I've been studying some problems in sequences of functions in "Berkeley Problems in Mathematics" chapter 1.6, and I have some difficulties understanding the solution to the following problem:
We are given a sequence of functions: $f_n(x)=\cos nx$. Prove that this sequence has no uniformly convergent subsequence.
To prove it, the author assumes that such a subsequence exists $f_{n_{j}} \rightarrow f$ uniformly. This means that $f$ is continuous and so $f(0)=lim_{j \rightarrow +\infty}\cos 0 =1$.
Hence there exists $\epsilon>0$ such that $f(x)>\frac{1}{2}$ for $|x|<\epsilon$. ($f(0)=1$ and $f$ is continuous so in a certain neighborhood of $0$ $f(x)>\frac{1}{2}$ - is this the reason?) .
And this is what I don't understand:
$|f(x)-f_{n_{j}}(x)|<\frac{1}{2}$ for $x, \frac{\pi}{2n_j}<\epsilon$.
I know that uniform convergence of $f_{n_{j}}$ implies the inequality but why this assumptions about $x$ and $\frac{\pi}{2n_j}$?
Could you explain that to me?
Thank you.
First, you're right, by continuity of $f$ at $0$, there exists $\epsilon>0$ such that $|f(x)-f(0)|<1/2$ for $|x|<\epsilon$. Hence $$ f(x)=f(0)+(f(x)-f(0))\geq f(0)-|f(x)-f(0)|>f(0)-\frac{1}{2}=\frac{1}{2} \quad\forall |x|<\epsilon. $$
Now by uniform convergence, there exists $n_j>\frac{\pi}{2\epsilon}$ such that $\|f-f_{n_j}\|_\infty\leq \frac{1}{4}$. In particular $$ \left|f\left(\frac{\pi}{2n_j}\right) \right|=\left|f\left(\frac{\pi}{2n_j}\right) -f_{n_j}\left(\frac{\pi}{2n_j}\right)\right|\leq\frac{1}{4} $$ since $f_{n_j}\left(\frac{\pi}{2n_j}\right)=\cos\left(\frac{\pi}{2}\right)=0$. But as we took $n_j$ large enough to have $\frac{\pi}{2n_j}<\epsilon$, we also have $$ f\left(\frac{\pi}{2n_j}\right) >\frac{1}{2}. $$ Contradiction.