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I am attempting to look at some Liouville type theorems via a Fourier analysis approach and after looking at a baby problem I seem to be very confused. I assume this doesn't count as a research problem (but the original questions might be).

So here is the baby problem. Consider $ \Delta \phi(x,y)=0$ for $ (x,y) \in R^2_+=R \times (0,\infty)$ with $ \phi(x,0)=0$. I will impose the a gradient bound like $ | \nabla \phi(x,y)| \le C y^{-\alpha}$ where $ \alpha \in (0,1)$. So $ | \phi(x,y)| \le C_1 y^{1-\alpha}$.

So without the gradient bound one clearly has solutions like $ \phi(x,y)= C y$ and after imposing the bound one sees they must have $ C=0$. So with this explicit solution if we take a Fourier transform in $x$ we arrive at $ \hat{\phi}(\xi,y)= C y \delta_0(\xi)$ where $ \delta_0$ is the Dirac mass at $0$.

Okay we now move on to the equation and assume $ \phi$ is generic (so not the explicit solution above). Taking a Fourier transform in $x$ we arrive at $$ \partial_{yy} \hat{\phi}(\xi,y) = \xi^2 \hat{\phi}(\xi,y) $$ and solving this ode gives $$ \hat{\phi}(\xi,y)= A(\xi) e^{\xi y} + B(\xi) e^{-\xi y} $$ for $ \xi \neq0 $ and $ \hat{\phi}(0,y)=C_1 y + C_2$. (I am being quite careful here since I can't tell where the problem is coming from). So now if we impose the boundary condition we get $B(\xi)=- A(\xi)$ and $ C_2=0$. From this formula i don't see how one is getting the possible solution with the Dirac mass.

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  • $\begingroup$ The Fourier transform of $\phi$ need not be a pure function. $\endgroup$ Commented Dec 15, 2020 at 20:55
  • $\begingroup$ i assume you mean it can be 'worse' like the explicit example i gave. I still thought it should `pop' out of the formula ? No ? $\endgroup$
    – Math604
    Commented Dec 15, 2020 at 21:21
  • $\begingroup$ I bet there is a way, but I doubt it is very simple: it has to include all sorts of other "exotic" harmonic functions, such as $xy$, $3 x^2 y - y^3$ and so on. (And if you go beyond tempered distributions, then also $\operatorname{Im} \exp((x + i y)^2)$ and so on). $\endgroup$ Commented Dec 15, 2020 at 21:31
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    $\begingroup$ If you consider zero boundary condition only, you can extend your solution via $u(x,-y) = -u(x,y)$ and then use Fourier transform in all of $\mathbb{R}^2$. This tells you that all solutions that are tempered distributions (jointly) are harmonic polynomials. $\endgroup$ Commented Dec 15, 2020 at 21:33
  • $\begingroup$ thanks for your comments; it really helps me. $\endgroup$
    – Math604
    Commented Dec 15, 2020 at 21:37

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