A close snorbaki friend has posed a problem about uniform convergence of a sequence of functions, and I've noted it has a particularly well behaved sequence of Fourier transforms. Can I dazzle my friend with a proof that proceeds primarily in the Fourier domain?

That is to say, suppose we are given a sequence of square integrable functions $f_i : [0, 1] \to \mathbb{R} $, as well as their fourier transforms $\hat f_i : \mathbb{R} \to \mathbb{C}$. Is there a statement we can prove about $\hat f_i$, short of invoking the inverse Fourier transform per se, which is a necessary and sufficient condition for the uniform convergence of $f_i$?

My guess so far is that "$f_i$ is uniformly convergent" $\Leftrightarrow$ "$\frac{1}{x}\hat f_i$ is uniformly convergent". I don't have much rationale for this, apart from a vague notion that uniform convergence in the time domain should imply different rates of converge for high versus low frequencies. My next step is to test this.


1 Answer 1


I think it's a bit more subtle than that.

["$f_i$ is uniformly convergent" $\Leftrightarrow$ "$\frac 1 \omega {\hat{f}_i(\omega)}$ is uniformly convergent"] is not true. Think of $\frac 1 \omega {\hat{f}_i(\omega)}$ as the Fourier transform of the antiderivative of $f_i$, therefore you can't say that uniform convergence of a sequence is equivalent to that of the antiderivatives.

The other point to notice is that the Fourier transform is a duality operation in that, thanks to the Hausdorff-Young inequality, it tends to map $L^p$ to $L^q$ for $1\leq p\leq 2$ and $q$ being the Hölder conjugate of $p$: $$\frac 1 p + \frac 1 q = 1$$

The $L^\infty$ norm of the $f$ is bounded by the $L^1$ norm of $\hat f$, so at least a sufficient condition for the uniform convergence of $\{f_i\}$ is that $\hat{f}_i$ converges in $L^1$. I know it's not exactly what you wanted, sorry :-/


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