Baby Rudin Chapter 6, Problem 15 : Strict inequality

Question

Problem 15 in Chapter 6 of Principles of Mathematical Analysis by Walter Rudin:

Suppose $f$ is real, continuously differentiable on $[a,b]$, $f(a)=f(b)=0$, and $\int_a^b f^2(x)dx = 1$. Prove that

• $\int_a^b xf(x)f'(x)dx = -1/2$
• $\int_a^b [f'(x)]^2dx \cdot \int_a^b x^2f^2(x)dx > 1/4$

I am able to prove the first part using integration by parts and the second using Cauchy-Schwarz.

To get a strict inequality, I assumed that there was equality, in which case $xf(x) = \lambda f'(x)$ for all $x \in [a,b]$ for some constant $\lambda$. I am unable to get a contradiction with this (note that Rudin has not introduced the exponential and logarithmic functions and so I don't want to integrate both sides directly).

Answer 1

Clearly $\lambda \neq 0$. Note that $f$ is solution to the (linear) Cauchy problem $f’(x)=\lambda^{-1}xf(x)$ with the initial condition $f(a)=0$.

So by uniqueness of the solutions of linear first-order ODEs $f=0$.