7

Is it true that every bounded sequence in $C[0,1]$ with sup norm has convergent subsequence? I really feel this statement is true and the crux of the proof will lie in that $C[0,1]$ is complete. But I am unable to prove it mathematically by constructing a Cauchy subsequence for any random sequence ( which will eventually be convergent). Also is my observation correct that every Banach space has this property?

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
ogirkar
  • 2,681
  • 14
  • 27

2 Answers2

4

It is not true. The sequence $f_n(x)=\sin nx$ is a uniformly bounded sequence in $C([0,1])$, but it doesn't even have a pointwise convergent subsequence, let alone one in the sup norm.

To identify families where you have a positive result, you can use the Arzela-Ascoli theorem. That is, you just need to check that your family is pointwise bounded and equicontinuous.

cmk
  • 12,303
  • I think $f_{n}=x^{n}$ will also work na! – ogirkar Sep 04 '19 at 13:48
  • Also one quick question I also asked in above comment.." can we say if X is finite dimensional then it has BW property" – ogirkar Sep 04 '19 at 13:51
  • 1
    @believer indeed, this follows from a finite dimensional space being isomorphic to $\mathbb{R}^n$ for some $n$. See https://math.stackexchange.com/questions/144464/bolzano-weierstrass-theorem-in-a-finite-dimensional-normed-space for details. – cmk Sep 04 '19 at 14:12
3

Its not true. Consider $f_k(x) = e^{i2\pi kx}$. If $k< j$ are integers, then $$\|f_k-f_j\|_{\infty} = \sup_{x\in[0,1]} |1 - e^{i2\pi (j-k)x}| \ge |1 - e^{i2\pi (j-k)\frac{1}{2(j-k)}}| = 2.$$ Consequently there is no Cauchy subsequence.

Essentially the same example by thinking of Fourier coefficients - you can try the Banach spaces $\ell^p(\mathbb N)$ where now the sequence of elements $f_k \in \ell^p(\mathbb N)$ are defined by $f_k(n) = \delta_{nk}$.

Calvin Khor
  • 34,903