# Uniform convergence of $f_n(x)$

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 $$E?$$

• There is no $y$-dependence in the function. Is this intentional or a typo? – 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 $$[0,\infty)$$.

• 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 $\delta>0$. – AlessioDV Feb 8 at 17:19
• @AlessioDV Oops! You are right. I will modify my argument. – Julián Aguirre Feb 8 at 17:26
• Great answer, i was giving a generic response without trying to solve it . There is indeed no need to compute the maximum – 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 $$lim$$.]