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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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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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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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