most recent 30 from http://mathoverflow.net 2013-05-23T19:22:27Z http://mathoverflow.net/feeds/question/118687 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118687/rfc-for-definite-integral-connection-to-second-derivative Ohad Asor 2013-01-12T00:31:08Z 2013-01-12T16:32:20Z

<p>Hi,</p> <p>During my research I found an interesting fact, and I'd like to know if it's interesting for others as well. Find a function $g(x,t):[0,T]\times[0,T]\rightarrow[0,T]$ such that for any twice differentiable $f(x):[0,T]\rightarrow[0,T]$ such that $f(0)=f'(0)=0$, the equality $$ f(x)=\intop_0^Tf''(t)g(x,t)dt$$ holds. Note that $g$ is independent of $f$.</p> <p>I found such a $g$, and I'll post it as an answer soon. I'd like to know if this is simple/known/interesting.</p>

http://mathoverflow.net/questions/118687/rfc-for-definite-integral-connection-to-second-derivative/118720#118720 Carlo Beenakker 2013-01-12T10:24:29Z 2013-01-12T10:24:29Z

<p>this looks like a simple consequence, upon twice partial integration, of $f(x)=\int_0^T f(t)\delta(t-x)dt$, so your $g(x,t)=(x-t)\theta(x-t)$</p> <p>$\delta(x)=d\theta(x)/dx$ relates Dirac delta function and Heaviside step function.</p>

http://mathoverflow.net/questions/118687/rfc-for-definite-integral-connection-to-second-derivative/118726#118726 Gerald Edgar 2013-01-12T12:43:57Z 2013-01-12T16:32:20Z

<p>Hint ... write $u(x) = f''(x)$, so that the condition is $$ \int_0^T u(t) g(x,y)dt = \int_0^x\left[\int_0^y u(s) ds\right]dy $$</p>