Let $\{f_n\} \subset C((0,1))$ be a sequence of functions such that

$$\sup_{n \in \mathbb{N}}\{f_n(0) + f_n'(0)\} =:M<\infty$$

and there exists $\alpha \in (1, 2)$ such that $\forall n \in \mathbb{N}$ and $x \in (0, 1)$,

$$|f''_n(x)|\leq 1/(1-x)^\alpha$$

Show that there is a subsequence that converges to a continuous function uniformly on every compact subset.

So we must prove $\{f_n\}$ is equicontinuous and bounded.

My calculus is a bit rusty so I'm struggling with this.

Is pointwise boundedness really as simple as saying, for every $x \in (0,1)$:

$$|f_n(x)| \leq \iint 1/(1-x)^\alpha dx = \frac{1}{(\alpha -1 )(\alpha - 2)(1-x)^{\alpha -2}}$$

for all $n \in \mathbb{N}$, and the RHS is independent of n, and thus is a bound?

I think that works, but I'm not sure how to show equicontinuity. I would appreciate the help.

Soft supplemental question:

What is the general approach and tricks used to prove stuff like this? I have been struggling with another question of the same sort (for which I would also appreciate help btw) and I would like to know the general strategy to prove boundedness and equicontinuity given these kinds of conditions on the functions.

  • $\begingroup$ You are missing absolute value signs in the first condition. The way you have stated $f_n(x)=-n$ provides a counter-example. $\endgroup$ Aug 27 '20 at 5:36

Applying the Fundamental Theorem of Calculus, we have:

$f_{n}(s) = f_{n}(0) + f'_{n}(0)s + \int_{0}^{s} \int_{0}^{x} f''_{n}(t) dt dx$.

The boundedness follows by invoking the two conditions given after taking absolute value and applying the triangle inequality.

If you use the same formula above, you should be able to get:

$f_{n}(s) - f_{n}(w) = f'_{n}(0)[s - w] + \int_{w}^{s} \int_{0}^{x} f''_{n}(t)dtdx$ and the equicontinuity should be able to drop out from this.


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