Real Analysis - Uniform Convergence

Let $\displaystyle f_n(x):=\frac{1}{n\sqrt{2\pi}}\exp{{\frac{-x^2}{2n^2}}}$

and $\displaystyle f(x):=0$

$\displaystyle f_n(x) \rightarrow f(x) $ pointwise.

$\displaystyle \forall n\in\omega\int_{-\infty}^{\infty}f_n(x)dx = 1 $

and $\displaystyle \int_{-\infty}^{\infty}f(x)dx=0$

By The Uniform Limit of L.I. functions is L.I. and the limits are the same, we can conclude

$\displaystyle f_n(x)\not\rightarrow f(x) $ uniformly.

Although that's a straight forward proof, I'd like to prove this from the definition. That is

$\displaystyle \exists \varepsilon >0 \,\forall \delta >0 \,\exists x,y\in\mathbb{R}(|x-y|<\delta\wedge|f(x)-f(y)|\geq\varepsilon)$

But this is turning out to be more difficult. Can someone help?

Re: Real Analysis - Uniform Convergence

No, there is uniform convergence on the real line (because $\displaystyle |f_n(x)|\leqslant \frac 1{n\sqrt{2\pi}}$). The integral of the limit is not the limit of the integral in this case. In general, considering unbounded domains, uniform convergence is not enough in order to switch the limit and the integral.