Convergence of $f_n(x)=e^{-nx^2}$

Question

I was given the following problem. I have already submitted my work, but was curious about an aspect of it regarding the uniform convergence. Here is the work I submitted (my question is below):

Find the pointwise limit $f$ of the sequence $\{f_n\}$. Determine if $f_n \to f$ uniformly.

$f_n(x)=e^{-nx^2}$, $x\in \mathbb{R}$.

$$f(x)=\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty} \frac{1}{e^{nx^2}}= \left\{ \begin{array}{lr} 1, \hspace{8mm} \text{ if } x=0, \\[1ex] 0, \hspace{8mm} \text{ otherwise. } \end{array} \right.$$

Because each $f_n$ is continuous over our given interval but $f$ is not, then $f_n$ cannot converge uniformly to $f$ for all $x\in \mathbb{R}$.

However, let $|x|\geq 1$. Considering the first two terms of the Taylor expansion of $e^{nx^2}=1+nx^2+...$, we are able to discern that $e^{nx^2}>nx^2$. Thus, given $\varepsilon>0$, we find that by choosing $n>N=\frac{1}{\varepsilon}$,

$$\left|\frac{1}{e^{nx^2}}-0\right|=\left|\frac{1}{e^{nx^2}}\right|<\left|\frac{1}{nx^2}\right|\leq \left|\frac{1}{n}\right|<\varepsilon.$$

In other words, we find that $f_n\to f$ uniformly on the intervals $(\infty,-1]$ and $[1,\infty)$.

Question: Are my intervals for uniform convergence correct? It seems to me like given an $\varepsilon>0$ and $\alpha>0$, I can find a single $N\in \mathbb{N}$ such that for all $n>N$ and $x\in [\alpha,\infty)$,
$$|f_n(x)-f(x)|<\varepsilon.$$ However, I can see how changing $\alpha$ (i.e. an $x$) we would be changing our $N$ and thus defy uniform convergence in a way (because it is dependent on $x$). Can someone please help me clear this up? Thanks in advance.

Answer 1

You are correct to be concerned about your intervals of convergence. Indeed, this sequence converges uniformly on $(-\infty,-\alpha]\cup[\alpha,\infty)$ for all $\alpha>0$. So for any $\alpha$, we could find an even smaller value $\beta$ such that $f_n$ also converges uniformly on $(-\infty,-\beta]\cup[\beta,\infty)$. But, as you note, you must change the value of $N$ you choose. The reason this doesn't contradict uniform convergence is that uniform convergence is defined by

$f_n\to f$ uniformly on $A$ if for all $\epsilon>0$, there exists $N$ such that if $n\geq N$, then $|f_n(x)-f(x)|<\epsilon$ for all $x\in A$.

The definition does not require that if we choose a different set for uniform convergence that we have the same value for $N$. In other words, the magnitude of $N$ does not depend on $x$, just on $A$.