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Hi,

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 $f$.

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.

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2 Answers 2

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this looks like a simple consequence, upon twice partial integration, of $f(x)=\int_0^T f(t)\delta(t-x)dt$, so your $g(x,t)=(x-t)\theta(x-t)$

$\delta(x)=d\theta(x)/dx$ relates Dirac delta function and Heaviside step function.

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  • $\begingroup$ Hint: $g$ is continuous. $\endgroup$
    – Ohad Asor
    Commented Jan 12, 2013 at 11:31
  • $\begingroup$ This $g$ is continuous. $\endgroup$
    – Gerald Edgar
    Commented Jan 12, 2013 at 12:47
  • $\begingroup$ so maybe I didn't get it.. doesn't Dirac and Heavyside discontinuous? $\endgroup$
    – Ohad Asor
    Commented Jan 12, 2013 at 13:11
  • $\begingroup$ @Ohad: the product of $x-t$ and $\theta(x-t)$ is a continuous function with a discontinuous first derivative, $dg/dt=-\theta(x-t)$. $\endgroup$
    – Carlo Beenakker
    Commented Jan 12, 2013 at 14:32
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    $\begingroup$ $g(x,t)={\rm max}(x-t,0)$ $\endgroup$
    – Carlo Beenakker
    Commented Jan 12, 2013 at 15:24
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Hint ... write $u(x) = f''(x)$, so that the condition is $$ \int_0^T u(t) g(x,y)dt = \int_0^x\left[\int_0^y u(s) ds\right]dy $$

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  • $\begingroup$ I dont understand where $f$ comes into the picture $\endgroup$
    – Ohad Asor
    Commented Jan 12, 2013 at 13:12
  • $\begingroup$ Fixed ${}{}{}{}$ $\endgroup$
    – Gerald Edgar
    Commented Jan 12, 2013 at 16:32

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