MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Sorry, but I can't seem to fix the TeX... – John H Jan 22 '12 at 1:05
I think linear functions might break your inequality what with the zero second derivative and non-zero first derivative. – BSteinhurst Jan 22 '12 at 2:31
I fixed your LaTeX, but don't immediately see how to fix your inequality – Yemon Choi Jan 22 '12 at 2:45
You are totally right, that inequality is false. – John H Jan 22 '12 at 2:51
up vote 5 down vote accepted

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.

share|cite|improve this answer
This argument also shows that $f'$ is of Hölder class $C^{0, 1/2}$ and in particular is continuous. So the conclusion is that $f$ belongs to $C^{1, 1/2}$, and in particular is continuously differentiable. – Phil Isett Jan 22 '12 at 4:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.