0
$\begingroup$

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.

Proof

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

Contradiction

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ To be honest, I'm not quite sure what you're doing. $\endgroup$ – user223391 May 9 '16 at 5:30
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @Winther I agree. The first part of the question is to show point-wise convergence. $\endgroup$ – Mark Viola May 9 '16 at 5:40
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.