If $u, u'' \in L^2(0,1)$, is it true that $u' \in L^2(0,1)$?

Question

Let $u \in L^2(0,1)$. If $$u'' \in L^2(0,1)$$ is it true that $$u' \in L^2(0,1)?$$ Why yes/not? If $u, u'' \in L^2(0, 1)$ do not imply that $u' \in L^2(0,1)$, how can I show that $u' \in L^2(0,1)$?

Thank you!

Answer 1

As stated in the comments, this follows from a much more general inequality. Anyway, in the $L^2$ context things are pretty easy to bound by exploiting Fourier series. Assuming $u,u''\in L^2(0,1)$ it follows that: $$ u(x) = M + \sum_{n\geq 1}\left( a_n \sin(2\pi nx)+ b_n \cos(2\pi nx)\right), $$ $$ u'(x) = \sum_{n\geq 1}2\pi n\left( a_n \cos(2\pi nx)- b_n \sin(2\pi nx)\right), $$ $$ u''(x) = -\sum_{n\geq 1}(2\pi n)^2\left( a_n \sin(2\pi nx)+ b_n \cos(2\pi nx)\right), $$ as well as: $$ \|u\|_2^2 = M^2 + \frac{1}{2}\sum_{n\geq 1}\left(a_n^2+b_n^2\right), $$ $$ \|u'\|_2^2 = \frac{1}{2}\sum_{n\geq 1}4\pi^2 n^2\left(a_n^2+b_n^2\right), $$ $$ \|u''\|_2^2 = \frac{1}{2}\sum_{n\geq 1}16\pi^4 n^4\left(a_n^2+b_n^2\right), $$ so $u'\in L^2(0,1)$ just follows from the Cauchy-Schwarz inequality.

Answer 2

For any reasonable sense of the word "derivative", you should have, for some constant $c$ (think $c = u'(0)$): $$u'(x) = c + \int_0^x u''(t)\,dt \quad \text{for a.e. $x$}.$$ Since $u'' \in L^2(0,1) \subset L^1(0,1)$ (Cauchy-Schwarz), it follows that $u'$ is bounded (or more properly, is a.e. equal to a bounded function). So it is definitely in $L^2$. From the dominated convergence theorem, it also follows that $u'$ is (a.e. equal to) a continuous function. In fact, it is absolutely continuous.