Let $\mathscr{C}[0,1]$ denote the set of continuous functions with bounded supremum and let $K=\{f\in\mathscr{C}[0,1]|\int_0^1f(t)dt=1\}$. Then is $K$ compact in the space $\mathscr{C}[0,1]$? Typically how do we characterize the compact spaces in the space of continuous functions? Will Heine-Borel property work here?

I think Heine-Borel would work, as $[0,1]$ is a compact Hausdorff space. Then, by using a function similar to spikes, or, somewhat like Dirac-Delta function, I think the space $K$ is not compact. Is my argument true? Any hints? Thanks beforehand.

  • $\begingroup$ Have you heard of Arzela-Ascoli theorem? $\endgroup$ – Thomas Shelby Jan 8 at 11:34
  • $\begingroup$ Your heuristics is correct, but you still need a rigorous proof. You should produce an explicit example of a sequence in $K$ that has no limit points. Your idea will work. $\endgroup$ – Giuseppe Negro Jan 8 at 11:39
  • $\begingroup$ @ThomasShelby oh! I just saw the statement. It states that a subspace of continuous functions is compact iff the space is bounded and equicontinuous. In this case, the space is neither bounded nor equicontinuous, am I right $\endgroup$ – vidyarthi Jan 8 at 11:39
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    $\begingroup$ @ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer. $\endgroup$ – daw Jan 8 at 12:56
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    $\begingroup$ @ThomasShelby You're talking about different things here. Every element of $K$ is a bounded functions, as you said. But $K$ is not a bounded subset of the normed space $C[0,1]$. That's just two different meanings of boundedness (I guess the latter is exactly what you call uniformly bounded). $\endgroup$ – MaoWao Jan 8 at 13:30

Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+n\sin(2\pi x)$. We have $(f_n)_n \subseteq K$ but $$\|f_n\|_\infty \ge f_n\left(\frac14\right) = 1+n\sin\left(\frac\pi2\right) = 1+n $$

Hence $K$ isn't bounded so it cannot be compact.

An alternative argument: define a linear functional $\phi : C[0,1] \to \mathbb{R}$ as $\phi(f) = \int_0^{1/2}f(t)\,dt$. We have that $\phi$ is bounded and hence continuous with respect to the supremum norm.

If $K$ were compact, $\phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have $$\phi(f_n) = \int_0^{1/2}f_n(t)\,dt = \frac12 + \frac{n}\pi$$ which is a contradiction.


Let $f_n(t)=(n+1)t^n$. Then $f_n \in K$ for all $n$. Can you proceed ?

  • $\begingroup$ The sequence $f_n$ is not uniformly convergent( in fact , it diverges at $t=1$). So, does this imply that the limit $\lim_{n\to\infty}\int_0^1f_n(t)dt=0$ which is not in $K$. Is this the right way? $\endgroup$ – vidyarthi Jan 8 at 11:47
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    $\begingroup$ @vidyarthi No, just note that $\|f_n\|_{\text{sup}} = n+1$ so that $K$ is unbounded in the norm so cannot be compact. $\endgroup$ – Henno Brandsma Jan 8 at 13:01

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