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.

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  • $\begingroup$ 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. $\endgroup$ – Ethan Mark Oct 20 at 6:49
  • $\begingroup$ Otherwise $f(x)\sqrt {1 - x^2}$ is always positive or always negative in $(-1,1)$ and therefore its integral can't be zero. $\endgroup$ – Robert Z Oct 20 at 6:52
  • $\begingroup$ Ah. In other words we are trying to observe that $f$ cannot be always positive or negative in $(-1, 1)$, am I correct? $\endgroup$ – Ethan Mark Oct 20 at 7:07
  • $\begingroup$ @EthanMark Yes, that's correct! $\endgroup$ – Robert Z Oct 20 at 7:07
  • $\begingroup$ @EthanMark Are you able now to show that the roots are at least 2? $\endgroup$ – 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 $h(-1)=h(1)=0$.

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

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