The discussion about $\int_a^b P^2(x)f(x)dx=0$

Question

$f$ is Riemann integrable in $[a,b]$,and $\int_a^b f(x)dx>0$. If the polynomial $P(x)$ satisfies $\int_a^b P^2(x)f(x)dx=0$. Prove $P(x)=0$.

@robjohn The problem seems a little weird.Actually,I think the function of P(x) depends on the root of P(x) and f(x). — 89085731, Apr 21 '12 at 03:06
Your problem is a variation of this one,
http://math.stackexchange.com/questions/16831/nonzero-f-in-c0-1-for-which-int-01-fxxn-dx-0-for-all-n/92937#92937 — , Apr 21 '12 at 13:48
@EduardoSiva: Note that this is one particular $f$ and one particular $P$. — robjohn, Apr 21 '12 at 14:35

score 3 · Answer 1 · answered Apr 21 '12 at 04:22

3

You need more conditions on $f, P$.

Choose $f(x) = 1$ when $x \in [0,\sqrt[3]{\frac{2}{3}}]$, and $f(x)=-2$ when $x \in (\sqrt[3]{\frac{2}{3}}, 1]$. Let $P(x) = x$. Then I have $\int_0^1 f(x) dx > 0$, and $ \int_0^1 P^2(x) f(x) dx = 0$, but clearly $P \neq 0$.

answered Apr 21 '12 at 04:22

copper.hat

178,207

Ah, I misread the problem. I thought that somewhere, it said that $f$ was positive. – robjohn Apr 21 '12 at 04:29
Yes, I suspect that $f(x)\geq 0$ should be part of the problem statement. – copper.hat Apr 21 '12 at 04:36
If $f(x)\geq 0$ I think this result is true. – Siming Tu Apr 21 '12 at 06:03
It is true under this condition, as @robjohn pointed out earlier. – copper.hat Apr 21 '12 at 06:33

robjohn · Accepted Answer · 2012-04-21T18:28:03.673

If, in addition, $f$ is non-negative, here is a hint.

Hint: Polynomials that are not identically $0$ can vanish only on a finite set. Consider the open sets $$ U_k=\left\{x\in(a,b):|P(x)|>\frac1k\right\}\tag{1} $$ and let $$ F_k=\int_{U_k}f(x)\,\mathrm{d}x\tag{2} $$ Show that $$ \int_a^bf(x)\,\mathrm{d}x=F_1+\sum_{k=2}^\infty(F_{k}-F_{k-1})\tag{3} $$ and $$ \int_a^bP^2(x)f(x)\,\mathrm{d}x\ge F_1+\sum_{k=2}^\infty(F_{k}-F_{k-1})\frac{1}{k^2}\tag{4} $$ What conclusions can you draw from $(3)$ and $(4)$?

The discussion about $\int_a^b P^2(x)f(x)dx=0$

2 Answers2