Uniform and pointwise convergence on $f_n(x)=e^{-(x-n)^2}$

Question

I have been given the following assignment:

Suppose you have the following sequence of functions:

$f_n(x)=e^{-(x-n)^2}$

1. Show that the sequence is pointwise convergent to the zero-function on $\mathbb{R}$

2. Show that the sequence is uniform convergent on an interval $[a,b]$ where $a<b$

I have tried to examen the limit of the sequence, and see if I could get a result, where I could intuitively see that it must be the zero function, but all I end up with is this:

$\lim_{n \rightarrow \infty} f_n(x) = e^{-(x-n)^2} = e^{-x^2-n^2+2nx}$

I can't really see how I would be able to simplify this to something I can work with.

Can someone give a hint as to where I need to start?

Answer 1

Hint: $f_n(x) = g(x-n)$, where $g$ is the "Gaussian" function $g(x) = e^{-x^2}$. Hence the graph of $f_n$ is obtained shifting the one of $g$ to the right of a quantity $n$ (i.e., $f_n$ attains its maximum at $x = n$).

Answer 2

Write:

$$ f_n(x) = e^{-n^2(\frac{x}{n}-1)^2} $$

Now, take any value of $x$ and make $n \rightarrow \infty$ in the expression above. That will prove pointwise convergence.

To prove uniform convergence, compute the global (and only) maximum of the function. You'll see that for $n$ sufficiently large the function will reach this maximum out of $[a,b]$. @Rigel 's answer points that out.

That means that the maximum in $[a,b]$ of $f_n$ is $f_n(b)$ for $n$ sufficiently large. Then use one of the known criteria for uniform convergence.