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Let $f: [0,1] \rightarrow [0,1]$ be a continuous function with the property that,

$$\int_{0}^{1} f(x)x^n dx = \frac{1}{n+2}$$

for all $n = 0, 1, \dots$

Show that $f(x) = x$.

Well, I tried to use induction, but that didn't really work out because for the case $n=0$, you just get $\displaystyle\int_{0}^{1} f(x) dx = 1/2$.

I tried to use the Mean Value Theorem, and got that there is a $\xi \in [0,1]$ such that

$$\int_{0}^{1} f(x)x^n dx = f(\xi) \int_{0}^{1} x^n dx = f(\xi)\frac{1}{n+1} = \frac{1}{n+2}.$$

Therefore, there exists $\xi \in [0,1]$ such that $f(\xi) = \frac{n+1}{n+2}$ for all $n$.

I'm not really sure what to try next. This is also a question from an old qual exam and possibly uses techniques not covered in my course.

Did
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user389056
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1 Answers1

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Note that Stone-Weierstrass shows that polynomials are dense in the set of continuous functions on $[0,1]$ (with the usual sup norm). But then

$$\int_0^1(f(x)-x)x^n\,dx = 0$$

for all $n\ge 0$ implies that $f(x)-x$ is the $0$ function, because the $x^n$ are a (Schauder) basis for the space of continuous functions i.e. $f(x)=x$.

Adam Hughes
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  • Since we are assuming $ f \ge 0$, could't we have concluded by saying that the integral of a positive function is $> 0$, without using weierstrass? – Ant Dec 05 '16 at 22:46
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    @Ant sure $f(x)>0$ but $f(x)-x$ might not be. – Adam Hughes Dec 05 '16 at 22:47
  • We haven't learned Stone-Weierstrass, but thanks. – user389056 Dec 05 '16 at 22:50
  • @user389056 you don't need anything that big as the general theorem, if you know polynomials are dense in continuous functions (straightforward) the same proof works. If you don't, I'm not sure how you would do it, so I would ask especially if it's on old qualifying exams. – Adam Hughes Dec 05 '16 at 22:51
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    "${x^n}$ is dense in the set of functions on $[0,1]$" That's false in 3 ways. First you probably meant "the set of continuous functions". Even if you meant that, you need to say in what sense it is dense. There should be a metric space lurking somewhere. So really it should "dense in $C[0,1]$ with its usual sup-norm metric." Lastly, that's still not right, because you need to take the linear span of the $x^n$ for a correct statement. – zhw. Dec 05 '16 at 23:03
  • @zhw typo, should be continuous functions. The only thing that makes any sense is ones on the actual set we're dealing with – Adam Hughes Dec 05 '16 at 23:04
  • @AdamHughes Sorry but the points zhw. enumerates are not mere typos. – Did Dec 05 '16 at 23:13
  • @Did yes I know, I'm on mobile so editing the non typo details has to wait until I have a keyboard. – Adam Hughes Dec 05 '16 at 23:15