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Let $M_c := \{f\in C^1([0,1]):\int |f|^2+\int |f'|^2\le c\}\subset C([0,1])$. Show that $\overline{{M}_c}$ is compact.

So since we are in a complete space, it is sufficient to show that $M_c$ is totally bounded, which means we need to verify equicontinuity and uniform boundedness. Equicontinuity is no problem, we get directly from the fundamental theorem and the Hölder inequality that $|f(y)-f(x)|\le |y-x|^\dfrac{1}{2}c$. Also we get from the definition $|f(x)-f(0)|\le c$, with which we can find that $f(0)\le c + \sqrt{c}$ such that $f(x)\le\sqrt{c} + 2c$ (analogous to here). So we have a uniform upper bound for all $f$, but I don't see how one can arrive at a lower bound from that, which is needed for uniform boundedness.

David C. Ullrich
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blst
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    Arguing more or less as above gives $|f(x)|\le k$ for some constant $k$, which is exactly what you need. – David C. Ullrich Nov 22 '17 at 16:24
  • The point is, I dont see the argument that "is more or less as above", hence the question. – blst Nov 22 '17 at 18:00
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    Look at it this way. If you've proved that $f(x)\le k$ for every $f\in M_c$ then it follows that $-f(x)\le k$ (and hence $f(x)\ge -k$), since $-f\in M_c$ if $f\in M_c$. That does what you need - it also shows that a proof that $f(x)\le k$ could be modified to give a proof that $|f(x)|\le k$. – David C. Ullrich Nov 22 '17 at 18:39
  • Sorry - yes, I missed the bar. It's easier to see now... – David C. Ullrich Nov 22 '17 at 20:56
  • Thanks! I did not see that the set also contains $-f$, from which, as you said, the other bound follows analogously. – blst Nov 24 '17 at 10:33

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Because the functions $f$ are absolutely continuous, $$ xf(x) = \int_{0}^{x}(f+tf')dt \\ (1-x)f(x) = \int_{x}^{1}(-f+(1-t)f')dt \\ f(x) = xf(x)+(1-x)f(x). $$ These identities and $2|ab|\le |a|^2+|b|^2$ give a uniform estimate for $f$: $$ |f(x)| \le (\|f\|\|1\|+\|f'\|\|t\|)+(\|f\|\|1\|+\|f'\|\|1-t\|) \\ \le (\|f\|^2+\|f'\|^2)+C \le c^2+C. $$

Disintegrating By Parts
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