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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?

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    $\begingroup$ Actually, not at all. The only normed vector spaces whose unit balls are compact are finite-dimensional spaces. $\endgroup$ – Mindlack Sep 4 '19 at 12:54
  • $\begingroup$ means if X is finite dimensional then it will have BW property? $\endgroup$ – Believer Sep 4 '19 at 13:50
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    $\begingroup$ Any normed space $X$ has the BW property iff it is finite-dimensional. $\endgroup$ – Mindlack Sep 4 '19 at 16:05
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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.

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  • $\begingroup$ I think $f_{n}=x^{n}$ will also work na! $\endgroup$ – Believer Sep 4 '19 at 13:48
  • $\begingroup$ Also one quick question I also asked in above comment.." can we say if X is finite dimensional then it has BW property" $\endgroup$ – Believer Sep 4 '19 at 13:51
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    $\begingroup$ @believer indeed, this follows from a finite dimensional space being isomorphic to $\mathbb{R}^n$ for some $n$. See math.stackexchange.com/questions/144464/… for details. $\endgroup$ – cmk Sep 4 '19 at 14:12
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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 $f_k(n) = \delta_{nk}$.

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