$e^{-nx}$ converges uniformly on $(0,\infty)$ pointwise on $[0,\infty)$

Question

wts: $e^{-nx}$ converges uniformly on $(0,\infty)$, pointwise on $[0,\infty)$

I feel that $x \in (0,\infty)$ $\Rightarrow$ $f_n(x) \rightarrow 0$ and that $x=0 \Rightarrow f_n(x) \rightarrow 1$.

When I try to prove it converges uniformly on $(0,\infty)$, I know that I am looking for some $N \in \mathbb{N}$ s.t. $|f_n(x)-0| < \epsilon$. That is, I want to find when $e^{-nx} < \epsilon$, which is when $-nx < \ln(\epsilon)$ or $n>-\frac{\ln(\epsilon)}{x}$ for all $x$.

I know that I can't have my answer dependent upon $x$ for uniform convergence, though. I thought it might work if I said $N=0$ (as it is always bigger than $-\frac{\ln(\epsilon)}{x}$), but this yields $|f_n(x)-0|=e^{-nx} < 1$, which isn't strong enough. Similarly, letting $N$ be any other integer doesn't work out very well either.

For pointwise, I can see that at $N=0$ works (as $|f_n(x)-1|=|e^0-1|=0<\epsilon$).

Answer 1

Take $\epsilon =\frac 1 e$.There cannot be any $n_0$ such that $|f_n(x)-0|<\epsilon$ for all $x$ for all $n \geq n_0$ because you get a contradiction by puting $x=\frac 1 n$. Hence, the convergence is not uniform on $(0,\infty)$.

Pointwise convergence: For $x>0$ the limit is $0$ and for $x=0$ the limit is $1$.

Answer 2

As already pointed out, the convergence is not uniform on $(0, \infty)$ but is uniform on $[r, \infty)$ for every $r > 0$. Let us show the latter.

Fix $r > 0$. We show that the sequence of functions $f_{n}$ defined on $[r, \infty)$ by $f_{n}(x) := e^{-nx}$ converges uniformly to the zero function.

Indeed, since $$\lim_{n \to \infty} e^{-nr} = 0,$$ we see that given any $\varepsilon > 0$, there exists $N \in \Bbb N$ such that $|e^{-nr}| < \varepsilon$ for all $n \geqslant N$.

Thus, given $\varepsilon > 0$, choose an $N$ as above. Then, for $x \geqslant r$ and $n \geqslant N$, note that $$|f_{n}(x) - 0| = |e^{-nx} - 0| = |e^{-nx}| \leqslant |e^{-nr}| < \varepsilon.$$