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I have a question that is bugging me and I tried hard to solve it.

Does the functions $f_n(x)=\frac {nx}{1+e^{nx}}$ and $g_n(x)=\frac {\frac nx}{1+e^{\frac nx}}$ pointwise convergence?

For $f_n(x)$ we can prove that for all $x \ge 0$ the function pointwise convergence.

It is easy to prove that as $n$ approaches $\infty$ the function approaches $f(x)=0$

To prove that a function uniformly converges, we can prove that $b_n = |f_n(x)-f(x)| < \epsilon$

Or in other words, I must prove that supreme of $b_n$ tends to zero.

The problem is that I can not find a supreme of the function $f_n(x)$ , apparently the supreme does not exist. How would I go to find out if the function uniformly converges?

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  • $\begingroup$ The same problem occur for $g_n(x)$ $\endgroup$ – Rab Mar 1 '17 at 23:03
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You don't actually have to find an exact value for the supremum to show this sequence fails to converge uniformly.

Notice that at $x = 1/n$ we have $f_n(1/n) = 1/(1+e) > 0.$

Thus, $\sup_{x \in [0,\infty)}f_n(x) \geqslant 1/(1+e)$ and consequently $\lim_{n \to \infty} \sup_{x \in [0,\infty)}f_n(x) \neq 0$.

The same argument applies for $g_n$ using $x_n = n$. The non-uniform convergence for $f_n$ ($g_n$) is associated with $x \to 0$ ($x \to \infty)$.

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  • $\begingroup$ Yes -- if the domain was $[a,b]$ there would be no problem. $\endgroup$ – RRL Mar 1 '17 at 23:21
  • $\begingroup$ Thank you, the domain is all $R$ for g(x) though. $\endgroup$ – Rab Mar 1 '17 at 23:21
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    $\begingroup$ $g_n(x)$ diverges for $x \leqslant 0$. The only interesting part is showing it does not converge uniformly on an interval like$[a,\infty)$. $\endgroup$ – RRL Mar 1 '17 at 23:25
  • $\begingroup$ Oh, I see where I went wrong, i solved for $\frac{x}{n}$ instead of $\frac{n}{x}$ thank you, I understood where I went wrong. $\endgroup$ – Rab Mar 1 '17 at 23:32

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