Consider the Banach space $C([0,1])$ with $\| . \|_{\infty}$ maximum norm. Consider the linear functional $F_{n} : C([0,1]) \to \mathbb{R}$, $ F_{n} (f) = f(1) - f\big(\frac{2n - 2}{2n - 1}\big) + f\big(\frac{2n-3}{2n-1}\big) - f\big(\frac{2n-4}{2n-1}\big) + ... - f(0)$

Show that $C^{1}([0,1])$ is dense in $C([0,1])$.

I would like to prove this using those functionals. My teacher gave me this hint: Weierstrass Approximation Theorem.

W.A. Theorem: If $f$ is continuous real-valued function on $[a,b]$ and if given any $\epsilon >0$ , then there exists a polynomial $p$ on $[a,b]$ such that $|f(x) - p(x)|< \epsilon \; \;\forall x \in \; [a,b]$ .

But I guess I'm needing another hint there. I know each one of there $F_n$'s are continuous real valued functions on $[0,1]$, then I can approximate by polynomials each one of them.

I know that in Baire spaces the enumerable intersection of open dense $G_{n}$ is dense. This is a Banach space, then it is complete, then this is true. Would $C^1 ([0,1])$ be an enumerable intersection of open dense $G_{n}$'s?

My problem is using these ideas for the start of the proof.

Any help would be appreciated.

It appear that it can be proven without the $F_n$. I actually need them to prove this:

$\forall f \in C^1 ( [0,1]), \lim_{n \to \infty} F_n (f) = \frac{ f(1) - f(0)}{2}$

  • 2
    $\begingroup$ $C^1[[0,1]$ contains the set of all polynomials restricted to $[0,1]$, which is dense in $C[0,1]$. Hense also $C^1[0,1]$ is dense in $C[0,1]$. $\endgroup$
    – Alex
    Oct 19, 2016 at 11:43

1 Answer 1


Maybe it is sufficient to note that if A $\subset$ B then $\bar{A} \subset \bar{B}$ since polynomials are in $C^1([0, 1])$.

  • $\begingroup$ I can see that $cl(Pol([0,1])) = C([0,1])$, by W.S.Theorem, and then since $Pol([0,1]) \subseteq C^1([0,1]$, then $cl(Pol([0,1])) \subseteq cl(C^1([0,1]))$, then $ C([0,1]) \subseteq cl(C^1([0,1]))$. But why $cl(C^1([0,1]) \subseteq C([0,1])$? $\endgroup$
    – user286485
    Oct 19, 2016 at 12:34
  • 1
    $\begingroup$ Because $C([0, 1])$ is complete, hence closed. But to show that $C^1([0, 1])$ is dense in $C([0, 1])$ you only need the reverse inclusion. $\endgroup$ Oct 20, 2016 at 0:58

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