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$\newcommand{\d}{\,\mathrm{d}}\newcommand{\c}{\mathcal{C}}$I need to show the following:

Let $[a,b]$ be a compact interval in $\Bbb R$; for any two $f,g\in\c[a,b]$, $f=g$ on $[a,b]$ if and only if: $$\int_a^b x^nf(x)\d x=\int_a^b x^ng(x)\d x$$For all $n\in\Bbb N$.

This comes as an exercise after a proof of the Stone-Weierstrass and Weierstrass approximation theorems.

My strategy:

One direction is trivial. For the other, the space of all real polynomials on $[a,b]$ (hereafter denoted by some $p$) is dense in $\c[a,b]$, so by basic integral properties we obtain:

$$\int_a^bp(x)f(x)\d x=\int_a^bp(x)g(x)\d x$$

Since $[a,b]$ is a normal space, for all $t\in(a,b)$ I can choose for some small $h\in(0,\frac{1}{2}(b-t))$ so that $[a,t]$ and $[t+h,b]$ are two disjoint closed sets and there is a continuous $\varphi$ which is $1$ on the first set, $0$ on the second. There is some $p$ which can uniformly approximate $\varphi$ by $\epsilon$ for arbitrary positive $\epsilon$, whence:

$$\left|\int_a^b\varphi(x)f(x)\d x-\int_a^b\varphi(x)g(x)\d x\right|\lt\epsilon(b-a)\\\left|\int_a^tf(x)\d x-\int_a^tg(x)\d x\right|\lt\epsilon(b-a)+h(\|f\|+\|g\|)$$

$\epsilon$ and $h$ both may approach $0$ at any rate, and I can conclude then that:

$$\int_a^tf(x)\d x=\int_a^tg(x)\d x$$

For all $t\in(a,b)$. The fundamental theorem of calculus gives immediately then that $f(x)=g(x)$ for all $x\in(a,b)$, and I believe also for all $x\in[a,b]$ by continuous extension (not sure about this point).

  1. Is this right? I feel as if my conclusion from $\epsilon,h$ being arbitrarily small maybe isn't quite right.
  2. Can someone elaborate on whether or not the continuous extension idea holds?
FShrike
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    I think it's fine. You can make your writing more precise by saying first you are fixing $t$ and then maybe just choose $h = \frac 1n = \varepsilon$ for clarity. There is a much more straightforward way to do this by applying Bernstein theorem to $h = f-g$ to obtain $|h|_{L^2}<\varepsilon$ for any $\varepsilon. $ – dezdichado Feb 01 '22 at 23:04
  • I think you're making this harder than it needs to be. I would: 1. Prove the equivalent statement: If, for all polynomials $p(x)$, $\int_a^b p(x) f(x) , \mathrm{d}x = 0$, then $f(x) =0$, 2. Approximate $f(x)$ by polynomials $p(x)$, so that $\int_a^b p(x) f(x) , \mathrm{d}x \approx \int_a^b f(x)^2 , \mathrm{d}x$, 3. Use the fact that if $g(x) \ge 0$ and $\int_a^b g(x) , \mathrm{d}x = 0$, then $g(x) = 0$ for all $x \in [a, b]$. – Theo Bendit Feb 01 '22 at 23:05
  • @dezdichado There are a few Bernstein theorems listed on Wikipedia, I'm not sure which one you mean – FShrike Feb 01 '22 at 23:07
  • @TheoBendit Yes this is simpler, Urysohn's lemma and fundamental theorem of calculus not needed. Thank you – FShrike Feb 01 '22 at 23:07
  • I guess I meant Weierstrass approximation theorem. Bernstein first provided an explicit proof by constructing Bernstein polynomials. – dezdichado Feb 01 '22 at 23:16
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    Best not to dance around with inequalities and $\epsilon$s. There is a sequence of polynomials ${p_n}$ that converges uniformly to $f-g$. That means the sequence ${(f-g)p_n}$ converges uniformly to $(f-g)^2$. Integrals behave nicely with uniformly converging sequences of continuous functions. Etc. No $\epsilon$s. – B. S. Thomson Feb 01 '22 at 23:57

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