Let $f_n(x) = \frac{x}{n^2} e^{-\frac{x}{n}}$, I want to show $f_n(x) \rightrightarrows 0$, i.e., $\{f_n\}$ converges uniformly to $0$.
Here is my trial:
Apparently, from l'Hôpital's theorem, I know $f_n(x) \rightarrow 0$.
So from the definition of uniformly convergent, I need to find $N$ such that $\forall x$, $|f_n(x)| <\epsilon$, $\forall n \geq N$. i.e., $|\frac{x}{n^2}e^{-\frac{x}{n}}| < \epsilon$ and I stuck with this issue.
The answer in textbook says : Use $\lim_{n\rightarrow \infty} \int_0^{\infty} f_n(x) dx=1$
I Know $f_n(x)$ is continuous(Since it is a product of two continuous function), so $f_n(x)$ over a closed interval is integrable. Note that $\int f_n(x)= – \frac{e^{-\frac{x}{n}}(n+x)}{n}$, so I have $\lim_{n\rightarrow \infty} \int_0^{\infty}f_n(x)dx=1$.
But this does not tells $f_n(x)$ converges uniformly to $0$, furthermore if $f_n(x)$ converges uniformly, then limit and integral can exchange and that means $1=\int_0^{\infty} \lim_{n\rightarrow \infty} f_n(x) dx = \int_0^{\infty} 0 dx$ but integral over $0$ is $0$ so I got confused.
Best Answer
It appears that you are talking about uniform convergence on $[0,\infty)$.
The function $ye^{-y}$ is a bounded continuous function on this interval. (It is continuous and tends to $0$ as $ y \to \infty$).
Let $ye^{-y} \leq C$ for all $y \geq 0$. Put $f=\frac x n$ and conclude that $0 \leq f_n(x) \leq \frac C n $. This proves uniform convergence of $f_n$ to $0$ on $[0, \infty)$.