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Let $U$ be an open subset of $\mathbb{R}^n$ with smooth boundary $\partial U$. Consider a function $f$ in $L^2(u)$. Suppose that for every $h$ in the Sobolev space$ H^2_0(U)$ it holds that $$\int_U f \Delta h=0.$$ Where $\Delta$ is the Laplacian operator.n

Can we conclude that $f=0$ almost everywhere on $U$?

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Let $u: \overline {U}\to \mathbb{R}$ be a harmonic function, i.e. $$\tag{1}\Delta u=0\ in\ U$$

Mutliply $(1)$ by $h\in H_0^2(U)$ in both sides and then integrate: $$\tag{2}\int_U\Delta u\cdot h=0,\ \forall\ h\in H_0^2(U)$$

Use the generalized Green identity to conclude from $(2)$ that $$\tag{3}0=\int_U\Delta u\cdot h=-\int_U\nabla u\nabla h=\int_Uu\Delta h,\ \forall\ h\in H_0^2(U)$$

From $(3)$ we have your claim but $u$ does not need to be zero almost everywhere.

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I think so. Assume that $f$ is nonnegative and smooth in some open set $S$. Now test against $\Delta h$ where $\Delta h=1$ in an open set $\Omega\subset S$ with homogeneous Dirichlet boundary conditions. Is this ok? Now, one can use the density of $C_c^\infty$ in $L^2$ and some approximation argument, isn't it?

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It looks good. I was thinking of finding a suitable test function sequence $\{\eta_n\}\subset C^\infty_0$ with the property that it's equal to one on some open set $\Omega\subset U$ such that $\Omega$ has compact neighborhood in $U$. Choose the sequence $\eta_n$ such that it converges in the appropriate sense to the indicator function $1_{\Omega}$. I now multiply this with the smooth function $g(x)=\sum_{i=1}^n \frac{x_i^2}{2}$. Then clearly on $\Omega$ the Laplacian of these functions $g\cdot\eta_n$ is one. – Aris May 20 '13 at 19:34
I think that the technical problem is now to choose this open set $\Omega$ good enough such that each $g\cdot\eta_n$ is an actual element of $H^2_0$. I believe that if $f\neq 0$ a.e. then there exists some $n$-dimensional cube $J$ on which $\int_J f\neq 0$. If we choose $\Omega$ equal to this $J$ then I think the sequence $\{g\eta_n\}$ converges to $1\_{\Omega} g$. Which shows that the latter is in $H^2_0$. – Aris May 20 '13 at 19:50
with $\Delta (g 1_{J}) = 1$ – Aris May 20 '13 at 19:55
Well, if you approximate $f$ by smooth functions (write $f_n$) then the level sets are smooth (right?). Then, assuming that $f_n$ is away from zero, you can consider $omega$ as the inner subdomain. Right? – guacho May 20 '13 at 20:04
Notice that the test function I said before is not in $H2_0$ :-(. – guacho May 20 '13 at 20:09

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