Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
Hint: $g$ is continuous. –  Ohad Asor Jan 12 '13 at 11:31
    
This $g$ is continuous. –  Gerald Edgar Jan 12 '13 at 12:47
    
so maybe I didn't get it.. doesn't Dirac and Heavyside discontinuous? –  Ohad Asor Jan 12 '13 at 13:11
    
@Ohad: the product of $x-t$ and $\theta(x-t)$ is a continuous function with a discontinuous first derivative, $dg/dt=-\theta(x-t)$. –  Carlo Beenakker Jan 12 '13 at 14:32
1  
$g(x,t)={\rm max}(x-t,0)$ –  Carlo Beenakker Jan 12 '13 at 15:24

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

share|improve this answer
    
I dont understand where $f$ comes into the picture –  Ohad Asor Jan 12 '13 at 13:12
    
Fixed ${}{}{}{}$ –  Gerald Edgar Jan 12 '13 at 16:32
    
Thanks Gerald I appreciate your help –  Ohad Asor Jan 12 '13 at 17:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.