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If $f, g\in \mathcal{C}([a,b],\mathbb{R})$ and $$\int^b_af(x)\,x^k \, dx=\int^b_ag(x)\,x^k \, dx, \text{ for all } k \in \mathbb{N}$$ Prove that $f=g$.

I'm trying to prove this but I don't know how to proceed. Any suggestions?

gen-ℤ ready to perish
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João Costa
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3 Answers3

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Let $h:= f-g$. The hypothesis states that $ \int h p = 0$ for any polynomial $p$.

To begin, we apply Stone-Weierstrass to get that the polynomials are uniformly dense in $\mathcal{C}([a,b],\mathbb{R})$.

In particular, for $\epsilon >0$, we can find a polynomial $p$ such that $ \| h-p \|_\infty < \varepsilon $.

Thus, $$ \left| \int h^2 \right| \leq \left| \int h p \right| + \left| \int h (h-p) \right| \leq \varepsilon \left| \int h \right|. $$

Since $h$ is continuous, $| \int h | < \infty $ and hence $ \int h^2 = 0$.

But for any continuous function, this is precisely $h=0$ or $f=g$.

Michael Hardy
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Luke Peachey
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  • Not accurate as written. The hypothesis states that $\ldots$ for any polynomial $p(x)$ that does not have a constant term. And these polynomials are not dense by Stone-Weierstrass as it does not contain a non-zero constant function. The trick is to do it as in the other answer. Also Stone-Weierstrass is overkill here as Weierstass approximation theorem will do. – Winther Dec 21 '17 at 09:52
  • Ah, i see. I didn't realise he wasn't including constant polynomials - I assumed he had included $ 0 \in \mathbb{N}$. But yes, it easy to tweak this by considering $xh$ instead. However, I think the point about the Weierstrass approximation theorem is kind of redundant, as SW is a stronger version with an almost identical proof. To me, the Weierstrass approximation theorem is only useful as a motivating example for SW. – Luke Peachey Dec 21 '17 at 21:39
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$\int_a ^{b} xf(x) x^{k} dx=\int_a ^{b} xf(x) x^{k} dx$ for $k \geq 0$. Hence, by standard Weierstarss approximation $xf(x)=xg(x)$ for all x which gives what you want.

Kavi Rama Murthy
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Maybe an overkill. One knows by tweaking Stone-Weierstrass that polynomials are dense in $C[a, b] $ with $\Vert \cdot \Vert_2$ norm. Let $\mathcal P$ be the set of all polynomials. What you have written is equivalent to $$<f-g, p>=\int_a^b (f(x) - g(x) ) p(x) = 0$$ for every polynomial $p\in \mathcal P$. Hence $f(x)-g(x) $ is in the annihilator of $\mathcal P$, namely $\mathcal {P} ^\perp$ and since $\mathcal P$ is dense we get $f(x) - g(x) \in \mathcal {P} ^\perp=\{0\} $. Hence $f=g$.

Edit/Remark. the same idea also works if $0\notin\mathbb{N}$. That proof is left to the reader.

Shashi
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  • Shashi, the equation $<f-g,p>=0$ is available only when there is no constant term. The whole point of this question seems to be that k=0 is not included in the hypothesis. My solution shows how to get over this difficulty. – Kavi Rama Murthy Dec 26 '17 at 07:46
  • @KaviRamaMurthy I see that yours can be done if $0\notin \mathbb N$ but the OP still did not specify whether $0\in\mathbb{N}$ or not. I added some remarks to make the post complete. Thanks! – Shashi Dec 26 '17 at 07:53