Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The exact question is:

$f:[0,1]\rightarrow\mathbb{R}$ is a continuous function such that $\int_0^1f(x)=\int_0^1xf(x)=0$. Prove that $\exists a,b \in [0,1], a<b$ such that $f(a)=f(b)=0$.

By using mean value theorem of integration, we get some $a,b$ s.t $f(a)=f(b)=0$, but how can we show that $a\neq b$.

share|improve this question
add comment

1 Answer

up vote 15 down vote accepted

We are given that $\displaystyle \int^1_0 (ax+b)f(x) dx =0 .$ Suppose $f$ is not identically zero for otherwise the result is trivial. The condition $\displaystyle \int^1_0 f(x) dx = 0 $ implies there is at least one sign-changing root, say at $ m$ (that is, the function has different signs after passing through $m$, or more formally, $f(m+\epsilon)f(m-\epsilon)< 0 $ for all sufficiently small $\epsilon>0.$) Suppose this is the only sign-changing root. Then $ (x-m)f(x) $ does not change signs and is not identically $0$ either, so $\int^1_0 (x-m) f(x) \neq 0 $, contradicting the first statement. Thus there are at least $2$ distinct sign-changing roots.

This idea can be employed in proving the generalized result: If $ f\in C( [0,1], \mathbb{R} ) $ is such that $ \displaystyle \int^1_0 x^k f(x) dx = 0 $ for all $ k= 0, 1,2,\cdots, n-1 $ then $f$ has at least $n$ distinct sign-changing roots in $[0,1].$

share|improve this answer
    
Pls clarify.when $x<m,(x-m)<0$ and viceversa for $x>m$.So we get same sign for $(x-m)f(x)$. How can we argue for $xf(x)$ or for $(x-m)^2f(x)$ –  bharath Feb 22 '12 at 18:02
    
Since $(x-m) f(x) $ has the same sign throughout $[0,1]$ and is not identically zero, we must have $ \left| \int^1_0 (x-m)f(x) dx \right| >0.$ But this contradicts $ \int^1_0 (x-m) f(x) dx = \int^1_0 x f(x) dx - m \int^1_0 f(x) dx = 0-m\cdot 0 = 0.$ –  Ragib Zaman Feb 23 '12 at 0:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.