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Continuous $f$ has $≥2$ roots if $int_{-1}^{1} f(x)sqrt {1 - x^2} mathrm{d}x = int_{-1}^{1} xf(x) mathrm{d}x = 0$?

Mathematics Asked on December 14, 2020

Let $f$ be a continuous function on $[-1, 1]$ such that $$int_{-1}^{1} f(x)sqrt {1 – x^2} mathrm{d}x = 0 = int_{-1}^{1} xf(x) mathrm{d}x .$$

Prove that the equation $f(x) = 0$ has at least two real roots in $(-1, 1)$.

I am not sure where to begin, but I am thinking that I need to squeeze out the integral of $f(x)$ on $[-1, 1]$, although I am not sure if that is relevant to this problem. I was also taught that if I needed to prove "at least (insert number) real roots", one would usually use the Intermediate Value Theorem, but I am not sure how to apply that here. Perhaps, is it possible/wise to determine what $f(x)$ is and proceed from there?

Any help will be greatly appreciated!

2 Answers

Since $sqrt{1-x^2}$ is positive in $(-1,1)$ then $int_{-1}^{1} f(x)sqrt {1 - x^2} dx = 0$ implies that the continuous function $f$ has at least a zero $ain (-1,1)$ (otherwise the product $f(x)sqrt {1 - x^2}$ has the same sign over $(-1,1)$ and, recalling that if $Fgeq 0$ is continuous and $int_a^b F(x),dx=0$ then $F=0$ everywhere in $[a,b]$, we have a contradiction.

Assume that $a$ is the unique root of $f$ in $(-1,1)$, then $f$ should be positive on one side of $a$ and negative on the other side. Moreover $$int_{-1}^{1} f(x)g(x),dx=0$$ where $g(x)=(xsqrt{1 - a^2}-asqrt {1 - x^2})$ is a continuous function which is negative in $[-1,a)$ and positive in $(a,1]$. Hence the product $fg$ has the same sign on $(-1,1)$, and, since its integral is zero, we have a contradiction.

Correct answer by Robert Z on December 14, 2020

You can subtract both integrals to get $$int_{-1}^1 f(x)(x-sqrt{1-x^2})dx=0,$$ define $$h(t):=int_{-1}^t f(x)(x-sqrt{1-x^2})dx$$ and notice that $h(-1)=h(1)=0$.

Answered by José Umaña on December 14, 2020

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