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

Question

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!

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 03:55
Igor, the question puts no stipulations on the value of $n$. — Tyler Clark, Dec 20 '13 at 04:01

score 6 · Accepted Answer · answered Dec 20 '13 at 03:53

6

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 $g \equiv 0$.

answered Dec 20 '13 at 03:53

Prahlad Vaidyanathan

31,554

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 04:26
Why do we need that integral condition? continuity of $f$ is sufficient to have a fixed point. – Myshkin Dec 20 '13 at 05:52
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
Ah yes, of course. I was thinking that it relied on the fact f(x)=x but it just relies on the fact that if f(x)=x then the integral is the same which is different. – Chris Dec 20 '13 at 11:39
I misssed it sorry.....then he should have written $f(x)\equiv x$ – Myshkin Dec 20 '13 at 11:54

score 3 · Answer 2 · answered Dec 20 '13 at 03:53

3

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.

answered Dec 20 '13 at 03:53

Igor Rivin

25,994
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score 2 · Answer 3 · answered Dec 20 '13 at 03:55

2

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?

answered Dec 20 '13 at 03:55

Eleven-Eleven

8,806

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

3 Answers3