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.
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$\begingroup$ Sorry, but I can't seem to fix the TeX... $\endgroup$– John HCommented Jan 22, 2012 at 1:05
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1$\begingroup$ I think linear functions might break your inequality what with the zero second derivative and non-zero first derivative. $\endgroup$– BSteinhurstCommented Jan 22, 2012 at 2:31
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1$\begingroup$ I fixed your LaTeX, but don't immediately see how to fix your inequality $\endgroup$– Yemon ChoiCommented Jan 22, 2012 at 2:45
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$\begingroup$ You are totally right, that inequality is false. $\endgroup$– John HCommented Jan 22, 2012 at 2:51
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1 Answer
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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.
answered Jan 22, 2012 at 4:19
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3$\begingroup$ 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. $\endgroup$ Commented Jan 22, 2012 at 4:31