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Let $f:[a,b]\rightarrow \mathbb {R} $ a continious function such that $$ \int_{a}^{b} x^n\,f(x)\,dx=0$$ for all $n \in \mathbb N$.

Show that $f$ is identically $0$.

I notice that $f$ is bounded and it touches its bounds. But I don't know how to continue.

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    $\begingroup$ is it zero for one $n$ or for all positive integers $n$? $\endgroup$
    – Student
    Jun 22, 2018 at 21:11
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    $\begingroup$ Let $f(x) = 1$ if $x = a$, $0$ otherwise... $\endgroup$ Jun 22, 2018 at 21:11
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    $\begingroup$ Your question is vague. Is this for all n or for some fixed n? If it's the latter, then there are cases where this is false. Take, for example, n = 0, in which case you can't conclude that f is identically 0. $\endgroup$ Jun 22, 2018 at 21:12
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    $\begingroup$ Is $f$ continuous? $\endgroup$
    – lhf
    Jun 22, 2018 at 21:18
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    $\begingroup$ "I notice that $f$ is bounded and it touches its bounds. But I don't know how to continue." How could you notice that? You give no hypotheses on $f.$ $\endgroup$
    – zhw.
    Jun 22, 2018 at 22:54

2 Answers 2

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If $\int_{a}^{b} x^n f (x) \, dx=0$ for all $n \in \mathbb N$, then $\int_{a}^{b} p(x)f (x) \, dx=0$ for all polynomials $p$.

By the Weierstrass approximation theorem, this implies that $\int_{a}^{b} g(x)f (x) \, dx=0$ for all continuous functions $g : [a,b] \to \mathbb R$.

If $f$ is continuous, then taking $g=f$ gives $\int_{a}^{b} f(x)^2 \, dx=0$, which implies $f^2=0$, that is, $f=0$.

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  • $\begingroup$ Not so fast. Isn't $\mathbb N = \{1,2,\dots\}?$ $\endgroup$
    – zhw.
    Jun 22, 2018 at 22:58
  • $\begingroup$ It's ambiguous whether or not 0 is in N, so it's up to the author $\endgroup$ Jun 23, 2018 at 2:04
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Assuming you mean for all $ n \geq 0 $ rather than some fixed $ n $, this is true. One possible proof is to first note that since this result holds for all $ n \geq 0 $, for any polynomial function $ p: [a,b] \to \mathbb R $, $ \int_a^b p(x) f(x) dx = 0 $. Using Stone-Weierstrass, we then have that for any continuous function $ g: [a, b] \to \mathbb R $, $ \int_a^b g(x) f(x) dx = 0 $. If $ f $ is itself continuous, then simply take $ g = f $ to conclude. If f is not continuous, then you can only conclude that $ f = 0 $ a.e., so I only know how do to this with Lebesgue integration.

In this case, we use the fact that for a function $ f \in L^1([-\pi, \pi]) $, if every Fourier coefficient $ a_n = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) e^{-inx} dx = 0 $ then $ f = 0 $ a.e. For a reference, this is Theorem 3.1 in chapter 4 of Stein and Shakarchi's Real Analysis. As $ \cos(nx) $ and $ \sin(nx) $ are continuous for all $ n \in \mathbb Z $, and as $ e^{inx} = \cos(nx) + i \sin(nx) $, we have that every Fourier coefficient of $ f $ is 0, so $ f = 0 $ a.e. by the theorem cited.

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  • $\begingroup$ An alternative approach for $f \in L^1([a,b])$ is to use that there exists a sequence of step functions, $\psi_n$ , that converge upward to $f$ e.g. ($\psi_n \nearrow f$). Then we can linearize these step functions to make them continuous and preserve the convergence upward. Now as these linearized step functions are continuous, and converge upward to $f$. We may apply the DCT with $f^2$ to obtain the claim. $\endgroup$
    – user518441
    Jun 23, 2018 at 20:11

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