • But why $|f(x)| \leq |g|_{G^*} |x|$? because we don't know if $f(x) >0$ – g.pomegranate Mar 25 '15 at 20:52
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    Maybe you should state your actual version of the Hahn-Banach theorem. (I have always heard it stated as this corollary, that there exists an extension with the same norm.) It seems to me that your problem goes away by just looking at $-x$: when $f(x)<0$, $f(-x)=-f(x)>0$, and now $f(-x)<p(-x)=p(x)$, so $|f(x)|<|p(x)|=p(x)$. – Ian Mar 25 '15 at 20:59