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Let $f$ be a continuous function on the interval $[1,2]$. It follows from Stone-Weierstrass theorem that if $\displaystyle \int_1^2x^nf(x) \, dx=0$ for integers $n=0,1,2,\ldots$, then $f$ must be identically zero. My question is,

Does there exist a non-zero continuous function $f$ on the interval $[1,2]$, and a positive constant $M$ such that $\displaystyle \left|\int_1^2x^nf(x)\,dx\right|\leq M$ for all integers $n=0,1,2,\ldots$?

If such a function exists, it must be an oscillating function which attains both positive and negative signs. I really appreciate for any answers, comments or suggestions.

Michael Hardy
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user70310
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    Have you already managed to prove through such assumptions that $f(2)=0$? – Jack D'Aurizio Apr 21 '18 at 18:21
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    No I did't. Thank you for your suggestion, To disprove the statement it is sufficient to prove that $f$ must vanish on a dense subset of $[1,2]$, from which it follows that $f$ must be identically zero. – user70310 Apr 21 '18 at 19:05
  • Could it be the case that for any continuous $f \not \equiv 0$, $\liminf_{n \to \infty} \frac{|\int_1^2 x^nf(x)dx|}{t^n} > 0$, where $t := \sup{x \in [1,2] : f(x) \not = 0}$? – mathworker21 Oct 26 '19 at 16:00
  • An idea perhaps: have you tried to determine this for functions of the form $\sin(nx)$ and $\cos(nx)$? Then for sums of the suggestive form $a_0/2+\sum (a_n\sin(nx)+b_n\cos(nx))$? And maybe an analog of Weierstrass (though much harder) such as Carleson's theorem? –  Oct 26 '19 at 16:50
  • @RobertWolfe I've been trying to get the result for functions in the Wiener algebra. I don't understand with the end of your comment though, since it's (theoretically) possible for $\sup_n |\int_1^2 x^n f(x)dx| = + \infty$, $f$ is approximated by its fourier series pointwise, and still $\sup_n |\int_1^2 x^n g(x)dx| < \infty$ for each $g$ a truncated fourier series. – mathworker21 Oct 26 '19 at 21:33
  • @mathworker21 I think you deleted an answer to this question (which I thought was correct!). What was the mistake you identified? – MathematicsStudent1122 Oct 26 '19 at 22:51
  • @MathematicsStudent1122 Chris Cutler identified the mistake. It was that the following is false. If $f(t) = 0$, then there is some $\delta > 0$ so that either $f(x) > 0$ for $x \in [t-\delta,t)$ or $f(x) < 0$ for $x \in [t-\delta,t)$. To see that it is false, consider $t=0$ and $f(x) = x\sin(\frac{1}{x})$. If you get 75 more reputation, you can see the deleted answer as well as his comment :) – mathworker21 Oct 26 '19 at 22:53
  • if $\int_1^{p_n}x^n|f(x)|=\int_{p_n}^2x^n|f(x)|$ then $p_{n+1}>p_n$. If $p=\sup_n p_n$ then $f\equiv0$ on $[p,2]$; $f$ oscillates,takes $\pm$ on any $(p-l,p),l>0$. We may replace $f$ with any $g_m(x)=x^mf(x)$ (where $m\in\Bbb Z$ if $M>C=\sup_{x\in[1,2]}|f(x)|$). We may replace $f$ with any$\sum_{n=0}^\infty\frac{x^nf(x)s_n}{2^n},s_n\in[-1,1]$ (doubling $M$). In particular, with $h(x)=f(x)-\sum_{n=1}^\infty\frac{x^nf(x)}{2^n}, h(1)=0,h\not\equiv0.$ If $f(r)>0$ and $1\le q<r$ then $\lim_{n}\int_q^rx^nf(x)=\infty$, $\lim_{n}\int_r^2x^nf(x)=-\infty.$ If $g(x)=\int_1^xt^nf(t)$ then $g'(x)=x^nf(x).$ – Mirko Oct 27 '19 at 04:40
  • @Mirko (1) I am unsure what the conclusion(s) of your comment is (are). (2) I don't see why you can replace $f$ with any $\sum_{n=0}^\infty \frac{x^n f(x)s_n}{2^n}$. I agree you can do so with finite sums of that form, but things might get weird with infinite sums of alternating signs. – mathworker21 Oct 27 '19 at 07:44
  • @mathworker21 (1) observations only,(2) didn't think of convergence problems. For any $1<b<2$ we have absolute and uniform convergence on $[1,b]$ by $|\sum_{n=0}^\infty\frac{x^nf(x)s_n}{2^n}|\le\sum_{n=0}^\infty(\frac b2)^nC$ (notation from earlier comment). Also,$\sum_{n=0}^\infty\frac{x^nf(x)s_n}{2^n}=0,p\le x\le2$. If $p<2$ we are lucky,absolute and uniform convergence on $[1,2]$; else sum need not be continuous at $2$. Take instead $h(x)=\frac{f(x)}2-\sum_{n=1}^\infty\frac{x^nf(x)}{3^n},h(1)=0$. Wanted to put a copy of $h$ on $[2,3]$(how?) get new $f$ on $[1,3]$,get contradiction at $2$. – Mirko Oct 27 '19 at 14:06
  • @mathworker21 Just tinkering, no clear idea of what I am after. Previous comment shows how to get $h(1)=0$. Then subst $x=2v,l(v)=h(2v),|\int_{1/2}^1v^nl(v)dv|\le\frac M{2^{n+1}}$,but I don't seem to get a contradiction (which was supposed to be that one cannot get $f$ on $[1/2,2]$ that simultaneously works for $[1/2,1]$ and $[1/2,2]$ (as well as $[1,2]$), $f$ being $l$ on $[1/2,1]$ and $h$ on $[1,2]$). Tried to think of an example,$\int_{1/2}^{2/3}x^n(\frac23-x)(x-\frac12)\sin(\frac1{(\frac23-x)^2})dx$ (function$=0$ on $[2/3,1]$) unable to estimate bounds (need $\frac{\mathrm{const}}{2^n}$). – Mirko Oct 27 '19 at 17:31
  • Does anyone know if there is a nonzero $L^2$ function that works? – mathworker21 Oct 28 '19 at 04:10

2 Answers2

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No, there does not exist such a non-zero $f.$

Suppose we have such a continuous function $f.$ Define $$g(z)=\int_0^{\ln 2} e^{-itz}f(e^t)dt$$ for $z\in\mathbb C.$ Then:

  • $g$ is an entire function of exponential type at most $\ln 2,$ specifically $g(z)\leq \|f\|_1 e^{(\mathrm{Im} z)\ln 2}\leq \|f\|_1 e^{|z|\ln 2}$
  • $g(in)=\int_1^2 x^{n-1}f(x)dx$ is bounded in absolute value: by assumption for $n\geq 1,$ and by boundedness of $f$ for $n\leq 1$
  • $|g(r)|\leq \|f\|_1$ for real $r$

Theorem 2 of Cartwright "On certain integral functions of order one" (https://academic.oup.com/qjmath/article-abstract/os-7/1/46/1587148?redirectedFrom=fulltext) states that if an entire function of exponential type less than $\pi$ is bounded on $\mathbb Z$ then it is bounded on $\mathbb R.$ Applying this theorem to $z\mapsto g(iz)$ shows that $g$ is bounded on the imaginary axis.

The Phragmén–Lindelöf principle states that a function of (arbitrary) exponential type that is bounded on the real and imaginary axes is bounded on the whole complex plane.

So $g$ is bounded. By Liouville's theorem, $g$ is constant. This forces $f$ to be a non-zero multiple of the Dirac delta $\delta_1,$ which is not a function let alone a continuous function.

Dap
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  • Who are you?? May you please let me know what stage of your mathematical career you are in? – mathworker21 Nov 01 '19 at 22:49
  • +1 of course. some nitpicks: (1) you only mention "$C$" once; (2) you forgot an "if" before "an entire function"; (3) I would say "nonzero multiple of the Dirac delta". – mathworker21 Nov 01 '19 at 23:25
  • (1) do you think there's any way this proof can be adapted to show that $\int_1^2 x^nf(x)dx$ must grow exponentially in $n$? Theorem 2 of Cartwright only concerns bounded functions rather than other growth rates. (2) your argument rules out any nonzero $L^\infty$ function, right? I think the defined $g$ is entire whenever $f$ is in $L^1$, but maybe I'm wrong. (3) it seems your argument does not work if $[1,2]$ is replaced with $[1,100]$. do you think the inability to apply theorem 2 of Cartwright is a true obstruction? I'd highly doubt it... – mathworker21 Nov 02 '19 at 04:29
  • @mathworker21: I like being anonymous, sorry. I've made those corrections. This argument will work for distributions supported on $[1,2],$ with the conclusion being that $f$ has to be a non-zero multiple of a Dirac delta; this includes $L^\infty$ and $L^1.$ I haven't thought about extensions - but would recommend starting with the more general results in Boas's book "Entire functions", in particular p. 191, 10.5.1, "Theorem of Duffin and Schaeffer" should be applicable. – Dap Nov 04 '19 at 11:24
  • thank you. you get 100 bounty for (1) the beauty of your solution; (2) the high quality questions you ask on this site; (3) responding to my comment. 100 might be too low, but whatever. – mathworker21 Nov 04 '19 at 13:18
  • I don't see how p. 191, 10.5.1 is applicable. The issue seems to be the exponential type $\pi$ rather than the subsequence along which $g$ is bounded. Anyways, I wonder whether any entire function of finite exponential type that is bounded on $i\mathbb{Z}$ and $\mathbb{R}$ is constant. – mathworker21 Apr 14 '20 at 07:36
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Here are some thoughts, not an answer.

For a polynomial $p$, let $||p||_{[1]}$ denote the sum of the absolute value of its coefficients.

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Claim: If $\{p \text{ polynomial} : ||p||_{[1]} \le \frac{1}{2}\}$ is dense in $L^2([1,2])$, then there is no nonzero $L^2$ function $f$ with $|\int_1^2 x^n f(x)dx| \le 1$ for each $n \ge 0$.

Proof: Suppose otherwise; let $f$ be a counterexample with $||f||_2 = 1$. Then $|\int_1^2 p(x)f(x)dx| \le ||p||_{[1]}$ for each polynomial $p$. So, if the hypothesis of the claim were true, we may take some polynomial $p$ with $||p||_{[1]} \le \frac{1}{2}$ and $||f-p||_2 < \frac{1}{2}$ to obtain $1 = |\int_1^2 f(x)^2dx| \le |\int_1^2 f(x)[f(x)-p(x)]dx|+|\int_1^2 f(x)p(x)dx| \le ||f||_2||f-p||_2+||p||_{[1]} < 1$, a contradiction.

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I'm not sure whether $\{p \text{ polynomial} : ||p||_{[1]} \le \frac{1}{2}\}$ is dense in $L^2([1,2])$. I'm not even sure whether the constant function $1$ is in the closure.

Here are some thoughts for trying to show that there is no such $f$ in the Wiener algebra (defined below).

By multiplying $f$ by $x-1$, we may assume $f(1) = 0$. Therefore, since obviously $f(2) = 0$, $f$ is $1$-periodic. Hence, the fourier transform $\hat{f}(k) := \int_1^2 f(x)e^{-2\pi i kx}dx$ is meaningful, e.g. we get $f(x) = \sum_{k \in \mathbb{Z}} \hat{f}(k)e^{2\pi ikx}$. If $f$ is in the Wiener algebra, i.e., if $\sum_{k \in \mathbb{Z}} |\hat{f}(k)| < \infty$, the condition $|\int_1^2 x^n f(x)dx| \le 1$ for all $n \ge 0$ translates to $|\sum_{k \in \mathbb{Z}} \hat{f}(k)I_{n,k}| \le 1$ for all $n \ge 0$, where $I_{n,k} = \int_1^2 x^ne^{2\pi i kx}dx$ (we needed $\sum_{k \in \mathbb{Z}} |\hat{f}(k)|$ to interchange the integral and the series).

Here's where I'm getting a bit out of my league. I'm not sure if there is some matrix $(B_{k,n})_{\substack{k \in \mathbb{Z} \\ n \ge 0}}$ such that $(B_{k,n})(I_{n,k}) = Id$, i.e., $\sum_{n \ge 0} B_{k,n}I_{n,k'} = \delta_{k=k'}$. If there is, then to disprove the existence of $(\hat{f}(k))_{k \in \mathbb{Z}} \in l^1(\mathbb{Z})$ with $|\sum_{k \in \mathbb{Z}} \hat{f}(k)I_{n,k}| \le 1$ for each $n \ge 0$, it suffices to show that whenever $(a_n)_{n \ge 0}$ lives in $[-1,1]^{n \ge 1}$, it holds that $(B_{k,n})(a_n) \not \in l^1(\mathbb{Z})$.

So, does anyone know how to invert the matrix $(\int_1^2 x^n e^{2\pi i kn}dx)_{\substack{n \ge 0 \\ k \in \mathbb{Z}}}$, if at all possible?

mathworker21
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