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Let $f: [0,1] \to \mathbb{R}$ be injective. Does $ \displaystyle\sum_{n=1}^{\infty} c_n \left( f(x) \right) ^n = 0\ \forall x\in [0,1] \implies c_n = 0\ \forall n\in\mathbb{N}\ ? $

Maybe something related to (i.e. a more general version of) Stone-Weierstrass could be helpful?

Adam Rubinson
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  • @AdamRubinson A mistake on my part, I assumed that $f^n$ is the $n$th iterate of $f$, but it's obviously the $n$th power of $f$. – Sarvesh Ravichandran Iyer Nov 23 '23 at 12:53
  • @AdamRubinson This is not what linearly independent means. – Jakobian Nov 23 '23 at 12:55
  • Also consider adding some more context e.g. if you already know that this is true for some $f$ with proof that'd be great. – Sarvesh Ravichandran Iyer Nov 23 '23 at 12:56
  • @Jakobian I thought it did in this context. What does linearly independence mean here then? – Adam Rubinson Nov 23 '23 at 12:56
  • Is it because the sum should be from $n=0$ not $n=1$? – Adam Rubinson Nov 23 '23 at 12:57
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    A set of functions $A$ is linearly independent if for any finite amount of its elements $f_1, ..., f_n\in A$ and $c_1, ..., c_n$ if $c_1f_1(x)+...+c_nf_n(x) = 0$ for all $x$ then $c_1 = ... = c_n = 0$. You're considering infinite sums, this should be a finite sum. – Jakobian Nov 23 '23 at 12:57
  • @Jakobian excellent point. I need to re-think how I am going to amend the question now. – Adam Rubinson Nov 23 '23 at 12:58
  • Well, obviously the set must be (finitely) linearly independent, right? – Adam Rubinson Nov 23 '23 at 13:00
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    If $h(z) = \sum c_n z^n$ has a positive radius of convergence, and the range of $f$ lies within this radius, then all $c_n$ must be zero: the range of $f$ is uncountable and thus has a limit point, and the holomorphic function $h$ then has an accumulation point of zeros, which implies it is the zero function. Does this seem correct to you? – Olius Nov 23 '23 at 13:02
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    It suffices to assume that $f$ takes infinitely many different values and use the Vandermonde determinant to show that $c=0$ is the only solution to any given postulated finite linear dependency. – Michal Adamaszek Nov 23 '23 at 13:05
  • @AdamRubinson do you assume that $f$ is continuous? Because as far as I can see only under this condition Stone-Weierstrass is related. – freakish Nov 23 '23 at 13:15
  • @freakish maybe there are variations/more general versions of of Stone-Weierstrass that do not require continuity. – Adam Rubinson Nov 23 '23 at 18:21

1 Answers1

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If you suppose that the infinite sum $\sum c_n \bigl(f(x)\bigr)^n$ is well-defined for every $x \in [0,1]$, then the range of $f$ must lie within the radius of convergence of $$ h(z) = \sum_{n=0}^\infty c_n z^n \text. $$ Then, as $[0,1]$ is an uncountable set and $f$ is injective, its range is an uncountable subset of $\mathbb R$, and any such set has uncountably many accumulation points. Supposing $h\bigl(f(x)\bigr) = 0$ for every $x \in [0,1]$ thus implies that the holomorphic function $h$ has an accumulation point of zeros which lies strictly withing the radius of convergence (thanks to freakish in the comments), and this forces it to be the zero function, i.e., all $c_n = 0$.

Olius
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