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I'm reading Jones' book Lebesgue integration on Euclidean space. Let $u(x, y)$ be a harmonic function on the half space $\mathbb{R}^n \times (0, \infty)$, with boundary condition $f(x) = u(x, 0)$. On p. 341, he concludes that

$$\displaystyle f(x) = (2\pi)^{-n} \int_{\mathbb{R}^n} e^{ix \cdot \xi} \hat{f}(\xi) \; d\xi$$ $$\displaystyle u(x, y) = (2\pi)^{-n} \int_{\mathbb{R}^n} e^{-|\xi|y + ix \cdot \xi} \hat{f}(\xi) \; d\xi$$

He then asks, "Do you see what has happened? We have essentially discovered that it is very easy to solve [the Dirichlet problem] in case the given function on $\partial G$ is $e^{ix \cdot \xi}$ for some fixed $\xi \in \mathbb{R}^n$; the solution is exactly $e^{-|\xi|y + ix \cdot \xi}$."

I don't see what happened. I substitute $f(x) = e^{ix \cdot \xi}$ into the above and I get nowhere with this.

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Take $\hat{f}$ to be a Dirac delta at $\xi$. –  Robert Israel Feb 17 '12 at 0:04
So $\hat{f} = 0$ except at $\xi$, where it is 1. But then the integral $\int_{\mathbb{R}^n} e^{ix \xi} \hat{f}(\xi) \; d\xi = 0$. So $f(x) = 0$? But we assume $f(x) = e^{ix \cdot \xi}$. I am confused. –  user980123 Feb 17 '12 at 0:06
If the Dirac delta is confusing, forget about. Simply check that the given solution indeed satisfies the equation and has the right boundary condition. No integrals needed. –  user53153 Feb 11 '13 at 13:22

1 Answer 1

I guess this question might be solved or abandoned right now but Dirac delta is infinity at xi and zero otherwise.

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