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Applying the Müntz–Szász theorem on $[0,1]$ repeatedly, we can represent $$ x= \sum_{n\geq 2} c_n x^n $$ as a uniformly convergent series (edit: only over some subsequence, see edits below) on $[0,1]$ of higher powers $x^n$ for $n\geq 2$. What can one say about the coefficients? Is there an explicit choice of $c_n$?

Edit: Comments below suggest that this is not possible. What is wrong with the following argument? Take $\epsilon>0$ and approximate $x$ by a finite combination of higher powers and a constant uniformly with an error $\epsilon/2.$ Plugging $x=0$ we see that the constant is smaller than $\epsilon/2$ so dropping it we get an approximation up to $\epsilon$ by a finite sum $\sum_{n=2}^{N_1} c_n x^n.$ Next, consider $x-\sum_{n=2}^{N_1} c_n x^n$ and approximate it by a linear combination $\sum_{n=N_1+1}^{N_2} c_n x^n$ up to an error $\epsilon/2.$ This gives us an $\epsilon/2$-approximation $\sum_{n=1}^{N_2}c_n x^n.$ Continue this construction repeatedly.

Edit II: Theorem holds for $a=0$ if we include constants, but Robert Israel's comment below contained the main point: the series only converges over some subsequence $(N_k)_{k\geq 1}$ as in the above construction. Let me rephrase the question accordingly:

Is there anything interesting one can say about $c_n$? Can one choose the subsequence and $c_n$ in a way that $(c_n)\in\ell^p$ for some $p$, or uniformly bounded?

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  • $\begingroup$ can you perhaps shed some light onto how this might work near $x=0$? $\endgroup$
    – Carlo Beenakker
    Feb 7, 2017 at 15:13
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    $\begingroup$ @CarloBeenakker It won't work - the theorem only holds on intervals $[a,b]$ with $a>0$. $\endgroup$
    – Dirk
    Feb 7, 2017 at 15:16
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    $\begingroup$ You can't get a convergent series, either: only some subsequence of the partial sums converges. $\endgroup$
    – Robert Israel
    Feb 7, 2017 at 15:24
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    $\begingroup$ As has been pointed out, a convergent series would define a holomorphic function near $x=0$, which cannot have two distinct power series representations. The problem with your argument is that the $c_n$'s will depend on $\epsilon$, so you're not constructing a series at all but a more general sequence of polynomials. $\endgroup$
    – Christian Remling
    Feb 7, 2017 at 17:34

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It can't be in $\ell^p$ or bounded, in fact you can't have $|c_n| = O(t^{-n})$ for any $t > a$ where the subsequence converges on $[a,1]$. This is because if $|c_n| = O(t^{-n})$, $\sum_n c_n z^n$ is analytic in $|z|<t$, and by uniqueness...

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[Sorry -- too long for a comment]. It has already been pointed out that if you want 0 in your domain then you'll have to use constant functions too. But even then the theorem does not say what you seem to think it says: it only says that we can approximate $f(x)=x$ arbitrarily well by such a power series -- and as the approximation gets better and better the terms might jump around like crazy. In particular the coefficients are not really well-defined. As one way of seeing this, take one expansion which is very close to $x$ and let's assume the coefficient of $x^7$ is non-zero. Then remove 7 from your set of allowable powers of $x$ and the theorem again says that one can approximate $x$ arbitrarily well; this time the coefficient of $x^7$ is forced to be zero.

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  • $\begingroup$ What is wrong with the argument I sketched in my edit? $\endgroup$
    – C. Eratosthene
    Feb 7, 2017 at 15:40
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    $\begingroup$ @C.Eratosthene As Dirk points out, the theorem holds only on intervals $[a,b]$ with $a>0$. Your arguments actually show why this condition is necessary. $\endgroup$
    – Oleg Eroshkin
    Feb 7, 2017 at 16:02
  • $\begingroup$ Robert Israel's objection is the point -- your resulting infinite series is not guaranteed to converge. You only know that it's well-behaved if you sum up to $N_1$, $N_2$ and so on. The intermediate sums will presumably be all over the place. $\endgroup$
    – Kevin Buzzard
    Feb 7, 2017 at 16:37
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    $\begingroup$ In fact one can I guess see this by fixing $N$ and then looking at behaviour for $x=\delta$ chosen so small that $x$ is far larger than $\sum_{i=1}^Nc_ix^i$. This then shows that some term beyond the $N$th term has to be gigantic. $\endgroup$
    – Kevin Buzzard
    Feb 7, 2017 at 16:40

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