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

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 $$a\in (-1,1)$$ (otherwise the product $$f(x)\sqrt {1 - x^2}$$ has the same sign over $$(-1,1)$$ and, recalling that if $$F\geq 0$$ 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 $$g(x)=(x\sqrt{1 - a^2}-a\sqrt {1 - x^2})$$ 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.

• May I know how you to the conclusion that $f$ has at least one zero in $(-1, 1)$? I am not sure I follow your logic for the first sentence, so I am unable to move on. – Ethan Mark Oct 20 at 6:49
• Otherwise $f(x)\sqrt {1 - x^2}$ is always positive or always negative in $(-1,1)$ and therefore its integral can't be zero. – Robert Z Oct 20 at 6:52
• Ah. In other words we are trying to observe that $f$ cannot be always positive or negative in $(-1, 1)$, am I correct? – Ethan Mark Oct 20 at 7:07
• @EthanMark Yes, that's correct! – Robert Z Oct 20 at 7:07
• @EthanMark Are you able now to show that the roots are at least 2? – Robert Z Oct 20 at 8:19

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 $$h(-1)=h(1)=0$$.

• That would show that $h$ has two zeros, but how do I infer the same of $f$? – Ethan Mark Oct 20 at 6:50
• Then $h$ is zero at $-1$ and $1$. How is this related with the two roots of $f$ in $(-1,1)$? – Robert Z Oct 20 at 6:55
• @Alearner Sorry, but how did you reach that conclusion? I am still not following... – Ethan Mark Oct 20 at 14:39
• @Ethan Mark use Rolle's theorem. – A learner Oct 20 at 15:12
• @Alearner Rolle's Theorem gives one root the question is asking for 2. – N. S. Oct 20 at 15:58