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.
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. 


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 $$ \leftf'(x)  f'(x_0)\right = \left \int_{x_0}^x f''(y) dy \right \leq \ f'' \_2 \phantom. xx_0^{1/2}, $$ using CauchySchwarz in the last step. Since $\ f'' \_2$ is a finite constant and $xx_0$ is bounded, so is $\leftf'(x)  f'(x_0)\right$, and we are done. This kind of argument is of course wellknown, 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. 

