One of the fundamental theorems in functional analysis is that if $X$ is a Banach space (say over $\Bbb C$) with a compact closed unit ball, then $X$ is finitely dimensional.

The usual proof is by assuming $X$ is infinite dimensional, and constructing by induction a sequence of vectors on the unit sphere which are not only linearly independent, but also have distances $>\frac12$ from one another.

But you can also prove this from automatic continuity. Namely, if every linear functional $f\colon X\to\Bbb C$ is continuous then $X$ has a finite dimension. If $\{v_n\mid n\in\Bbb N\}$ are linearly independent and lie the unit sphere, the function $f(v_n)=n$ can be extended to a linear functional on $X$. It is unbounded and therefore not continuous.

Can you prove directly from the assumption that the closed unit ball (equiv. the unit sphere) is compact that every linear functional is continuous?

  • $\begingroup$ Probably I am missing something, but I think that the proof that linear functionals defined on finite-dimensional spaces are continuous works in this case, namely: if $f:X\to\mathbb C$ is a linear functional then $\max_{\|v\|=1}f(v)$ exists, say $C$, so for any $w\ne0$ in $X$ we have $f\bigl(w/\|w\|\bigr)\leq C$, and thus $f(w)\leq C\|w\|$. $\endgroup$ Commented Dec 15, 2015 at 6:22
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    $\begingroup$ It's really hard for me to imagine such a proof that doesn't essentially go through finite-dimensionality (how else could you possibly show that every functional is continuous?). $\endgroup$ Commented Dec 15, 2015 at 6:25
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    $\begingroup$ @MatemáticosChibchas: You are assuming $f$ is already continuous to get that that max exists. $\endgroup$ Commented Dec 15, 2015 at 6:25
  • $\begingroup$ @EricWofsey You are right, sorry. $\endgroup$ Commented Dec 15, 2015 at 6:29
  • $\begingroup$ @Eric: I couldn't either, which is why asked here... $\endgroup$
    – Asaf Karagila
    Commented Dec 15, 2015 at 6:44

2 Answers 2


This is definitely not an answer to the question. But it's a possibly amusing proof of the result - I was thinking about proving that automatic continuity and found I'd proved that $X$ is finite-dimensional.

Say $B(x,r)$ is the closed ball in $X$ about $x$ of radius $r$. Suppose $B(0,1)$ is compact. Then there exists a finite set $F\subset B(0,1)$ such that $$B(0,1)\subset\bigcup_{y\in F}B(y,1/2).$$

Hence, by the normed-vector-spaceness of $X$, for every $x\in F$ we have $B(x,1/2)\subset\bigcup_{y\in F}(x+B(y/2,1/4))$, so that $$B(0,1)\subset\bigcup_{y_0,y_1\in F}B(y_0+ y_1/2,1/4).$$Etc. It follows that for every $x\in B(0,1)$ there exist $y_0,y_1\dots\in F$ with $$x=\sum_{n=0}^\infty2^{-n}y_n.$$

And now if you regroup the terms in that sum it follows that $x$ is a linear combination of the elements of $F$, qed. (In fact $x/2$ is in the convex hull of $F$.)


Edit: As pointed out by Daniel Fisher in a comment, my first attempt was incomplete. It does, however, work if we know that the unit ball in every finite-dimensional normed space is compact. I admit that this is not a satisfactory answer.

Let $f\colon X\to\mathbb{K}$ be a linear functional. To show continuity it suffices to prove that $\ker f$ is closed. Now, $\ker f$ is a locally compact subspace of $X$ (since it is finite-dimensional). In particular, $\ker f$ is a locally compact subgroup of the Hausdorff topological group $X$. By general theory, locally compact subgroups of Hausdorff groups are closed.

(I don't have a reference for the last statement besides Wikipedia. In Dieudonné's Elements d'Analyse it's proven in the metrizable case - for Banach spaces that additional assumption is obviously satisfied).

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    $\begingroup$ How does the compactness of the unit ball imply the local compactness of $\ker f$? $\endgroup$ Commented Dec 15, 2015 at 13:45
  • $\begingroup$ You're right. I thought I could tweak Dieudonné's argument to make it fit the op's question, but it seems like it doesn't work. I will think about it a little more and place a warning in my answer. $\endgroup$
    – MaoWao
    Commented Dec 15, 2015 at 14:05

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