Given a sequence of functions $f_n(x)={{(x-1)^n}\over{1+x^n}} \arctan({n^{x-1}}).$
I have studied its pointwise convergence and I found as set of convergence $E=[0,+\infty)$ and as limit function the null function.
But I have a problem with the uniform convergence: how can I calculate $\;\sup|f_n(x)|\;$ in $E?$

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    $\begingroup$ There is no $y$-dependence in the function. Is this intentional or a typo? $\endgroup$ – Klaus Feb 8 at 12:38

There is no need to compute the maximum (or minimum) of $f_n$; it is enough to have inequalities. We have $$ |f_n(x)|\le\frac\pi2\,\frac{|x-1|^n}{1+x^n}\le\frac\pi2\,\,\Bigl|1-\frac1x\Bigr|^n\quad\forall x\ge0. $$ Let $A>3/2$. If $3/4\le x\le A$, then $$ |f_n(x)|\le\frac\pi2\,\Bigl|1-\frac1A\Bigr|^n, $$ proving that $f_n$ converges uniformly to $0$ on $[3/4,A]$.

Let's treat now the case $0\le x\le3/4$. Then $n^{x-1}\le n^{-1/4}$, $\arctan(n^{x-1})\le n^{-1/4}$ and $|f_n(x)|\le n^{-1/4}$, proving uniform convergence on $[0,3/4]$.

What happens on $[0,\infty)$? The convergence is not uniform. To prove it, we bound from below $f_n(n)$, $n\in\Bbb N$. First of all, $\arctan(n^{n-1})\ge \pi/4$ for all $n\in\Bbb N$. Next, $$ \frac{|n-1|^n}{1+n^n}\ge\frac{|n-1|^n}{2\,n^n}=\frac12\,\Bigl|1-\frac1n\Bigr|^n\to\frac1e>0. $$ This shows that $f_n(n)\ge c$ for some constant $c>0$, proving that $f_n$ does not converge uniformly to $0$ on $[0,\infty)$.

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    $\begingroup$ It seems to me that the proposed argument is not enough. To conclude that $\left|1-\frac1A\right|^n\rightarrow 0$ as $n\rightarrow +\infty$ you must have $\left|1-\frac1A\right|<1$, that holds if and only if $A>\frac 12$. Thus the proposed estimate shows uniform convergence in $\left[\frac 12 +\delta, \frac 12 + \delta'\right]$ for all $\delta>0$ and $\delta'>\delta$. An extra effort is needed to prove uniform convergence on $[0,\delta]$ for every $\delta>0$. $\endgroup$ – AlessioDV Feb 8 at 17:19
  • $\begingroup$ @AlessioDV Oops! You are right. I will modify my argument. $\endgroup$ – Julián Aguirre Feb 8 at 17:26
  • $\begingroup$ Great answer, i was giving a generic response without trying to solve it . There is indeed no need to compute the maximum $\endgroup$ – Milan Feb 8 at 17:27

In general you can find the supremum of function by finding its maximum.I dont recommend differentiating the whole function.

You should use the fact that $ arctan $ is bounded by $ \pi/2 $. Then find $max$ of $ {{(x-1)^n}\over{1+x^n}} $ ( the maximum will be a function of n).

[ By multiplying $ \pi/2 $ and $max$ you might get something that is bigger than the actual but it doesnt because matter $ \pi/2 $ can get infront of $lim$.]


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