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I'm studying functional analysis and I'm having a "lot of problem with this problem". The question is the following:

Let $E$ be a normed space. Is it true that for all countinuous map $\varphi:E\to \mathbb{R}$, $\varphi(S_E(0,1))$ is bounded?

Here $S_E(0,1)=\{x\in E; \|x\|=1\}$.

Well, I know that in finite dimensional the unit sphere is compact, but in infinite dimensional it can be. Thus, I think the solution is to find a exemple that the map $\varphi$ is unbounded.

Someone has any suggestion to do this?

Yiorgos S. Smyrlis
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DiegoMath
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1 Answers1

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There is the following theorem:

A metric space $M$ is compact iff every continuous $f: M\to\mathbb R$ is bounded.

Hence, if $E$ is an infinite dimensional normed space, its unit sphere is not compact, and thus there is a continuous $f: S_E(0,1)\to\mathbb R$ that is not bounded.

For example, let $E=\ell^2(\mathbb N)$. Then $\{e_n\}_{n\in\mathbb N}\subset S(0,1)$, and $\|e_i-e_j\|=\sqrt{2}\,\delta_{ij}$. Define $f:S(0,1)\to\mathbb R$ as $$ f(x)=\left\{\begin{array}{cll} n\big(\tfrac{1}{2}-\|x-e_n\|\big) &\text{if}\,\,\, x\in B\big(e_n,\tfrac{1}{2}\big)\cap S(0,1), \\ 0 & \text{otherwise}. \end{array} \right. $$ It is not hard to verify that $f$ is continuous and unbounded.

Yiorgos S. Smyrlis
  • 83,933