I showed the pointwise convergence of
$$f_n:[1,\pi] \rightarrow \mathbb{R}, \, f_n(x):= \frac{\cos(x/n)}{1-e^{-xn}}$$
with
$$f_n(x)=\frac{\cos(x/n)}{1-\frac{1}{e^{xn}}}
\xrightarrow{n \, \rightarrow \, \infty}
f(x):=\frac{\cos(0)}{1-0}=1.$$
How can I now show whether or not
$f_n$
converges uniformly?
Would
$$0 \le \left| f_n(x)-f(x) \right| = \left| \frac{\cos(x/n)}{1-e^{-xn}}-1 \right| \le \left| \frac{\cos(\pi/n)}{1-e^{-1 \cdot n}}-1 \right| \xrightarrow{n \, \rightarrow \, \infty} |1-1|=0$$
be right to show that it converges uniformly?
Thanks in advance!
Best Answer
To show uniform convergence you need to show that:
$$ \underset{x\in[1,\pi]}{\sup} \Big\vert f_n(x)- f(x) \Big\vert \rightarrow 0 $$
But since $f(x)\equiv 1$, you need to show that:
$$ \underset{x\in[1,\pi]}{\sup} \Big\vert f_n(x)- 1 \Big\vert \rightarrow 0 $$
It would be enough to find a sequence $0\leq a_n\rightarrow0$ such that:
$$ \underset{x\in[1,\pi]}{\sup} \Big\vert f_n(x)- 1 \Big\vert\leq a_n $$
Hint: Note that $\vert f_n(x)-1\vert$ is bounded from above for all $x$ by:
$$ \frac{\vert 1-\cos(x/n)\vert+\vert e^{-xn}\vert}{\vert 1-e^{-xn}\vert} $$