1
$\begingroup$

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.

2 Answers 2

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 $[0, \infty)$?

  • 0
    Well in your hint, do we have to consider the negative root of x ? – 2017-01-21
  • 0
    No. The $\sqrt n$ is positive for $n\ge 1$ – 2017-01-21
  • 0
    No I completely understand it. Thank you very much!! – 2017-01-22
  • 0
    You're welcome. My pleasure. – 2017-01-22
1

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.

  • 0
    Okay. Why is it that if the point wise limit $f(x)$ is continuous, then the sequence is uniformly convergent? – 2017-01-21
  • 0
    @overachiever That's not correct. If it uniformly converges, the the limit function of a sequence of continous functions is continuous. The logic does not go the other way. But if the limit function is discontinuous, then the sequence cannot converge uniformly. – 2017-01-21