# Show that a continuous function with a certain integral property must be f(x)=x.

I am studying for the quals in January and am working through the following problem:

Let $f:[0,1] \to [0,1]$ be a continuous function with the property that $\displaystyle\int_0^1 f(x)x^n \, \textrm{d}x = \dfrac{1}{n+2}$. Show that $f(x)\equiv x$.

I was thinking to do integration by parts, but this does not seem to yield anything. Any suggestions would be much appreciated!

-
For what range of $n?$ –  Igor Rivin Dec 20 '13 at 3:51
note that you don't have any assumptions about differentiability so, strictly speaking, integration by parts isn't available to you –  crf Dec 20 '13 at 3:55
Igor, the question puts no stipulations on the value of $n$. –  Tyler Clark Dec 20 '13 at 4:01

Consider $g(x)=f(x) -x$, then $$\int g(x)p(x)dx = 0$$ for every polynomial $p$. Now by Weierstrass Approximation, and the continuity of the integral, it follows that $$\int g^2(x)dx = 0$$ Since $g$ is continuous, it follows that $g \equiv 0$.

-
Thank you! My second thought was Weierstass, but I wasn't sure if there was a better route to go. –  Tyler Clark Dec 20 '13 at 4:26
Why do we need that integral condition? continuity of $f$ is sufficient to have a fixed point. –  Une Femme Douce Dec 20 '13 at 5:52
Where does the first line come from? –  Chris Dec 20 '13 at 11:34
@Tojamaru : Not sure what you mean, but the question does not ask for a fixed point, but for $f$ to be identically equal to $x$. –  Prahlad Vaidyanathan Dec 20 '13 at 11:36
@Chris : The given condition says that $$\int f(x)x^ndx = \int x\cdot x^ndx$$ –  Prahlad Vaidyanathan Dec 20 '13 at 11:36

Perhaps this helps... $$(n+2)\int_0^1f(x)x^ndx=1$$ Then $$\int_0^1(n+2)f(x)x^ndx=1$$ You can see the power rule embedded in there now, except $x^n$ would need to be $x^{n+1}$. So what must $f(x)$ be?

-

I would say, (a) Prove it for $f$ polynomial. (b) use Stone-Weierstrass to approximate by polynomials, and then use (a) to show that the coefficients other than the linear term are arbitrarily small.

-