Closed unit ball in $(C^0[0,1],\mathbb{R})$ equipped with the sup norm is not sequentially compact.

I consider $f_n(x) = x^n$ where $0≤x≤1$ Then $f_n(x)$ is clearly a subset of the closed unit ball. Also $f_n(x)$ converges pointwise to $f(x) = 0$ for $0≤x<1$ and $1$ for $x=1$. $f$ is not continuous, so $f_n$ does not converge uniformly to $f$? Then I am stuck?

A subset of a normed space is sequentially compact if every sequence with all terms in the subset has at least one convergent subsequence with its limit in the subset.

  • $\begingroup$ There's something off with the "$<$" and "$\le$" signs. $f_n(x)$ should be $x^n$ for all $0\le x\le 1$ (instead of $0\le x<1$), while $f(x)$ should be $0$ for $0\le x<1$, instead of $0<x\le 1$. $\endgroup$ – user228113 Jan 29 '16 at 16:27
  • $\begingroup$ Corrected those thanks $\endgroup$ – Gerniant Jan 29 '16 at 16:29

The argument should be as it follows:

Suppose there exists a subsequence $f_{n_k}\to g\in C^0[0,1]$ (with $\lVert g\rVert_\infty\le1$) uniformly on $[0,1]$. Then, $f_{n_k}\to g$ pointwise. But we know that $(f_{n})_{n\in\Bbb N}$ converges pointwise to a non-continuous function - and so must $(f_{n_k})_{k\in\Bbb N}$. But $g$ is continuous by definition. Absurd.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.