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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 $a\neq b$.

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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 $[0,1].$

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  • $\begingroup$ 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 $(x-m)^2f(x)$ $\endgroup$ – bharath Feb 22 '12 at 18:02
  • $\begingroup$ 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.$ $\endgroup$ – Ragib Zaman Feb 23 '12 at 0:09

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