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Is this a valid reason why the sequence of functions $f_n(x)=\frac{e^{nx}-e^{-nx}}{e^{nx}+e^{-nx}}$ does not converge uniformly?

$\text{lim}\frac{d}{dx}f_n(x) \neq \frac{d}{dx}\text{lim}f_n(x)$?

Also I do not think a reason could be $\text{lim} \int_{-1}^{1}f_n \neq \int_{-1}^{1}\text{lim}f_n$

since the latter statement appears untrue.

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  • $\begingroup$ What does "valid" mean here? Are you asking for a simple one-line explanation of why we know your $f_n$ do not converge uniformly? $\endgroup$
    – kimchi lover
    Dec 19, 2020 at 21:22
  • $\begingroup$ I know a reason is the limit is discontinuous I want to know if another reason is the first statement $\endgroup$
    – Scott Frazier
    Dec 19, 2020 at 21:24
  • $\begingroup$ @QC_QAOA I noticed that the derivative of the limit is zero, however this derivative is undefined at $0$. Also the limit of the derivative seems to be zero, but not undefined at $0$, this is correct reasoning, for why the derivative condition is satisfied in this situation correct? $\endgroup$
    – Scott Frazier
    Dec 19, 2020 at 22:07
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    $\begingroup$ I disagree. The derivative condition is not enough to prove that $f_n$ is not converging uniformly to $f$. You can prove that you have a sequence of functions that is uniformly convergent to $f$ and still the interchange between limit and derivative cannot be performed. I give an example: consider the functions $f_n : [0,2\pi] \rightarrow \mathbb{R}$ be $f_n(x)= n^{-1/2}sin(nx)$ and $f(x)=0$. Here by the sqeeze thm $f_n$ converges uniformly to $f$ but you cannot interchange lim & derivative. So, from the condition you mentioned you cannot conclude that it is not uniformly converging. $\endgroup$
    – Crash Bandicoot
    Dec 19, 2020 at 23:16
  • $\begingroup$ If this is true I wonder who has been giving QC_QAOA likes, because who ever it is thinks the condition is enough. $\endgroup$
    – Scott Frazier
    Dec 19, 2020 at 23:51

1 Answer 1

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You see that the sequence of functions you have written is $f_{n}(x)= \tanh(nx)$. If you take the limit for $n \rightarrow \infty$ you can see that it converges pointwise to a discontinuous function, hence it doesn't converge uniformly.

For the interchange of limit and derivative, it is not enough to prove that $f_n$ is not converging uniformly to $f$. You can prove that you have a sequence of functions that is uniformly convergent to $f$ and still the interchange between limit and derivative cannot be performed. I give an example: consider the functions $f_n : [0,2\pi] \rightarrow \mathbb{R}$ be $f_n(x)= n^{-1/2}$sin$(nx)$ and $f(x)=0$. Here by the squeeze thm $f_n$ converges uniformly to $f$ but you cannot interchange lim & derivative. So, from the condition you mentioned you cannot conclude that it is not uniformly converging.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson ♦
    Dec 20, 2020 at 13:01

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