Proving smoothness for a sequence of functions.

Question

Say I have a function $f_0(t)$ that is pointwise discontinuous in finite number of points and continuous in all other points. Now let the sequence of functions $f_1,f_2,\cdots$ be generated by the process:

$$f_n(t) = \frac{1}{2\epsilon}\int_{t-\epsilon}^{t+\epsilon}f_{n-1}(\tau)d\tau$$

For some $\epsilon > 0$. Can we prove how many times $f_k(t)$ will be continuously differentiable?

Answer 1

Let $f:I\rightarrow \mathbb{R}$ be a integrable function and $F(t):=\int_a^tf(x) d x$, ($a\in I)$ ; Then $F:I\rightarrow \mathbb{R}$ is continuous and differentiable on each point $t_0\in I$ where $f$ is continuous (then $F'(t_0)=f(t_0)$).

Using that you will have that $f_n(t)$, as you define it, is $\mathcal{C}^{n - 1}$ and $f_n^{(n - 1)}(t)$ is piecewise differentiable.

Answer 2

The derivative of $f_{n+1}$ is differentiable as many times as $f_n$: \begin{align} f^{'}_{n+1}(t) &= \frac{d}{dt} \left( \frac 1 {2 \epsilon}\int_{t-\epsilon}^{t+\epsilon}f_{n}(\tau)\,d\tau\right) \\ &= \frac 1 {2 \epsilon} \bigg( f_{n}(t+\epsilon)-f_{n}(t-\epsilon) \bigg) \end{align}

Consequently, if $f_n$ is $C^x$, then $f_{n+1}$ is $C^{x+1}$. $f_1$ is $C^0$ (continuous but not differentiable) and so it follows that $f_n$ is $C^{n-1}$ (i.e. differentiable $n-1$ times).