I am writing a paper on the compactness of closed balls in Banach spaces, with particular attention paid to the following theorem

Let $V$ be a Banach space over $\mathbb R$ or $\mathbb C$. The closed unit ball in $V$ is compact if and only if $V$ is finite-dimensional.

I am looking for some consequences or applications of this theorem (probably mostly related to the part which asserts that: if $V$ is infinite-dimensional, then the closed unit ball is not compact). I do prove the immediate corollary of this theorem, which is basically replacing "the closed unit ball" with "the closed ball of radius $r>0$ around $x_0\in V$" in the statement of the theorem. I have also been looking at the notion of weak convergence, and how this can allow for compactness (in the weak sense) in infinite-dimensional spaces. Other than those two, I am looking for some other applications of this theorem. In particular, are there any specific interesting examples one can look at that follow from this theorem?

Any feedback is appreciated.


1 Answer 1


One of the simplest consequences is that continuous function doesn't attain their minimum on the ball, fact that is always true in finite dimensional spaces for Weierstrass theorem. Consider for example the integral functional $\int_0^1|\cdot|:(C([0,1]),\Vert\cdot\Vert_{\infty})\to\mathbb{R}$, whose infimum over the ball is zero but zero is never attained on the ball.

Another interesting thing is that is easy to find an infinity of elements that lie on the ball which are $\epsilon$-separated. For an example consider the characteristics functions $\{f_n=\chi_{[n,n+1]}(x)\}_{n\in\mathbb{N}}\subset L^\infty(\mathbb{R})$. One can easily check that they belong to the unit ball of $L^\infty$ and that are 2-separated, i.e. $\Vert f_n-f_m\Vert_\infty=2$ for each $n\ne m$.

  • $\begingroup$ Yeah the second point you have is something similar to what I had in mind: we can show that if $V$ is a Hilbert space then $\Vert e_n-e_m\Vert=2$ for any orthonormal sequence $(e_n)_{n\geq 1}$. But for your first point: wouldn't the integral functional equal zero with the zero function, and also the minimum would be $-1$ which is attained with the constant function at $-1$, no? $\endgroup$
    – Dave
    Nov 17, 2018 at 23:27
  • $\begingroup$ You are right! I had in mind the example with the absolute value, which is not linear but still works as a continuous functional. I’ll edit right now $\endgroup$
    – Paul
    Nov 17, 2018 at 23:32
  • 1
    $\begingroup$ but wouldn't the minimum still be attained with the zero function in $C[0,1]$? $\endgroup$
    – Dave
    Nov 17, 2018 at 23:37
  • $\begingroup$ Yes, on the entire space, but the zero function doesn’t belong to the unit ball. $\endgroup$
    – Paul
    Nov 18, 2018 at 8:27
  • $\begingroup$ The standard is to define the unit ball as $\{x\in X: \|x\|\le 1\}$ which surely contains $0$. Perhaps @Marco means the unit sphere $\{x\in X: \|x\|=1\}$. $\endgroup$
    – Jochen
    Nov 18, 2018 at 13:15

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