Problem 2

Exercise 2.2 (Arzelà-Ascoli on smooth functions)   Let $\{f_n : [0,1]\to\mathbb{R}\}$ be twice differentiable such that $f_n(0)=f_n'(0)=0$ and $\vert f_n''(x)\vert\leq 1$ uniformly. Then there exists a subsequence which converges uniformly.

Proof. This is a direct application of Arzela-Ascoli. Include the derivatives in an ambient space

$$\{f_n'\}_{n=1}^\infty \subseteq C([0,1])$$ (2.8)

If we show the sequence is bounded and equicontinuous, then the uniformly converging subsequence will be summoned by Ascoli himself.

For boundedness, compute the sup norm of each derivative

$$\Vert f_n'\Vert= \sup_{x\in[0,1]}\vert f_n'(x)\vert \leq \sup_{x\in[0,1]} \int_0^x \vert f_n''(y)\vert dy \leq 1$$ (2.9)

For equicontinuity, we can show the sequence of derivatives is Lipschitz. Bound the derivative

$$\vert f_n'(y) - f_n'(x)\vert \leq \int_x^y \vert f_n''(z)\vert dz \leq y-x$$ (2.10)

Therefore,

$$\sup_{x\neq y} \frac{\vert f_n'(x)-f_n'(y)\vert}{\vert x-y\vert} \leq 1$$ (2.11)

Therefore, we may select $f_{n_k}'$ a subsequence converging uniformly to $f'$, which we now prove equals the derivative.

Define a function

$$f(x) = \int_0^x f'(y)dy$$ (2.12)

We will show that $f_{n_k}\to f$ uniformly.

The definition of integration in $\mathbb{R}$ begets

$$f_n(x) = \int_0^x f_n'(y)dy \quad$$ (2.13)

Now for the limit

$$\sup_{x\in[0,1]}\vert f_{n_k}(x) - f(x)\vert = \sup_{x\in[0,1]}\left\vert \int_0^x f'_{n_k}(y) - f'(y)dy \right\vert \leq \sup_{x\in[0,1]} \int_0^x \vert f'_{n_k}(y)-f'(y)\vert dy$$ (2.14)

$$\leq \int_0^1 \vert f'_{n_k}(y)-f'(y)\vert dy \leq \Vert f'_{n_k} - f'\Vert$$ (2.15)

Therefore, the convergence is uniform. $\qedsymbol$