Showing that a function is in $L^1$

Question

I need to prove the following statement or find a counter-example: Let $u\in L^1\cap C^2$ with $u''\in L^1$. Then $u'\in L^1$.

Unfortunately, I have no idea how to prove or disprove it, since the $|\bullet|$ in the definition of $L^1$ is giving me huge problems. I found counter-examples if either $u\notin L^1$ or $u''\notin L^1$, but none of them could be generalized to a example that satisfies all the conditions.

Answer 1

It is useful to recall the following lemma:

If $f$ is a twice differentiable function over $I=[a,b]$ and $$M_0=\sup_{x\in I}|f(x)|,\quad M_1=\sup_{x\in I}|f'(x)|,\quad M_2=\sup_{x\in I} |f''(x)|,$$ then $M_1^2\leq 4M_0 M_2$.

It is a well-known exercise from baby Rudin's: you can find a proof of it here.

By the Cauchy-Schwarz inequality, it gives:

$$\begin{eqnarray*} \int_{-M}^{M}|u'(t)|\,dt &\leq& \sum_{k=0}^{N-1}\sup_{t\in I_k=\left[-M+k\frac{2M}{N},-M+(k+1)\frac{2M}{N}\right]}|u'(t)|\cdot\mu(I_k)\\&\leq& 2\sum_{k=0}^{N-1}\sqrt{\mu(I_k)\sup_{t\in I_k}|u(t)|\cdot \sup_{t\in I_k}|u''(t)|\mu(I_k)}\\&\leq&2\sqrt{\sum_{k=0}^{N-1}\sup_{t\in I_k}|u(t)|\mu(I_k)\cdot \sum_{k=0}^{N-1}\sup_{t\in I_k}|u''(t)|\mu(I_k)},\end{eqnarray*} $$ so, by letting $N\to +\infty$, we have that the $L^1$-norm of $u'(t)$ on $[-M,M]$ is bounded by twice the geometric mean of the $L^1$-norms of $u(t)$ and $u''(t)$ on the same interval. This proves the claim.