Is the limit uniform on $R$? Is the limit uniform on $[0,1]$?

Question

a)Find the pointwise limit $\frac{e^{x/n}}{n}$ for $x\in R.$

For given $x\in R$ and $\varepsilon>0,$ let $M \in N$ be such that $M=\frac{x}{\ln(\varepsilon)}$.

I claim that the limit is $0$. Then, for $n\ge M,$ we have $$|\frac{e^{x/n}}{n}|\le e^{x/n} \le e^{x/M} \le e^{\ln (\varepsilon)}=\varepsilon. $$

b) Is the limit uniform on $R$?

Suppose the limit is uniform on $R$. Then, I have to find some $x$ such that given $\varepsilon>0$, for $n\ge M,$ $$\frac {e^{x/n}}{n} \not < \varepsilon.$$

Since it contradicts to the definition, the limit is not uniform on $R$.

c) Is the limit uniform on $[0,1]$?

Could you give some hint for b) and c)?

Answer 1

For $b)$ you have that $$\sup_{x \in \Bbb{R}}|\frac{e^{x/n}}{n}| \geq |\frac{e^{\frac{n^2}{n}}}{n}|=\frac{e^n}{n} \to +\infty$$

For $c)$ note that $$e^{x/n} \leq e,\forall x \in [0,1] \Rightarrow \sup_{x \in [0,1]}|\frac{e^{x/n}}{n}| \leq \frac{e}{n}$$

Answer 2

Correct me if wrong :

b) Assume uniform convergence on $\mathbb{R}$.

Given $\epsilon >0$.

Assume there is a $N(\epsilon) \in \mathbb{Z^+}$ such that

for $n \ge N$ and $x \in \mathbb{R}$

$|e^{x/n}/n| \lt \epsilon.$

Then:

$e^{x/n}/n \le e^{x/N}/N \lt \epsilon$, or

$e^{x/N} \lt N\epsilon.$

Choose $x > N \log (2N\epsilon),$

a contradiction.