# Cauchy-Schwarz-type inequality

Let $f\in C^1[0,\infty)$ such that $f(x)\to 0$ as $x\to\infty$. Prove that $$\int_0^\infty f(x)^2\, dx\le \Biggl(\int_0^\infty x^2f(x)^2\, dx\Biggr)^{1/2}\Biggl(\int_0^\infty f'(x)^2\, dx\Biggr)^{1/2}\,.$$ Hint: write $f(x)^2=-\int_x^\infty \Bigl(f(t)^2\Bigr)'\, dt$.

So I tried doing this without the hint, as follows:

For ease, all integrals are taken from $0$ to $\infty$. Consider $xf(x)f'(x)$ and integrate by parts with $u=xf(x)$ and $dv=f'(x)dx$. This gives $$\int xf(x)f'(x)=\underbrace{xf(x)^2\Bigr|_0^\infty}_A-\int f(x)^2-\int xf(x)f'(x)\,.$$ But the term $A$ does not necessarily go to zero, which is a total buzzkill.

I've putzed around with the hint a bit, but it's been fruitless. Any help is appreciated, thanks

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The question is a bit weird. I don't think $f\in C^1$ and $f(x)\rightarrow 0$ is sufficient to guarantee $f^2$ is integrable... –  Potato Aug 24 '12 at 2:36
@Potato: since $f$ must be continuous at any finite positive point, you will get $\infty\leq\infty$ if $f^2$ is not integrable. –  Alex R. Aug 24 '12 at 2:40
@Sam That's true. –  Potato Aug 24 '12 at 2:44

Cauchy-Schwarz tells us that the right side of the inequality is greater than

$$\int_0^\infty xf(x)f'(x)\ dx.$$

Using the hint, note

$$\int_0^\infty f(x)^2 \ dx= \int_0^\infty -\int_x^\infty \left(f(t)^2\right)'\, dt\ dx= \int_0^\infty \int_x^\infty -2f(t)f'(t)\, dt\ dx.$$

We can use Fubini's theorem to interchange the order of integration. It is helpful to draw out the region of integration on paper when doing this. You get the sector bounded by the line $x=t$ and the $t$ axis.

$$\int_0^\infty \int_x^\infty -2f(t)f'(t)\, dt\ dx = -\int_0^\infty \int_0^t 2f(t)f'(t)\ dx \ dt= -2\int tf(t)f'(t) \ dt.$$

At this point we are off by a sign and a constant, but hopefully the suggestion to switch the integrals was helpful. I will edit this if I find my mistake.

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It's generally a good idea to take the hints you are given. Let \eqalign{J &= \int_0^\infty f(x)^2\ dx = - \int_0^\infty \int_x^\infty (f(t)^2)'\ dt\ dx \cr &= 2 \int_0^\infty \int_x^\infty (t^{1/2} f(t)) (-t^{-1/2} f'(t)) \ dt \ dx\cr} Now use Cauchy-Schwarz: $$J \le 2 \left( \int_0^\infty \int_x^\infty t f(t)^2\ dt \ dx\right)^{1/2} \left( \int_0^\infty \int_x^\infty t^{-1} f'(t)^2\ dt\ dx \right)^{1/2}$$ Interchanging the order of integration $$\int_0^\infty \int_x^\infty t f(t)^2\ dt \ dx = \int_0^\infty \int_0^t \ dx\ t f(t)^2 \ dt = \int_0^\infty t^2 f(t)^2\ dt$$ $$\int_0^\infty \int_x^\infty t^{-1} f'(t)^2\ dt\ dx = \int_0^\infty \int_0^t \ dx\ t^{-1} f'(t)^2 \ dt = \int_0^\infty f'(t)^2\ dt$$ We're just off by that pesky factor of $2$. But in fact Cauchy-Schwarz is an equality when the two factors are equal. If $t ^{1/2} f(t) = - t^{-1/2} f'(t)$, i.e $f'(t) = - t f(t)$, then $f(t) = c e^{-t^2/2}$. In that case $J = \int_0^\infty c^2 e^{-t^2}\ dt = c^2 \sqrt{\pi}/2$ while $\int_0^\infty t^2 f(t)^2\ dt = c^2 \sqrt{\pi}/4$ and indeed the inequality (with the factor of $2$) is an equality. We conclude that the correct inequality should in fact be

$$\int_0^\infty f(x)^2\ dx \le 2 \left(\int_0^\infty t^2 f(t)^2\ dt\right)^{1/2} \left( \int_0^\infty f'(t)^2\ dt \right)^{1/2}$$

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