1

How can i demonstrate this inequality?

$(\int{f(x)g(x)h(x)dx})^2\le \int{f(x)g^2(x)dx}\int{f(x)h^2(x)dx}$

with,
$g, h$ arbitrary scalar functions and $0 \le f(x)$

I tried to use cauchy inequality to 2 functions:

$(\int{f(x)g(x)h(x)dx})^2 \le \int{f^2(x)g^2(x)dx}\int{h^2(x)dx}$
or
$(\int{f(x)g(x)h(x)dx})^2 \le \int{g^2(x)dx}\int{f^2(x)h^2(x)dx}$, but im stuck

user123
  • 11
  • 1
    Welcome to [math.se]. Can you please [edit] your post and write your attempts at solving the problem? If your question is clear and focused on your specific difficulty and you show your effort in solving the problem, it's more likely to get good and helping answers. By the way, take the opportunity to take the [Tour], if you haven't done it already. See also some tips on [ask], on formatting help and on writing down equations using LaTeX / MathJax. – Ertxiem - reinstate Monica Nov 14 '19 at 03:11

1 Answers1

1

As far as I'm not missing anything, splitting $f$ into two square roots should work:

Let $J\subseteq \mathbb R$ measurable. Assuming $f,g,h\colon J\to \mathbb R$ are (Lebesgue-)integrable and $0\leq f(x)$, we have that $\sqrt f g$ and $\sqrt f h$ are square-integrable, hence we can apply Cauchy-Schwarz as follows: \begin{align*} {\left( \int_J f(x)g(x)h(x)\mathrm dx \right)}^2 &= {\left( \int_J \sqrt{f(x)}g(x)\sqrt{f(x)}h(x)\mathrm dx \right)}^2 \\ &\leq \int_J {\left( \sqrt{f(x)}g(x)\right)}^2 \mathrm dx \int_J {\left( \sqrt{f(x)}h(x)\right)}^2 \mathrm dx\\ &= \int_J f(x)g(x)^2 \mathrm dx\int_J f(x)h(x)^2 \mathrm dx\\ \end{align*}

Note that in general, the square root is not integrable, but we only need the square integrability, since $(a,b)\mapsto \int_J a(x)b(x)\mathrm dx$ is a scalar product on $L^2(J,\mathbb R)$.

Lukas Juhrich
  • 3,365