Suppose that $D\subset \mathbb R^2$ is the closed unit disk and that $f\in C^{\infty}(D)$. Assume that for all $\lambda \in (1,\infty)$ there holds $$ \left|\int_D f(x^1,x^2)\, e^{\lambda (x^1+ix^2)}\,dx \right|\leq e^{-\frac{1}{2}\lambda}.$$ Does it necessarily follow that $f$ must vanish near the point $(1,0)$?
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No, of course not. Essentially, you want to bound a $2$-dimensional object (the complex Fourier transform. I talk about the complex dimension here) from the at most $1$-dimensional bound (that is, even if we allow your bound for all $\lambda\in \mathbb{C}$). Dimensions do not match, so this must be false.
Indeed, we will make your integral zero for all $\lambda\in \mathbb{C}$. So, we want $\hat{f}(\xi_1, \xi_2)$ to be zero for $\xi_2 = i\xi_1$, and the easiest way to do so is to just consider $\hat{g} = \hat{f}(\xi_2 - i\xi_1)$ (if you don't like complex numbers, multiply by $(\xi_2^2 +\xi_1^2)$). On the function side this will corrsepond to some differential operator, which does not increase the support, and if the function is generic enough it does not change it either. The second operator is actually just a Laplacian.
So, the recipe is: pick any function $F\in C^\infty(D)$ you like, then put $f = \Delta F$. Now, you don't even need fancy words like Fourier transform, integration by parts shows that the integral is zero for all $\lambda\in \mathbb{C}$. And of course we can have the support of both $F$ and $f$ to be the whole disk.
answered Apr 18 at 13:56
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1$\begingroup$ In hindsight, this was trivial indeed. The key observation is perhaps that, as you pointed out, there is a local differntial operator, the Laplace operator in this case, that annihilates the exponential weight, $e^{\lambda z}$ for each $\lambda\in \mathbb C$. $\endgroup$– AliCommented Apr 18 at 16:18