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Let g be continuous, not negative, and strictly increasing in [a, b]. Prove that if $f$ is continuous and $$\int_{a}^{b}f(x)g^n(x)dx=0, \quad \forall n \in \mathbb{N},$$ then $f\equiv 0$.

With a change of variable I have arrived here but I could not continue: $$\int_{g(a)}^{g(b)}f(u)u^n \frac{du}{u'}=0, \quad \forall n \in \mathbb{N},$$

Two particular cases already resolved by the community are:

$g(x)=x$ and $x \in [0,1]$: $\quad \int_{0}^{1}f(x)x^ndx=0, \quad \forall n \in \mathbb{N}, \; $ then $f\equiv 0$.

$g(x)=e^x$: $\quad \int_{a}^{b}f(x)e^{nx} dx=0, \quad \forall n \in \mathbb{N}, \; $ then $f\equiv 0$.

u0tf1s
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    This is basically saying that ${g^n(x)}$ is dense in $L^2$ for any continuous, non-negative, strictly increasing $g$ in $L^2[a,b]$. That equates to the following : given a function $f \in L^2$, there exists a sequence in the span of ${ g_n(x)}$ that converges to $f$ in $L^2$. Try to prove the weaker fact that there exists such a sequence that converges to $f$ pointwise a.e. (which is implied by, but does not imply $L^2$ convergence). A counterexample here, will be a counterexample for the larger case as well. – Sarvesh Ravichandran Iyer Apr 30 '17 at 05:51
  • Your change of varaible leads to the well-known $x^n$ case, but needs differentiability. – Hagen von Eitzen Apr 30 '17 at 06:01
  • Could you give an answer and explain in more detail? – u0tf1s Apr 30 '17 at 06:37
  • How would it be? – u0tf1s Apr 30 '17 at 06:44
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    @астонвіллаолофмэллбэрг ${g^n}$ has not to be dense in $L^2$. For example take $g(x) =x$ then the sequence ${x^n}$ convergences to $h(x)=0$ in $L^2$ so it can't be dense in $L^{2}$ – Red shoes Apr 30 '17 at 08:00
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    @nonlinearthought I meant the span of ${ g^n}$ should be dense in $L^2$, not ${g^n}$ itself. The span of ${ x^n}$ is indeed dense in $L^2[a,b]$,since given a function in $L^2[a,b]$, we can find a polynomial (which is in the span of $x^n$) that is arbitrarily close to it in the $L^2$ sense. – Sarvesh Ravichandran Iyer Apr 30 '17 at 08:13
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    @астонвіллаолофмэллбэрг can you please prove that, or put here a reference about this density, since this is way stronger than the question here. – Red shoes Apr 30 '17 at 08:32
  • @nonlinearthought I thought this question was easier than the one given here , anyway : https://math.stackexchange.com/questions/548713/polynomials-are-dense-in-l2 – Sarvesh Ravichandran Iyer Apr 30 '17 at 08:40
  • I suggest you take a look at the Stone-Weierstrass theorem. – Michał Miśkiewicz Apr 30 '17 at 10:46
  • Is $\mathbb N = {1,2,\dots }$? – zhw. Apr 30 '17 at 19:11
  • @zhw I'm sorry for not clarifying, in this case $\mathbb N = {0,1,2,\dots }$, that is, $n = 0,1,2, ...$. – u0tf1s Apr 30 '17 at 21:45

4 Answers4

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Per the problem statement, $g$ is a homeomorphism $[a,b]\to [c,d]$. The continuous function $f\circ g^{-1}\colon [c,d]\to\Bbb R$ can be approximated uniformly by polynomials. So for $\epsilon>0$, we find $p\in\Bbb R[X]$ with $|f(g^{-1}(t))-p(t)| <\epsilon$ for all $t\in[c,d]$, or equivalently $|f(x)-p(g(x))|<\epsilon$ for all $x\in[a,b]$. Then $$\begin{align}\int_a^bf(x)^2\,\mathrm dx&=\int_a^b f(x)p(g(x))\,\mathrm dx+\int_a^bf(x)(f(x)-p(g(x))\,\mathrm dx \\ &\le\rlap{\qquad0}\hphantom{\int_a^b f(x)p(g(x))\,\mathrm dx+}\llap{+\;\epsilon}\int_a^b|f(x)|\,\mathrm dx.\end{align}$$ As $\epsilon$ was arbitrary $>0$, we conclude $\int_a^bf(x)^2\,\mathrm dx\le 0$ and therefore $f\equiv 0$.

Hagen von Eitzen
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  • If $\mathbb N = {1,2,\dots }$ there's a little more to do. – zhw. Apr 30 '17 at 19:22
  • @zhw I'm sorry for not clarifying, in this case $\mathbb N = {0,1,2,\dots }$, that is, $n = 0,1,2, ...$. – u0tf1s Apr 30 '17 at 21:45
  • @Hagen von Eitzen For $g$ to be a homeomorphism, only $g$ need to be bijective and both $g$ and $g^{−1}$ need to be continuous. Where do you use that $g$ is non-negative? – u0tf1s May 02 '17 at 08:32
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For simplicity let assume $[a,b] = [0,1]$ and first suppose $g(x)=x$ taking into account the particular case, we have $\int_{0}^{1} f(x) p(x) = 0 $ for each $P(x) = polynomial$.

now since $f(x)$ is continuous on $[0,1]$ then there exists a sequence of polynomials, say $P_n$ such that $P_n \rightarrow f$ uniformly. This implies that $$ 0= \int_{0}^{1} f(x) p_n (x) \rightarrow \int_{0}^{1} |f(x)|^2 $$ Hence, $\int_{0}^{1} |f(x)|^2 =0$, and because $f$ is continuous we arrive $f=0.$
Now let W.L.O.G assume $g:[0,1] \rightarrow [0,1]$ is onto (otherwise you can scale this g) this implies $0=g(0),~ 1=g(1)$ now by substitution $y=g(x)$ the condition of question becomes $$ \int_{0}^{1} F(y) y^n dy =0 $$ where $F(y) = f(g^{-1} (y)) (g^{-1} (y))' $ hence according to my pervious argument $F=0$ which implies $f=0$

Some Clarifications :

Note that since $g$ is strictly increasing so $g^{-1}$ is. And it is a well-known fact that every increasing function has almost countable non-differentiable points, thus $g^{-1}$ is differentiable a.e. Therefor it does not impact on the integrability and the value of integral. Thus this makes $F=0$ a.e and then $f=0$ a.e and since $f$ is continuous $f=0$ every where!.

P.S for those who have hard time of scaling $g$, if $[g(0) , g(1)] \neq [0,1]$ you can scale this $g$ in this way $\frac{g(x)-g(0)}{g(1)-g(0)}.$

Red shoes
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  • People who voted down this answer, Please tell why? put comment here.

    If you cant understand my solution that doesn't meant it is wrong !!!!

     – Red shoes Apr 30 '17 at 06:05
  • where your function $g^n$ here ? – Empty Apr 30 '17 at 06:07
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    The question presumes some arbitrary, fixed function $g(x)$. It doesn't say that the equality holds for any function $g(x),$, so you don't get to choose the $g(x),$. – dxiv Apr 30 '17 at 06:09
  • Look at bottom of question. and what I write here "taking into account the particular case" ! – Red shoes Apr 30 '17 at 06:09
  • @divx look at my solution again , I added few lines to it, is it clear for you? – Red shoes Apr 30 '17 at 06:46
  • even if you can assume that $g(x)$ is onto (which I don't think your scaling argument is correct), $g^{-1}$ does not have to be a well-defined function. You need injectivity to assume the inverse function. – dezdichado Apr 30 '17 at 07:01
  • @divx all I am saying is that first we solve problem for special case when $g(x)=x$. Then for general case where $g$ is an arbitrary function , by substitution of $y=g(x)$ we can turn problem into particular case again ! – Red shoes Apr 30 '17 at 07:01
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    @dezdichado do you know what strictly increasing means ?!! – Red shoes Apr 30 '17 at 07:03
  • I make it more clear like 3 times!! I don't care if people are still voting down my solution, since they can't understand it ! I myself love my solution! – Red shoes Apr 30 '17 at 07:19
  • @nonlinearthought I removed my downvote since your later clarifications, though I am not entirely convinced that this approach must work. I still don't see why $g^{-1}$ would have to be differentiable, which was one of the first comments posted under OP's question. – dxiv Apr 30 '17 at 07:25
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    @dxiv thanks, since $g$ is strictly increasing so $g^{-1}$ is. And it is well-known fact that every increasing function has almost countable non differentiable points, so it is almost every where differentiable so it does not impact on integral. so this make $f=0$ almost every where and since $f$ is continuous $f=0$ every where! – Red shoes Apr 30 '17 at 07:33
  • Differentiability a.e. is not enough for the purposes of changing variables in the integral. In general, using an increasing substitution $x \mapsto y$ fails if you only use the derivative a.e. (unless the substitution is absolutely continuous). You need to consider the distributional derivative (which is a finite positive measure) and prove that $F=0$ as a measure. – Michał Miśkiewicz Apr 30 '17 at 10:04
  • @MichałMiśkiewicz Here $g$ is indeed an absolutely continuous function, since it is continuous and increasing. Plus I suggest you to look at https://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral – Red shoes Apr 30 '17 at 17:48
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    @miguel Note that at bottom of your question you yourself considered the interval [0,1], that actually holds true for any interval [a,b]. (just follow the first part of my argument with [a,b] instead [0,1]), then for general case $g:[a,b] \rightarrow [g(a),g(b)]$ is onto and all things likewise follow from the first argument if you consider interval $[g(a),g(b)]$ instead $[0,1]$ for second integral – Red shoes Apr 30 '17 at 17:56
  • No, $g$ does not have to be absolutely continuous (and the comment section is not the right place to convince you). Yes, I guess using the Riemann-Stielstjes could lead to the proof, as it is equivalent with the Lebesgue integral (if $g$ has bounded variation). – Michał Miśkiewicz Apr 30 '17 at 18:50
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Hagen von Eitzen above has indeed given an elegant solution valid when the given g is strictly increasing but no assumption that g be non-negative is needed or used. However it is not necessary assume the integral (1) I[a,b]f(x)(g(x))^n dx =0 for all non negative integers n= 0.1,2,3... . It suffices to assume this for integers n >= M where M is any positive integer and by increasing M we assume M is odd .We use the Stone Weierstrass theorem instead of the Weierstrass Theorem .Note that (1) implies that I[a,b]f(x) (g(x)^M (p(g(x))dx =0 for all polynomials If g(x) is never 0 in [a,b] then g(x)^M is strictly increasing hence the algebra of functions of the form (2)q(x)= (g(x)*M (p(g(x)) separates the points of [a,b] and don't all vanish at any one point and so dense in the continuous functions for the mac norm .Using such a sequence converging to f(x) uniformly we see that I(f(x)^2 ) =0 so f vanishes identically as before What happens if g(c) = 0 for some c in [a,b] (exactly one ,g is strictly increasing . Then the functions of the form C + q (q as in (2),C a real constant ) are a dense set . But if h(c)=0 H continuous then a sequence of functions c+q oonverges uniformly to h and all q(c)=0 hence the sequence of constants c tends to 0 so the the sequence of q 's =c+q -c tends to h uniformly . Take h = f T where T(x) is between 0 and 1 ,T(c) =0 .and T=1 off and interval of length e containing c , T continuous then see that I (f^2T) =0 Let e tend to 0 to get I(f*2)=0 and hence f=0 . Stuart M.N.

user439545
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Okay, re-attempt. Assume $f$ is not identically zero. Then there exists at least one point $x_0 \in [a,b]$ such that $f(x_0) \neq 0$ and say w.l.o.g. that $f(x_0) > 0$ (otherwise just take the negation). Since $f$ is continuous, there is a small neighborhood around $x_0$ where $f>0$ in that interval. Since $g$ is non-negative and increasing, this implies that

$$ \int f(x)\,g^n(x)\,dx > 0 $$

Which is a contradiction to the original statement. Thus $f \equiv 0$.

MasterYoda
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