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Rudin 6.15 asks one to show that, for $f$ a real, continuously differentiable function on $[a,b]$, $f(a)=f(b)=0$, and $\int_a^b f^2(x)dx=1$, $\int_a^b xf(x)f'(x)dx=-\frac{1}{2}$. This is a simple application of integration by parts.

But I am having trouble with the next part, namely to prove that

$\int_a^b [f'(x)]^2dx \cdot \int_a^b x^2f^2(x)fx > \frac{1}{4}$.

I can see how Hölder's Inequality gets us $\int_a^b [f'(x)]^2dx \cdot \int_a^b x^2f^2(x)fx ≥ \frac{1}{4}$. But whence the strict inequality?

Wolfram MathWorld gives a condition for equality, namely that $|f'(x)|=c|xf(x)|$. Would it be valid to say that the only function with this property is $e^\frac{x^2}{2}$, which never equals zero (whereas it is required that $f(a)=f(b)=0$)?

Any suggestions would be appreciated. Thanks!

mww
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  • There are a few such functions, namely $Ae^{cx^2/2}$, not just one. But I think your argument is valid nonetheless. – Greg Martin May 18 '13 at 02:33
  • Show that the only function which satisfies the equality condition and $f(a) = f(b) = 0$ is the zero function, and hence doesn't satisfy the initial conditions. Hence we must have strict inequality. – Calvin Lin May 18 '13 at 02:43

1 Answers1

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One slightly different approach is to note that the condition

$$|f'(x)| = cx |f(x)|$$

is exactly a setup for which we can use the differential form of Gronwall's inequality without actually solving any differential equation. In particular, we have that

$$|f(x)| \le |f(0)| \exp\left\{\int_0^x ct \, dt\right\} = 0$$

where all we needed was finiteness of the integral. One advantage of this method is that it doesn't require knowing anything about the signs of $f'$ or $f$.