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Let $X$ be a normed space, $f\in X^*\setminus\{0\}$ (the continuous dual), $E:=\{x\in X\mid f(x)=\|f\|\}$. Prove that $E$ is a nonempty closed set and that $\inf \{\|x\|\mid x\in E\}=1$.

I have no idea how to prove that $E$ is non empty and that $\inf \{\|x\|\mid x\in E\}=1$. $E$ is closed because it is the inverse image of a point ($\|f\|$) and $f$ is continuos, so inverse images of closed sets are closed.

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  • $\begingroup$ The last claim seems wrong. Did you mean $x \in E$? $\endgroup$ Aug 16, 2018 at 18:48
  • $\begingroup$ Yes of course, bad typo, sorry $\endgroup$
    – Algebra
    Aug 16, 2018 at 18:50
  • $\begingroup$ You may wanna state that $\inf\{||x|| : x \in \color{red}{E}\} = 1$. The answer is somewhat here. $\endgroup$
    – metamorphy
    Aug 16, 2018 at 18:50
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    $\begingroup$ Is here $X^*$ the continuous dual of $X$? $\endgroup$
    – qualcuno
    Aug 16, 2018 at 18:50
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    $\begingroup$ @GuidoA. But the infimum would be $0$... $\endgroup$
    – user251257
    Aug 16, 2018 at 19:07

2 Answers 2

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If $f=0$ identically, then $\|f\|=0$ and $E=X$ which is nonempty and closed and clearly contains $x$ with $\|x\|=1$. So suppose $f(x) \neq 0$ for some $x\in X$. Then $f(\alpha x)=\|f\|$, where $\alpha=\|f\|/f(x)$. Therefore, $E$ is nonempty. $E$ is clearly closed since it is the inverse image of the closed set $\{\|f\|\}$.

Next, let $x\in E$ and so $f(x)=\|f\|$. It follows from the definition of the norm that $\|f(x)\| \leq \|f\| \|x\|$. By substitution, we have $1\leq \|x\|$.

Finally, by the definition of the norm, we have a sequence $x_i \in X$ such that $f(x_i)/\|x_i\| \rightarrow \|f\|$. By normalizing, we can assume $\|x_i\|=1$. So $f(x_i) \rightarrow \|f\|$. Now for $i$ large enough, $f(x_i) \neq 0$ and so $\alpha_i x_i \in E$, where $\alpha_i=\|f\|/f(x_i)$. One has $$\|\alpha_i x_i\|=\|f\|/f(x_i) \rightarrow 1,$$ and so $\inf \{\|x\|: x\in E\}=1$.

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Since $f \neq 0$ we have $f(x^*) \neq 0$ for some $x^*$, so then $f({\|f\| \over f(x^*)} x^*) = \|f\|$ so $E$ is not empty.

Since $f$ is continuous, $f^{-1} ( \{ \|f\| \})$ is closed.

Note that $f(x) = \|f\|$ for all $x \in E$. Since $\|f\| = f(x) \le \|f\| \|x\|$ we see that $\|x\| \ge 1$.

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  • $\begingroup$ this only proves $\inf \{\|x\|: x\in E\}\geq 1$. $\endgroup$
    – Marco
    Aug 16, 2018 at 19:32
  • $\begingroup$ I understand, it is fairly straightforward to finish using the definition of norm. $\endgroup$
    – copper.hat
    Aug 16, 2018 at 19:37

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