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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).

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1 Answers1

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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 $f=0$.

Aphelli
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    Rudin does not go into differential equations. Is there some simpler argument that is tailored for this problem. For example, something using the mean value theorem maybe. –  Sep 20 '19 at 14:46
  • Then you can just mimic the proof of the uniqueness for solutions of first-order linear ODE: let $M_1$ be the sup norm of $f$ over $[a,b]$, $M_2$ be the max of $|a|$ and $|b|$, and show by induction over $n$ that $|f(t)| \leq n!^{-1}M_1M_2(t-a)^n$... – Aphelli Sep 20 '19 at 15:29
  • Is there some easy argument using some basic theorems? I don't think Rudin expects us to do this. –  Oct 05 '19 at 22:25