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?