# Prove that the functional sequence has no uniformly convergent subsequence -check

#### Arnold

##### New member
SOLVED Prove that the functional sequence has no uniformly convergent subsequence -check

$$\displaystyle n \in \mathbb{R}, \ \ f_n \ : \ \mathbb{R} \rightarrow \mathbb{R}, \ \ f_n(x) =\cos nx$$

We want to prove that $$\displaystyle {f_n}$$ has no uniformly convergent subsequence.

This is my attempt at proving that:

Suppose that $$\displaystyle f_{n_{j}} \rightarrow f$$ uniformly. This means that $$\displaystyle f$$ is continuous(because cosnx is) and $$\displaystyle f(0) = \lim _{j \rightarrow +\infty}\cos 0=1$$.

Therefore there exists a $$\displaystyle \delta > 0$$ s.t. $$\displaystyle f(x) > \frac{1}{2}$$ for $$\displaystyle |x|< \delta$$.

Hence for $$\displaystyle j$$ large enough, the uniform convergence of $$\displaystyle f_{n_{j}}$$ tells us that $$\displaystyle |f(x) - f_{n_{j}}(x)|<\frac{1}{2}$$

But for one such $$\displaystyle j$$ and $$\displaystyle x=\frac{\pi}{2n_{j}}$$ we have

$$\displaystyle \frac{1}{2} < f(x) \le |f(x) - f_{n_{j}}(x)| + |f_{n_{j}}| < \frac{1}{2} + f_{n_{j}} = \frac{1}{2} + 0 = \frac{1}{2}$$ and we have a contradiction.

My question is - is everything here correct? I'm just beginning to study functional sequences and I would really appreciate any help.

Thank you.

Last edited:

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Re: SOLVED Prove that the functional sequence has no uniformly convergent subsequence -check

Hence for $$\displaystyle j$$ large enough, the uniform convergence of $$\displaystyle f_{n_{j}}$$ tells us that $$\displaystyle |f(x) - f_{n_{j}}(x)|<\frac{1}{2}$$
It's nice to specify, "for all $x$".

But for one such $$\displaystyle j$$ and $$\displaystyle x=\frac{\pi}{2n_{j}}$$ we have

$$\displaystyle \frac{1}{2} < f(x) \le |f(x) - f_{n_{j}}(x)| + |f_{n_{j}}| < \frac{1}{2} + f_{n_{j}} = \frac{1}{2} + 0 = \frac{1}{2}$$ and we have a contradiction.
You need $j$ large enough not just so that $$\displaystyle |f(x) - f_{n_{j}}(x)|<\frac{1}{2}$$ for all $x$, but also so that $$\displaystyle \frac{\pi}{2n_{j}}<\delta$$.

Otherwise, the proof looks very nice.

#### Arnold

##### New member
Ok, I'll do that. Thanks.