Uniform Convergence on exponential function.

Question

The question:

Let $f_{n}: \mathbb{R} \rightarrow \mathbb{R}$ be a sequence of functions which $f_{n} (x) = \exp[-nx^2]$. Determine whether or not this sequence converges uniformly.

The attempt:

I drew some pictures of these types of functions and I am claiming the sequence of functions do not converge uniformly. I am provided that this sequence converges point-wise to $0$ if $x \neq 0$ and to $1$ if $x = 0$. I am using the $\epsilon-N$ definition of uniform convergence.

Let's look at the case when $x \neq 0$ (This is my scratch work). We choose $\epsilon = 1/2$. Then, for all $N \geq 1$,we need to find a $n \geq N$ and an $x \in \mathbb{R}$, such that $| \exp[-nx^2]| \geq \frac{1}{2}$. Since the function is positive, we need to find an $n$ and an $x$ such that $\exp[-nx^2] \geq \frac{1}{2}$. The problem I am having is the algebra. Am I on track or not?

Thank you for the feedback.

Answer 1

HINT:

To answer the question in the OP, yes, you are on track.

For $\epsilon=1/2$, simply take any $n\ge 1$ and take $x=\sqrt{\log(2)/n}$.

Can you now conclude that the sequence fails to converge uniformly on $[0, \infty)$?

Answer 2

The pointwise limit function is not continuous, so the convergence is not uniform.

Of course, if $(f_n)$ converges uniformly, then the uniform limit function is a pointwise limit as well. Then it is enough to investigate theproperties of a pointwise limit.