Show that the sequence $f_n(x) = \frac{e^x}{n}$ converges pointwise but not uniformly on $[0,\infty)$. Show that $f_n(x)$ converges uniformly on $[0,10]$.

My solution was not accepted for full credit.


Consider $f_n(x) = \frac{e^x}{n}$ converges pointwise.

Choose $\epsilon = 1$ $\exists N$ s.t. $\forall n>N$ $\forall x \in [0,\infty)$

$\Rightarrow |\frac{e^x}{n} - 0| < 1$. That is, $|e^x| < n$

We have, $|e^x| < N+1$ hence unbounded


Thus not uniformly convergent.

The problem with this part of the problem was that my professor did not accept my unbounded argument of less than N+1.

Then for $x\in [0,10]$

We have, $|f_n(x) - 0| = |\frac{e^x}{n} - 0| \leq \frac{e^{10}}{n} \rightarrow 0$

Thus $f_n(x)$ is uniformly convergent.

My professor said i did not explain enough details such as the limit function, and i am not sure what he meant by that.

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    $\begingroup$ You should point out what the pointwise limit-function is: $f_n(x) \to 0$. In your proof of this that ends with "hence unbounded" is unclear. What you want to show is that for any given $x$ (not $\forall$ as you state) and any given $\epsilon > 0$ (not just $\epsilon = 1$) that you can find an $N$ such that $|e^x/n-0| < \epsilon$ for all $n > N$. $\endgroup$ – Winther May 9 '16 at 5:25
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    $\begingroup$ To be honest, I'm not quite sure what you're doing. $\endgroup$ – user223391 May 9 '16 at 5:30
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    $\begingroup$ @Winther The OP is attempting to show that $f_n$ fails to uniformly converge on $[0,\infty)$ by finding an $\epsilon>0$ for which ... $\endgroup$ – Mark Viola May 9 '16 at 5:33
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    $\begingroup$ @Dr.MV That might be. If that is the case then a big piece is missing: a proof of the pointwise convergence as the question asks for. $\endgroup$ – Winther May 9 '16 at 5:35
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    $\begingroup$ @Winther I agree. The first part of the question is to show point-wise convergence. $\endgroup$ – Mark Viola May 9 '16 at 5:40

@Winther left a comment that explains how to show that $f_n(x)$ converges point-wise.

To show that $f_n(x)=\frac{e^x}{n}$ fails to uniformly converge on $[0,1)$, we find an $\epsilon>0$, such that for all $N$ there exists an $x\in [0,\infty)$ and an $n>N$ for which

$$|f_n(x)|\ge \epsilon$$

Take $\epsilon=1$. Then using $e^x\ge 1+x$ for all $x$, we assert that for $x=n$

$$\begin{align} |f_n(x)|&=\left|\frac{e^x}{n}\right|\\\\ &\ge \frac{1+x}{n}\\\\ &=\frac{1+n}{n}\\\\ &\ge 1 \end{align}$$

And therefore, $f_n(x)$ fails to uniformly converge.

To show that $f_n(x)$ uniformly converges on $[0,10]$, note that we have for any given $\epsilon>0$

$$\begin{align} |f_n(x)|&=\left|\frac{e^x}{n}\right|\\\\ &\le \frac{e^{10}}{n}\\\\ &<\epsilon \end{align}$$

whenever $n>N=\lfloor \frac{e^{10}}{\epsilon}\rfloor +1$

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