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?

  • $\begingroup$ The same problem occur for $g_n(x)$ $\endgroup$ – Rab Mar 1 '17 at 23:03

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 $x \to 0$ ($x \to \infty)$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ $g_n(x)$ 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

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.