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Existence of roots

Let f:[0, 1] \rightarrow \mathbb{R} be a continuous function such that

(1)   \begin{equation*} \int_0^1 f(x) \, \mathrm{d}x = \int_0^1 x f(x) \, \mathrm{d}x =0 \end{equation*}

Prove that there exist a, b \in [0, 1] such that a<b and f(a)=f(b)=0.

Solution

The existence of a, b follows from Mean Value Theorem of Integrals. To prove that a \neq b we proceed as follows:

We are given that  \int^1_0 (ax+b)f(x) \, \mathrm{d}x =0. Suppose f is not identically zero for otherwise the result is trivial. The condition \int^1_0 f(x) \, \mathrm{d}x = 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)\, \mathrm{d}x \neq 0, contradicting the first statement. Thus there are at least 2 distinct sign-changing roots.

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