# Valid reason the sequence of functions does not converge uniformly

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.

• 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? Dec 19, 2020 at 21:22
• I know a reason is the limit is discontinuous I want to know if another reason is the first statement Dec 19, 2020 at 21:24
• @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? Dec 19, 2020 at 22:07
• 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. Dec 19, 2020 at 23:16
• If this is true I wonder who has been giving QC_QAOA likes, because who ever it is thinks the condition is enough. Dec 19, 2020 at 23:51

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.