Edit: A previous version of this question claimed that $\widehat{\phi_N}$ converged uniformly; I meant to say locally uniformly. It has been updated.
The context is the following problem: given $\phi \in L^1(\mathbb{R})$ and $\int_{\mathbb{R}} \phi = 1$, define $\phi_N(x) = N\phi(Nx)$ and prove that $\widehat{\phi_N}$ converges locally uniformly (i.e. uniformly on any finite interval) to the function which is identically 1 on $\mathbb{R}$. (And I suppose an answer to this problem would maybe almost answer this post.)
Using dominated convergence theorem, one can show that $\widehat{\phi_N}(\xi) \to 1$ pointwise everywhere. It comes down to showing that
$$
|\widehat{\phi_N}(\xi) – 1| \leq \int_{\mathbb{R}} |e^{i\xi y/N} – 1||\phi(y)|\, dy,
$$
which then as $\phi \in L^1$ goes to zero as $N \to \infty$. This does NOT, as far as I have gone, show uniform continuity. So, my questions are:
- How to show uniform convergence on a finite interval here?
- On a more abstract level, I find it hard to see how dominated convergence theorem could ever give uniform convergence when the integrand depends on a second parameter, since applying DCT "ignores" any hard quantitative estimates and just says the limit is zero. Does anyone know if DCT can be applied to give uniform convergence? Or the opposite?
Best Answer
This is false. The easiest way to prove this is to put $\xi=N$. Uniform convergence would imply that $|\int [e^{iy}-1] \phi (y) dy |<\epsilon$ for every $\epsilon >0$ which need not be true.
Local uniform convergence: There exists $T$ such that $\int_{|x| >T} |\phi (x)| d x<\epsilon$. Now $\int_{-T}^{T} |e^{i\xi y/N}-1|dy \leq \frac {|\xi T|} N \int_{-T}^{T} |\phi (y)|dy $. Can you finish?
I have used the inequality $|e^{ia} -1| \leq |a|$ for all $a \in \mathbb R$.
How does one use DCT in this argument?: If $\int_{-T}^{T} [e^{i\xi y/N}-1] \phi (y) dy$ does not tend to $0$ uniformly on some bounded set then there exist $\epsilon >0$ and sequence $\xi_n, N_n$ such that $(\xi_n)$ is bounded and $|\int_{-T}^{T} [e^{e\xi y/N}-1] \phi )(y) dy |\geq \epsilon$ for all $n$. Now use the fact that the integrand is bounded by a constant times $|\phi|$ to get a contradiction to DCT.