For what values of $b$ does the sequence of functions:
for each $n\in\mathbb{N}$, let
$$f_n(x)=b\cos\left(\frac{x}{n}\right), \text{ } x\in[0,1]$$
converge uniformly in the space $C[0,1]$ equipped with the supremum metric $$d(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|$$
My working and my doubts:
I claim that the sequence $g_n(x)=\cos\left(\frac{x}{n}\right)$ converges uniformly to $g(x)=1$ on $x\in[0,1]$.
For any $\epsilon>0$, we need to show there is an $N$ such that for all $n\geq N$ implies $d(g_n,g)<\epsilon$.
As we know as $n\to\infty$, $\frac{x}{n}\to0$ and $\cos(\frac{x}{n})\to1$. And the convergence is independent of the $x$, as long as $x\in[0,1]$.
1. What got me stuck is the explicit value of $N$. I am not sure how to find an explicit value of $N$ for each $\epsilon$.
2. After we have shown that $g_n(x)=\cos\left(\frac{x}{n}\right)$ converges uniformly, can we just say that $b\cos\left(\frac{x}{n}\right)$ will still converge regardless of the value of $b$?
3. So is it correct that the solution to the original question is for all value of $b\in\mathbb{R}$, $f_n$ converges uniformly?
Helps are greatly appreciated! Thank you.
For the first question, note that $ \cos(x)$ is positive and decreasing for $x \in[0, \frac{\pi}{2}]$.
So $f_n(x)$ is also positive and decreasing on $[0,1]$, thus we have $$d(1,f_n) = \sup_{x \in[0,1]} 1 - \cos \left( \frac{x}{n} \right) = 1 - \cos \left( \frac{1}{n} \right)$$ So you are left to show that $\cos \left( \frac{1}{n} \right) \to 1$ as $n \to \infty$, which follows from the continuity of $\cos(x)$ in $0$.
Then if you multiply with $b$, the same trick holds for positive $b$, and for negative $b$ you must be careful with the sign, but the trick still holds.
So yes, $f_n$ converges uniform for every $b \in \mathbb{R}$.