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If necessary, recall that $$ \ell^2 = \{x=\{x_n\}_n\subset \mathbb{R} : \|x\|^2:=\sum_n |x|^2<\infty\} $$

and $ \overline{B_1(0)} $ is the closed unit ball with respect to that norm.

Can we find an explicit example of a continuous function $f: \overline{B_1(0)} \subset \ell^2\to \mathbb{R}$ which does not attain its maximum?

The point is that this ball is not a compact set.

Thank you.

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  • $\begingroup$ As per my previous comment, your question still doesn't make the most sense. Can you maybe rephrase the exact question of the exercise in the body ? $\endgroup$ – Rebellos Mar 1 at 18:50
  • $\begingroup$ No, sorry, I did a mistake in the question and now is correct. The point is to show a function satisfying the conditions mentioned in the question in order to convince by an example that it is not compact. $\endgroup$ – Rubén Fernández Fuertes Mar 1 at 18:50
  • $\begingroup$ So it comes down to showing that $\overline{B_1^{\ell_2}}$ is not compact ? $\endgroup$ – Rebellos Mar 1 at 18:51
  • $\begingroup$ I said no, and I say, again, no. I know by another proof that it is not compact. I repeat, the point is to give an example of this function. $\endgroup$ – Rubén Fernández Fuertes Mar 1 at 18:52
  • $\begingroup$ Note that $f : \overline{B_1(0)} \in \ell_2$ isn't a solid expression for a function and this is why I am struggling to understand. Do you mean that you want to construct a function with its domain being the Unit Ball of $\ell_2$ such that it is a sufficient example in showing that it cannot be compact ? That's pretty straightforward using the defintiion of a compact operator. $\endgroup$ – Rebellos Mar 1 at 18:54
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Try $$f(x) = \sum_{n=1}^\infty (1-1/n) x_n^2$$ Note that $f(x) < 1$ for all $x \in \overline{B_1(0)}$, and you can get arbitrarily close to $1$...

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