Consider the interval $[0,1]$ and let $\mu\_k(t)$ with $k=1,\ldots,n$ be continuous functions such that they are all strictly increasing on the interval $[0,1]$ and such that $\mu\_1(t)<\mu\_2(t)<\ldots<\mu\_n(t)$ for all $t \in [0,1]$.


Consider for each $k \geq 0$, the functions $$f\_k(t)=\sum\_{j=1}^n (\mu\_j(t))^k$$


Is it true that any continuous function on $[0,1]$ can be approximated uniformly by linear combinations of the functions $f\_0,f\_1,f\_2,f\_3,\ldots$?


The case $n=1$ clearly holds due to the Stone-Weierstrass theorem but I can't see the general case. Thanks for your help.



uniform approximation by a particular set of functions
Assume that the linear combinations of $\{f\_i\}$ are not dense in $C[0,1]$. Then by Hahn - Banach theorem and Riesz - Markov - Kakutani theorem there exists a non-trivial Borel finite sign measure $\eta$ on $[0,1]$ such that $0=\int f\_k(t)d\eta=\sum\_{i=1}^n \int (\mu\_i(t))^kd\eta(t)=\sum\_{i=1}^n \int\_{\mathbb{R}} x^k d\mu\_i^\*(\eta)$ (where $\mu\_i^\*(\eta)$ is the pullback of $\eta$). So the finite finitely supported Borel sign measure $\sum \mu\_i^\*(\eta)$ has zero moments, thus it is zero by the usual Weierstrass theorem on the segment. However if we denote by $a\in [0,1]$ the minimal element of the support of $\eta$, the point $\mu\_1(a)$ belongs to the support of $\mu\_1^\*(\eta)$, but not to the supports of $\mu\_i^\*(\eta)$, $i>1$. A contradiction.


