uniform approximation by a particular set of functions

Question

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.

Answer 1

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.

Answer 2

Let $a:=\mu_1(0)$ and $b:=\mu_n(1)$. The linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^nP\circ\mu_j : P\in\mathbb{R}[x]\big\}$, whose closure in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^nf\circ\mu_j : f\in C^0([a,b])\big\}$, just because polynomials are uniformly dense in $C^0([a,b]) $ and $f\mapsto f\circ\mu_j$ are (linear) continuous maps. So the question is: can any $\alpha\in C^0([0,1])$ be written in the form $$\alpha=\sum_{j=1}^nf\circ\mu_j$$ for some $f\in C^0([a,b]) $?

Let's define inductively a finite sequence in $[a,b]$ putting $c_0:=\mu_n(0)$ and $c_{k+1}:=\mu_n(\mu_{n-1}^{-1}(c_k))$ until we reach some $c_K$ out of the range of $\mu_{n-1}$, which happens in finitely many steps, because $$c_{k+1}-c_k=\mu_n(\mu_{n-1}^{-1}(c_k))-\mu_{n-1}(\mu_{n-1}^{-1}(c_k))\ge\delta:=\min_{0\ge t\ge1}\mu_n(t)-\mu_{n-1}(t)>0.$$ Since $\mu_n:[0,1]\to[c_0,b]$ is invertible, we can define arbitrarily $f$ on $[a,c_0]$, with the condition $$\alpha(0)=\sum_{j=1}^nf(\mu_j(0))$$ and state the functional equation for $f$ on $[c_0,b]$ equivalently as: $$f(y) =\alpha(\mu_n^{-1}(y))-\sum_{j=1}^{n-1}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[c_0,b].$$ But this equation is self-solving: the RHS gives the unique extension of $f$ to the interval $[a,c_{1}]$, then to $[a,c_{2}]$, till we cover $[a,b]$.

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edit. If you change the definition of $f_k$, keeping the assumptions on the $\mu_k$, and also assuming $\mu_n>0$, to $$f_k:=[a^k]{1\over(1-a\mu_1)(1-a\mu_2)\dots(1-a\mu_n)}$$ that is $$f_k=\sum_{j=1}^n\lambda_j\mu_j^k$$ with $$\lambda_j(t):=\prod_{1\le i\le n\atop i\ne j}{\mu_j(t)\over \mu_j(t)-\mu_i(t)},$$ Then the linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^n\lambda_j(P\circ\mu_j) : P\in\mathbb{R}[x]\big\}$, and the closure of this space in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^n\lambda_j(f\circ\mu_j) : f\in C^0([a,b])\big\}$. To show the latter space is the whole space $C^0([0,1])$, means solvability for $f\in C^0([a,b]) $ of the functional equation $$\alpha(t)=\sum_{j=1}^n \lambda_j(t)f(\mu_j(t)),\quad t\in[0,1]$$ for any datum $\alpha\in C^0([0,1])$. To this end we can argue as before: choose a continuous $f$ on $[a,c_0]$ satisfying the functional equation at $x=0$: $$\alpha(0)=\sum_{j=1}^n \lambda_j(0)f(\mu_j(0))$$

And extend it to a solution $f\in C^0([a,b])$ now iterating for $j=1,2\dots$

$$f(y) ={\alpha(\mu_n^{-1}(y))\over\lambda_n(\mu_n^{-1}(y))}-\sum_{j=1}^{n-1}{\lambda_j(\mu_n^{-1}(y))\over\lambda_n(\mu_n^{-1}(y))}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[a,c_j].$$