0

I have this problem:

Let $f(x)$ be continuous on the interval $[0,1] .$ Prove that if $$ \int_{0}^{1} f(x) d x=\int_{0}^{1} x f(x) d x=0 $$ then $f$ has at least two different roots in (0,1)

Is it better to start with parts integration on second one? or triying $$ \int_{0}^{1} f(x) d x=\int_{0}^{1} x f(x) d x=0\Rightarrow \int_{0}^{1} f(x) -x f(x) d x=0 ? $$

If I use mean value Theorem I coul get $$ \int_{0}^{1} (1-x) f(x) d x=(1-\xi)\int_{0}^{1} f(x) d x=0 $$ With $\xi\in[0,1]$

Therefore $$ \int_{0}^{1} f(x) d x=0\qquad or\qquad \xi =1 $$

But doesn't seems to help, I already know $$ \int_{0}^{1} f(x) d x=0 $$ and $\xi=1$ has no further use

So I'm kind of stuck right now

I'll appreciate any help or suggestion

Thanks in advance

Erestor
  • 31
  • 1
    See https://math.stackexchange.com/questions/112017/prove-that-exists-ab-s-t-fa-fb-0-when-int-01fxdx-int-01xfxdx and https://math.stackexchange.com/questions/360503/continuous-function-on-0-1-has-at-least-two-zeroes – Robert Z Dec 30 '20 at 15:47

2 Answers2

2

With $\int_0^1 f = 0$ we get that there is at least one root $x_0$. In fact, $f$ can't be always positive or always negative or the integral wouldn't be $0$ (i exclude the trivial case where $f=0$).

Let's suppose it is the only root. Then since $\int_0^1 f = 0$ and $\int_0^1 xf(x) dx = 0$, we have that $\int_0^1 \pm(x-x_0) f(x) dx = 0$. Let's say for instance that $f$ is positive before $x_0$, and negative after. Then $x\mapsto (x-x_0) f(x)$ is always positive. Since its integral is $0$, this means the function is $0$, which is contradictory.

So there are at least two roots.

math
  • 2,313
0

for a (non-zero) function over this range there must be two areas that "cancel out" i.e. there is a $0<x_0<1$ where $f(x_0)=0$ which suggests: $$0=\int_0^1f(x)dx=\int_0^{x_0}f(x)dx-\int_{x_0}^1f(x)dx$$ If there is one root in a domain then there must be a positive and negative area, and if there are two roots there must be three regions, 2 positive/negative and 1 the opposite.

On the other hand if we define a function $g(x)=xf(x)$ we know that $g(0)=0$ which means for the regions to cancel out again we must have 3 roots in the function $g$, one being $x=0$ which suggests that $f$ has 2 roots.

I am not sure how much sense this makes so I will try and define it slightly differently, lets say we have a function: $$\begin{matrix}f_n(x)=\prod_{i=1}^n(x-x_i) & 0\le x_i\le1\end{matrix}$$ this represents a polynomial with $n$ roots in the domain $x\in[0,1]$

If we assume that $f=f_1$ we have the expression: $$\int f(x)dx=\int_0^1x-x_idx=\left[\frac{x^2}{2}-x_ix\right]_0^1=\frac{1}{2}-x_i=0$$ so there is an $x_i$ which satisfies this solution. If we now try and do the second integral: $$\int xf(x)dx=\int_0^1x^2-x_ixdx=\left[\frac{x^3}{3}-\frac{x_i}{2}x^2\right]_0^1=\frac{1}{3}-\frac{1}{2}x_i=0$$ and whilst this does have a solution that makes this work it does not match the one from the first integral, suggesting that if this is a polynomial it must have at least two roots.

Henry Lee
  • 12,215