most recent 30 from http://mathoverflow.net 2013-05-21T10:36:34Z http://mathoverflow.net/feeds/question/86344 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86344/interpolation-of-derivatives John H 2012-01-22T01:03:01Z 2012-01-22T04:19:12Z

<p>If $U$ is an open interval of $\mathbb{R}$ and $f : U \to \mathbb{R}$ is an $L^2(U)$ function with <strong>second</strong> derivative $f'' \in L^2(U)$ (in the weak sense), is $f \in W^{1,1}(U)$?</p> <p>EDIT: Removed false inequality.</p>

http://mathoverflow.net/questions/86344/interpolation-of-derivatives/86353#86353 Noam D. Elkies 2012-01-22T04:19:12Z 2012-01-22T04:19:12Z

<p>I assume $U$ is a <em>finite</em> open interval, else the assertion is clearly false (let $f(x)=x$).</p> <p>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 $p$).</p> <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.</p> <p>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.</p>