Twice differentiable but not infinitely differentiable

Question

I want an example of a function satisfying:

(1) $f:\mathbb R\to\mathbb R$ twice continuously differentiable

(2) $f''(x)$ $\ge 0$ for every x$\in \mathbb R$

(3)$\int\limits_0^x{f(t)dt}$ is NOT infinitely differentiable with respect to x.

I find it very difficult to get a function satisfying conditions 2 and 3. If I try $e^t$ it violates 3. If I try a function involving $t^5$sin(1/t), it violates condition 2. Please help me with a correct example. Thank you.

Answer 1

This should do it: Define $g(s) = 1-s^2, |s|\le 1,$ $g=0$ elsewhere. Set

$$f(x) = \int_0^x\int_0^t g(s)\,ds\,dt.$$

Note $f''(x) = g(x),$ which is not differentiable at $\pm 1.$

Answer 2

How can $$G(t)=\int_0^t[1-s^2] ds = t(1-\frac13t^2)$$ be differentiable at both $1$ and $-1$ and yield $g(s)$? As a hobbyist, I have to work it out explicitly. Since $G(-1)=-\frac23$ and $G(1)=\frac23$, we define $$G(t)=\begin{cases} t(1-\frac13t^2), &|t|\le 1,\\ -\frac23, &t<1,\\ \frac23, &t>1 \end{cases}$$ Piecewise differentiation yields $g(t)=g(s)$. Then differentiating $$f(x)=\begin{cases} \frac{x^2}{12}(6-{x^2}), &|x|\le 1,\\ -\frac23x, &x<1,\\ \frac23x, &x>1 \end{cases}$$ yields $G(x)=G(t)$. Thanks!