Exercise 6.15

Applying integration by parts, applied to the functions $f^2$ and $1$, we get

which yields the first assertion. In the solution to Exercise 10 I showed that

with equality only if $F^2$ is a multiple of $G^2$. Letting $F(x)=f^{\prime }(x)$ and $G(x)=\mathit{xf}(x)$, then

Squaring both sides almost yields the second assertion, with $\ge$ instead of $>$. To get the strict inequality, note that the equality only holds if ${f{}^{\prime }}^2(x)$ is a multiple of $x^2{f}^2(x)$. Since ${\int }_a^b{f}^2(x)\mathit{dx}=1$, the continuous function $f(x)$ cannot have the constant value 0. Since $f(a)=f(b)=0$, there are points $c$, $d$ such that $a\le c<d\le b$, $f(c)=f(d)=0$ and $f(x)\ne 0$ for $c<x<d$. Then by Theorem 5.10 there is a point $x$ between $c$ and $d$ such that $f^{\prime }(x)=0$. If equality were to hold above, then we would have ${f{}^{\prime }}^2(x)=0=x^2{f}^2(x)\ne 0$, which is impossible.