Consider the integral $I(\lambda) = \sqrt {\frac {\lambda \mathbb{i}}{\pi}}^n \int_U \mathbb{e}^{-\mathbb{i} \lambda \|x\ - x_0|^2} f(x) \mathbb{d}x, \lambda>0$ and $U$ some open neighbourhood of $x_0 \in \mathbb{R}^n$. It is known that $\lim \limits_{\lambda \to \infty} I(\lambda) = f(x_0)$. Is this limit uniform with respect to $x_0$?
In general, $f$ is taken to be smooth with compact support. May I relax these hypotheses?
When you say "uniform with respect to $x_0$", I presume you're keeping $U$ and $f$ fixed. If $x_0$ is not in the closure of $U$, the limit must be $0$. A function that is $f(x_0)$ for $x_0$ in $U$ and $0$ for $x_0$ outside the closure of $U$ won't be continuous unless $f$ is $0$ on the boundary of $U$, and so it isn't the uniform limit of continuous functions. But $I(\lambda)$ is a continuous function of $x_0$. So the limit can't be uniform.
EDIT: OK, take $U = \mathbb R^n$ and suppose, more generally, that $f$ is a Schwartz function. Then $I(\lambda, x_0)$ (to include the dependence on $x_0$) is the convolution of $f$ with the bounded continuous function $ g_\lambda(x) = \sqrt{\lambda i/\pi} \exp(-i\lambda |x|^2)$. Its Fourier transform is the product of $\widehat{f}$ and $\widehat{g_\lambda}$: $$ \widehat{g_\lambda}(p) = \exp(i|p|^2/(4 \lambda))$$ Given $\epsilon > 0$, let $\chi_B$ be the indicator function of a ball $B$ large enough so that $$\|(1 - \chi_B) \widehat{f} \widehat{g_\lambda}\|_{L^1} = \|(1 - \chi_B) \widehat{f} \|_{L^1}< \epsilon/3$$ Take $\lambda$ large enough so $|g_\lambda - 1| < \delta$ on $B$, where $\delta \|\widehat{f}\|_{L^1} < \epsilon/3$, so that $$\|\widehat{f} (\widehat{g_\lambda} - 1)\chi_B\|_{L^1} < \epsilon/3$$ Then $$\|\widehat{f} - \widehat{f} \widehat{g_\lambda}\|_{L^1} \le \|\widehat{f} - \widehat{f} \chi_B \||_{L^1} + \| \widehat{f} \chi_B - \widehat{f}\widehat{g_\lambda} \chi_B\|_{L^1} + \|\widehat{f}\widehat{g_\lambda} \chi_B - \widehat{f} \widehat{g_\lambda}\|_{L^1} < \epsilon$$ and then $|f(x_0) - I(\lambda, x_0)| < \epsilon$ for all $x_0$.
EDIT: Looking back at this, it seems all we're actually using about $f$ is that its Fourier transform is in $L^1$,