Getting uniform convergence via a dominated convergence argument 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?

 A: 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.
A: To start with, uniform continuity isn't enough to get uniform convergence here since $\mathbb{R}$ is unbounded.  You also need some kind of "tightness," which is actually not possible.
For each $N \in \mathbb{N}$, $\hat{\varphi}_{N}(\xi) \to 0$ as $|\xi| \to \infty$, being the Fourier transform of an $L^{1}$ function.  Consequently, there is no way $(\hat{\varphi}_{N})_{N \in \mathbb{N}}$  converges uniformly to $1$.
An example: let $\varphi(x) = (2\pi)^{-\frac{1}{2}} e^{-2^{-1} x^{2}}$ be the Gaussian kernel.  It's Fourier transform is $\hat{\varphi}(\xi) = e^{-2^{-1} \xi^{2}}$.  Hence $\hat{\varphi}_{N}(\xi) = e^{-2^{-1} N^{-1} \xi^{2}}$ for all $N$.  These converge uniformly to $1$ in $[-B\sqrt{N},B\sqrt{N}]$ as $N \to \infty$ for any fixed $B > 0$, but not uniformly in $\mathbb{R}$, again precisely because they are far from $1$ for $|\xi| \gg \sqrt{N}$.
