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

Question

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!

Robert Z · Accepted Answer · 2020-10-20T15:57:59.243

5

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.

edited Oct 20 '20 at 15:57

answered Oct 20 '20 at 06:13

Robert Z

145,942

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 '20 at 06:49
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 '20 at 06:52
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 '20 at 07:07
@EthanMark Yes, that's correct! – Robert Z Oct 20 '20 at 07:07
@EthanMark Are you able now to show that the roots are at least 2? – Robert Z Oct 20 '20 at 08:19
May I know how you were able to “cook up” $g$? Is there some intuition behind it? – Ethan Mark Oct 20 '20 at 09:03
I just tried to exploit the same idea we used to show the existence of at least one root. – Robert Z Oct 20 '20 at 10:27
@EthanMark I included more details. I hope it is clear now! – Robert Z Oct 20 '20 at 11:12
Your logic is great, but why do you say $f=0$ a contradiction, it makes a possible case, i think you should edit there. Otherwise it looks very good(+1). – A learner Oct 20 '20 at 13:24
@Alearner Do you mean here: " implies =0 everywhere on [,] , we have a contradiction."? I edited my answer. Is it better now? – Robert Z Oct 20 '20 at 16:06

score 0 · Answer 2 · answered Oct 20 '20 at 05:43

0

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 Oct 20 '20 at 05:43

James Garrett

2,558

That would show that $h$ has two zeros, but how do I infer the same of $f$? – Ethan Mark Oct 20 '20 at 06:50
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 '20 at 06:55
@Ethan Mark use Rolle's theorem. – A learner Oct 20 '20 at 15:12
2

@Alearner Rolle's Theorem gives one root the question is asking for 2. – N. S. Oct 20 '20 at 15:58
@N. S. Yeah! You are right. This answer doesn't fulfill the requirement. – A learner Oct 21 '20 at 07:34

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

2 Answers2