# Detecting Uniform Convergence By Fourier Methods

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.

["$$f_i$$ 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 :-/