precompactness of integral of functions in $C^1[a,b]$

Question

Prove that the set of continuously differentiable functions $x \in C^1[a,b]$ such that $\int_a^b[x^2(t) + |x'(t)|^2]dt < k = $ const, is pre-compact in $C[a, b]$.

I've read that a family of continuously differentiable functions with uniformly bounded derivatives is equicontinuous, so if we have that and boundedness it is precompact. Here we are given boundedness so I think it only remains to show that the family of functions has uniformly bounded derivatives. How do I show this?

Answer 1

Let's call your set of functions $X.$ As David Ullrich showed, uniform equicontinuity holds for any sequence in $X.$ For precompactness, all we need to show in addition is that given a sequence $x_n$ in $X,$ there exists $t\in [a,b]$ along which $x_{n}(t)$ is bounded along some subsequence. This follows from Fatou's Lemma: We have

$$\int_a^b \liminf |x_n(t)|^2\,dt \le \liminf\int_a^b |x_n(t)|^2\,dt \le k.$$

Thus $\liminf |x_n(t)|^2 <\infty$ a.e., which gives a lot of $t$'s for which the desired subsequences exist.

Answer 2

"Here we are given boundedness"? Not as far as I can see. Your family is uniformly bounded, but as far as I can see that's going to follows after we prove equicontinuity.

The derivatives are not uniformly bounded. Luckily you don't need to show that. Cauchy-Schwarz shows that if $x<y$ then $$|f(x)-f(y)|\le\int_x^y|f'(t)|\,dt\le\left(\int_x^y 1\,dt\int_x^y|f'(t)|^2\,dt\right)^{1/2}\le k^{1/2}(y-x)^{1/2}.$$

That gives equicontinuity. Pointwise boundedness follows - it's actually simpler than I realized. Inspired by what zhw said, or more precisely by the part of his argument that he left unsaid:

Say $\alpha=\inf_{t\in[a,b]}|f(t)|$. Then $$k\ge\int_a^b|f|^2\ge(b-a)\alpha^2,$$so $$\alpha\le(k/(b-a))^{1/2}.$$There exists $t$ with $|f(t)|=\alpha$; now for every $x\in[a,b]$ we have $$|f(x)|\le|f(t)|+|f(x)-f(t)|\le\alpha+k^{1/2}|x-t|^{1/2} \le(k/(b-a))^{1/2}+(k(b-a))^{1/2}.$$

Original version: I don't see how to get pointwise boundedness except by a little trick: Note first that $$f(x)=\frac1{b-a}\int_a^b(f(x)-f(t))\,dt+\frac1{b-a}f(t)\,dt,$$so$$|f(x)|\le\frac1{b-a}\int_a^b|f(x)-f(t)|\,dt+\frac1{b-a}|f(t)|\,dt.$$You can use Cauchy-Schwarz again on one integral and use the fact that $|f(x)-f(t)|\le k^{1/2}|x-t|^{1/2}$ on the other integral.