# 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!

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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