functional-analysisreal-analysis
Let $C([0,1])$ be the set of all real continuous functions with the standard supremum norm. Let $C^1([0,1])$ be the set of all real continuosly differentiable functions on $(0,1)$ such that the derivative can be continuously extended to $[0,1]$.
Is $C^1([0,1])$ a dense subset of $C([0,1])$?
You probably mean $\|f\| =\max{\{\|f\|_{\infty},\|f'\|_{\infty}\}}$ where $\|f\|_{\infty} = \sup_{t \in [0,1]} |f(t)|$.
Hint: Recall the basic fact that if $f_{n}$ is a sequence of $C^{1}$ functions such that $f_{n}$ converges uniformly (or only pointwise) to a continuous function $f$ and $f_{n}'$ converges uniformly to a continuous function $g$ then $f \in C^{1}$ and $f' = g$. It's probably a good idea to try to prove this on your own without consulting your books.
That $\|T\| \leq 1$ is just the definition of the operator norm. Consider also $f(x) = x$.
The appeal to "density" is a little sloppy. I believe the following is what the author meant:
I don't have a suitable reference for (1), but this fact is not hard to prove directly: (a) smooth functions are in $c^\beta$; (b) $c^\beta$ is a closed subspace of $C^\beta$; (c) mollifying a $c^\beta$ function produces an approximating sequence that converges in $C^\beta$ norm; the key is that sufficiently small scales don't really contribute to the norm.
For reference, here is the relevant part of the source (page 25 in Selected Aspects of Fractional Brownian Motion by Ivan Nourdin).
The reference Young [67] is to the classical article An inequality of the Hölder type, connected with Stieltjes integration, Acta Math. (1936) 67, 251-282. Young actually works with larger spaces of functions of finite $p$-variation, and I couldn't locate a relevant density argument there. A more accessible source is a pair of blog posts by Fabrice Baudoin:
Best Answer
Yes. Although this might be overkill, this follows from Weierstrass' approximation theorem, because $\mathbb{P}([0,1])$ is a subset of $C^1([0,1])$. Here $\mathbb{P}([0,1])$ is the space of all polynomials on $[0,1]$.