# can we have $u=0$ from the integral value 0?

If u is a bivariate function and we have $\int_\theta^{\theta+1}{\int_\theta^y{u(x,y)(y-x)^{n-2}}dx}dy=0$ for all $\theta\in\mathbb R$, here $n>2$ is a constant, can we infer that $u=0$ a.e. on the area between two straight lines $y=x$ and $y=x+1$?

If yes, please give a proof; if no, please give a counterexample. Thanks!

PS: 1. This is not an exercise. 2. I have no idea about it.

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The integral only depends on $u(x,y)$ when $\theta\le x\le y\le\theta+1$, so any non-zero function that vanishes on this region is a counterexample. For example, $u(x,y)=\max(0,x-y)$.

Edit following question update:

You can find $u$ of the form $u(x,y) = f(y-x)$. We need $I(f)=0$: $$I(f) = \int_\theta^{\theta+1}{\int_0^{y-\theta}{f(t) t^{n-2}}dt}dy$$ $$= \int_0^1{\int_0^z{f(t) t^{n-2}}dt}dz$$ (so $I(f)$ does not depend on $\theta$)

Then take any two linearly independent $f_1$, $f_2$ with whatever smoothness condition you need. $I$ is a linear map from the span of $\{f_1,f_2\}$ to $\mathbb R$, so there must be some non-zero linear combination $f$ such that $I(f)=0$.

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Thank you for the useful reply. Your reply make me find a need to sharpen somewhat my question. – Zhou Heng Apr 23 '12 at 2:57
Updated the reply :) – Generic Human Apr 23 '12 at 4:30
Yes, $y-x-\frac{n - 1}{n + 1}$ is the case. – Zhou Heng Apr 23 '12 at 8:38