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

Question

wts: $e^{-nx}$ converges uniformly on $[r,\infty)$ when $r>0$

I feel that $x \in [r,\infty)$ $\Rightarrow$ $f_n(x) \rightarrow 0$.

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 $|f_n(x)-0|=e^{0} = 1$, which isn't strong enough. Similarly, letting N be any other integer doesn't work out very well either.

original booboo: here

Hello :) For $x\geq r>0$ and $\epsilon>0$ we have $\frac{|\ln\epsilon|}{r}\geq\frac{|\ln\epsilon|}{x}$ — Jochen, Sep 30 '21 at 08:21
Does this answer your question? [$e^{-nx}$ converges uniformly on $(0,\infty)$ pointwise on $0,\infty)$ — Brian Borchers, Oct 03 '21 at 03:53

score 0 · Accepted Answer · answered Sep 30 '21 at 10:07

0

We see that for $x\ge r>0$,$$ e^{nx} > nx \ge nr $$

So $$e^{-nx} < \frac{1}{nr}$$

Now you can see that neither $n$ nor $r$ are dependent on $x$ and you can easily prove the uniform convergence. Note that if $r=0$ the above method will not work.

answered Sep 30 '21 at 10:07

Infinity_hunter

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

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