Let $C([0,1], \mathbb R)$ denote the space of real continuous functions at $[0,1]$, with the uniform norm. Is the set $H=\{ h:[0,1] \rightarrow \mathbb R : h(x)= \sum_{j=1}^n a_j e^{b_jx} , a_j,b_j \in \mathbb R, n \in \mathbb N \}$ dense in $C([0,1], \mathbb R)$?
[Math] Dense subset of continuous functions
analysisreal-analysis
Related Solutions
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 $[0,1]$.
The appeal to "density" is a little sloppy. I believe the following is what the author meant:
- The closure of $\mathcal C^1$ with respect to $C^\beta$ norm is the little Hölder space $c^{\beta}$ (defined by the condition $|f(x)-f(y)| = o(|x-y|^\beta)$ as $x-y\to 0$)
- $C^{\beta+\epsilon}\subset c^\beta$ for every $\epsilon>0$.
- Given $f\in C^\alpha$ and $\beta$ such that $\alpha+\beta>1$, pick $\gamma <\alpha$ such that $\gamma+\beta >1$. Then $f\in c^\gamma$, hence the density argument applies and shows that the integral $\int fg'$ is bounded by $C\|f\|_\gamma\|g\|_\beta$. The latter is controlled by $C\|f\|_\alpha\|g\|_\beta$, as desired.
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, by the general Stone-Weierstrass Approximation Theorem. Indeed, $H$ is an algebra of continuous real-valued functions on the compact Hausdorff space $[0,1]$, and $H$ separates points and contains a non-zero constant function. That is all we need to conclude that $H$ is dense in $C([0,1],\mathbb{R})$.