# Fourier Transforms of Harmonic Functions

Suppose you are presented with the equation ($D=3$) \begin{equation} \nabla^2 A(x) = \nabla^2 B(x). \end{equation} Decompose $A$ and $B$ into their Fourier components, \begin{equation} A(x) = \int d^3k \ e^{ikx} \tilde A(k),\qquad B(x) = \int d^3k\ e^{ikx} \tilde B(k), \end{equation} to yield \begin{equation} \int d^3k \ e^{ikx}\left[-k^2 (\tilde A - \tilde B)\right]=0. \end{equation} Assuming the inverse Fourier transform of zero is uniquely zero (or based on the answers Fourier Transforming the Klein Gordon Equation), we conclude \begin{equation} \tilde A = \tilde B \quad\Rightarrow\quad A(x) = B(x). \end{equation} Yet we know $A(x)$ may vary up to arbitrary harmonic functions.

What gives?

Also consider this:
• $xf(x)=0$ does not imply $f(x)=0$. It implies that $f(x)=\sum_i c_i\delta^{(i)}(x)$ for some coefficients $c_i$. Jun 12, 2018 at 19:48
• @AccidentalFourierTransform what does the superscipt $i$ on the delta function represent? Yes I was thinking that $f(x) = g(x)\delta(x)$ provided $xg(x)|_0 = 0$. Jun 12, 2018 at 20:11
• $f^{(i)}(x)=\frac{\mathrm d^i}{\mathrm dx^i}f(x)$. Jun 12, 2018 at 20:12
• @AccidentalFourierTransform I see. Do you know where else I can read about decompositions such as $f(x) = \sum_i c_i \delta^{(i)}(x)$ ? Also it seems the expression for $f(x)$ is not exactly unique. Jun 12, 2018 at 21:22
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