Is $C^1[a,b]$ separable space with norm $||f||=\int_{a}^{b}\left | f(x) \right |dx$? with norm $||f||=(\int_{a}^{b} f(x)^{2}dx)^{1/2}$?

I have read with theorems about base but I am confused. Thanks.


1 Answer 1



The polynomials with real coefficients constitute a dense (albeit uncountable, but bear with me) subspace ${\cal P}$. (To see why it is dense in the $L^2$-norm, recall that the polynomials are dense in the $\sup$-norm by the Stone-Weierstrass Theorem, and that convergence in the $\sup$ norm implies convergence in the $L^2$-norm.)

In turn, the polynomials with rational coefficients constitute a countable subspace dense in ${\cal P}$, hence also in your space.

  • $\begingroup$ The polynomials are uncountable unless you take their coefficients from a countable set. $\endgroup$ Sep 27, 2016 at 20:59
  • $\begingroup$ @carmichael561 Good point, thanks. Are the polynomials with rational coefficients dense in those with real ones? If so, I can patch things up. Seems like they should be. $\endgroup$
    – user8960
    Sep 27, 2016 at 21:00
  • $\begingroup$ Yes, that should be fine. $\endgroup$ Sep 27, 2016 at 21:00
  • 1
    $\begingroup$ @user8960 on $[-1,1]$. as $ n \to \infty$ : $(1-x^2)^n$ is clearly a polynomial approximation of $1_{|x | < \epsilon}$, so its shifts are dense in all the common function spaces $\endgroup$
    – reuns
    Sep 27, 2016 at 21:11
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    $\begingroup$ @user8960 yes of course, for example shifts by $a /2^b$. And $(1- x^2/4)^n$ is easier to manage (since it is stricly decreasing on $[0,2]$) $\endgroup$
    – reuns
    Sep 28, 2016 at 18:32

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