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

Arnold

New member
Jan 11, 2013
16
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
Jan 30, 2012
2,492
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
Jan 11, 2013
16
Ok, I'll do that. Thanks.