# uniform approximation by a particular set of functions

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.

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.

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 $$[a,b]$$.

$$*$$

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 $$j=1,2\dots$$

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

• Thanks for the responses. Do you think any of these proofs generalizes if I were to change the definition of $f_k's$ to for example the coefficient of $a^k$ in $(1+a\mu_1+a^2\mu_1^2+\ldots)\ldots(1+a\mu_n+a^2\mu_n^2+\ldots)$?
– Ali
Apr 7 '19 at 18:50
• I'd say you can repeat the argument with minor modifications. In this case, you get some coefficients functions $\lambda_j(t)$ (depending on the $\mu_j$'s) in front of $f\circ\mu_j$ in the expression of the elements in the closure. The $\lambda_j$ have constant sign, and in particular, $\lambda_n>0$. So you only have to modify conveniently the of the iteration formula for $f$ adding these coefficients. Apr 7 '19 at 19:34
• (details added) Apr 7 '19 at 20:00
• Note: dividing by $\lambda_n$ requires it has constant sign, that is, $\mu_n$ has constant sign. I guess this assumption is not necessary, though. Apr 7 '19 at 20:16
• you made a very nice trick to symmetrize all the expressions again and therefore the same approach works. I just wonder if the same result must always be true. If I randomly put some positive coefficients $c_k's$ as follows $(1+c_1a\mu_1+c_2a^2\mu_1^2+\ldots)\ldots(1+c_1a\mu_n+c_2a^2\mu_n^2+\ldots)$ then the same approach can not work right?
– Ali
Apr 8 '19 at 8:45