Uniform convergence of $f_n(x)$

Question

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 $E?$

Answer 1

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 $[0,\infty)$.

Answer 2

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 $lim$.]