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Let $(p_n)_{n \in \mathbb N} \subset (0,\infty)$ be a sequence of distinct positive numbers such that their series of reciprocals diverges. By the theorem of Müntz-Szasz the closure of the span of the system $\{ x \mapsto x^{p_n} : n \in \mathbb N\}$ with respect to the supremum norm is equal to $C[0,1]$, the space of continuous functions. I was wondering if an extension of this result of the following form is possible:

Suppose that $f:[0,1] \to (0,\infty)$ is a smooth function which is strictly decreasing (or increasing). Then the closure of the span of system $\{ x \mapsto f(x)^{p_n} : n \in \mathbb N\}$ with respect to the supremum norm is equal to $C[0,1]$.

If this is not the case, can we strengthen the conditions on $f$ so that a statement of the above type holds?

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  • $\begingroup$ For Müntz-Szasz you also need the constant function. The desired generalization holds for $p_n=n$ (the Weierstraß case) by Stone-Weierstraß. Apart from this, I don't know. $\endgroup$
    – Jochen Wengenroth
    Jan 5, 2022 at 13:29
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    $\begingroup$ Just change variable $y=f(x)$. This tranforms $C[0,1]$ into $C[a,b]$ and the systemt $y^{p_n}$ into $f^{p_n}$. $\endgroup$
    – Giorgio Metafune
    Jan 5, 2022 at 14:05
  • $\begingroup$ To be specific the result holds for any such function (no need for the constant function since the values are in $(0,1)$), even under the assumption that the $p_n$ be complex numbers with positive real parts such that $$\sum \left(1-\left|\frac{p_n-1}{p_n-1}\right|\right)=\infty.$$ $\endgroup$
    – bathalf15320
    Jan 5, 2022 at 14:29
  • $\begingroup$ @bathalf15320 Unfortunately, you can't edit comments after 5 minutes. Perhaps, you may want to rewrite the last condition. in another comment. $\endgroup$
    – Jochen Wengenroth
    Jan 5, 2022 at 16:08
  • $\begingroup$ Thanks. The minus in the denominator should be a plus. $\endgroup$
    – bathalf15320
    Jan 14, 2022 at 10:42

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This is an attempt to salvage the damaged comment of bathalf15320. We consider the case where $f$ maps $[0,1]$ onto itself. Then a sufficient condition for the completeness of the given family of functions is that the sequence $(p_n)$ is a set of uniqueness for analytic functions which are bounded on the open right-half plane. Using a Möbius transformation to reduce to the corresponding question for the open unit ball and the theory of Blaschke products then shows that the condition $$\sum \left (1-\left |\frac {p_n-1}{p_n+1} \right| \right )=\infty$$ suffices. In this formulation there is no necessity to assume that the exponents are real—they can be arbitrary points in the open right half plane.

This shows that, just as in the classical Müntz-Szász theorem, there are many interesting ways for this to happen beyond the case of real exponents converging to infinity slowly enough, for example reals decreasing slowly to zero or complex exponents converging to the imaginary axis or infinity in more exotic ways.

The more general case can presumably be attacked by using a change of scale but I am unaware of any such work, even for Müntz-Szász with $x$ replaced by $ax+b$.

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