Fourier coefficients of real analytic functions on an n-dimension torus

Let $(\mathbf{R}^n,\langle\;,\; \rangle)$ be the n-dimensional euclidean space endowed with the standard inner product. For a lattice $L\subseteq \mathbf{R}^n$ we let $cov(L)$ denote the covolume of $L$ with respect to $\langle\;,\; \rangle$. We let $L^*$ denote the dual lattice of $L$ with respect to $\langle\;,\rangle$. Let $T:=\mathbf{R}^n/L$ be the corresponding real torus endowed with the Lebesgue measure dx. For every integrable function $f:T\rightarrow \mathbf{C}$ and each $\xi\in L^*$ we let $$a_{\xi}(f):=\frac{1}{cov(L)}\int_T f(x)e^{-2\pi i\langle x,\xi\rangle}dx$$ denote the $\xi$ Fourier coefficient of $f$. Assume now that $f$ is a real analytic function on $T$. Does this imply that there exists positive constants $c,C\in\mathbf{R}_{>0}$ such that $$|a_{\xi}(f)|\leq Ce^{-c||\xi||}\;\;$$ for all $\xi\in L^*$ ?

Here $||\xi||$ corresponds to the norm of the vector $\xi\in L^*$. If such a result is true I would like to have a reference. I seem to remember that this is true when $n=1$. I would also be interested to know if one can go in the other direction namely if $f$ is a smooth function (infinitely many times differentiable) such that its Fourier coefficients satisfy an inequality as above then $f$ is real analytic.

1 Answer

By the compactness of the torus, there is a uniform radius of convergence $r>0$ working for every point. You can extend $f$ to complex variables and use Cauchy's formula to find $|\partial^k f|\le Ck!(r/2)^{-k}$, with the constant $C$ only depending on $\sup_{|\Im z|\le r} |f|$. Then you get the same sort of bound for $|\xi|^k \hat f(\xi)$. Sum over $k$ with weights $c^k/k!$ where $c<r/2$ to get the desired estimate.

The converse is also true. Simply observe that the bound for $\hat f(\xi)$ allows you to extend the Fourier series to the region $|\Im z|<c$ where it still converges uniformly on compacts, to a limit that is analytic.

• What goes wrong here for the function $f(t)=t$? After all it’s Fourier series does not show exponential decay. – Ali Sep 26 '20 at 7:53