If $U$ is an open interval of $\mathbb{R}$ and $f : U \to \mathbb{R}$ is an $L^2(U)$ function with **second** derivative $f'' \in L^2(U)$ (in the weak sense), is $f \in W^{1,1}(U)$?


EDIT: Removed false inequality.



Interpolation of derivatives
I assume $U$ is a *finite* open interval, else the assertion is clearly false (let $f(x)=x$).


Then a standard estimate shows that $f'$ is bounded, and thus in $L^1(U)$, whence $f$ is in the Sobolev space $W^{1,1}(U)$ (in fact in $W^{1,p}(U)$ for all $p$).


Fix some $x\_0 \in U$, and write
$$
\left|f'(x) - f'(x\_0)\right| = \left| \int\_{x\_0}^x f''(y) dy \right|
\leq \| f'' \|\_2 \phantom. |x-x\_0|^{1/2},
$$
using Cauchy-Schwarz in the last step. Since $\| f'' \|\_2$ is a finite constant and $|x-x\_0|$ is bounded, so is $\left|f'(x) - f'(x\_0)\right|$, and we are done.


This kind of argument is of course well-known, and probably predates Sobolev himself, but is easier to write up than to look up. A reader better versed in the literature may be able to supply a canonical reference.


