5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @ThomasShelby $K$ is definitely not a bounded subset of $C[0,1]$, see Fred's answer. $\endgroup$ – daw Jan 8 at 12:56
  • 2
    $\begingroup$ @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
2
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.