# An exercise on an integral inequality

I need help with this exercise:

let $f,g,h : [0,1] \to [0,\infty]$ integrable functions. Prove that the following statements are equivalent:

i) $(f(x))^2 \leq g(x)h(x)$ almost everywhere.

ii) For every measurable set $E \subset [0,1]$ we have:

$$\left(\displaystyle\int_E f(x)dx\right)^2 \leq \displaystyle\int_E g(x)dx\displaystyle\int_E h(x)dx.$$

I proved that $i) \Rightarrow ii)$ with Holder inequality, but i have some problems with the other implication.

Consider the polynomial

$$P_E(t):=\left(\int_E h(x)dx\right)t^2+2\left(\int_E f(x)dx\right)t +\left(\int_E g(x)dx\right)$$

For each $E$, we have that the leading coefficient is non-negative with non-positive discriminant. Moreover, when $\int_E h(x)dx=0$ we have $\int_E f(x)dx=0$, so even in that case $P_E(t)=\int_Eg(x)dx\geq0$.

Then $$\int_E\left(t^2h(x)+2tf(x)+g(x)\right)dx\geq0$$ for all $E$. It follows that $$q_x(t):=t^2h(x)+2tf(x)+g(x)\geq0\ \ \ \ a.e.\ (\text{in }x)$$

Delicate point: So, for each $t\in\mathbb{Q}$ there is a set of measure zero $A_t$. This means that for each $x\in A:=[0,1]\setminus\bigcup_{t\in\mathbb{Q}}A_t$ and all $t\in\mathbb{Q}$ we have that $q(t)\geq0$. Notice that the measure of $[0,1]\setminus A=\bigcup_{t\in\mathbb{Q}}A_t$ is zero because it is a countable union of sets of measure zero..

Therefore, for each $x\in A$, the discriminant is non-positive, i.e. $$(f(x))^2\leq h(x)g(x)\;\text{ for }\;x\in A.$$

• Thank you, really nice proof! How could I even come up to apply polynomial theory to integrals? (You are a wizard :) ) – mrprottolo Feb 4 '15 at 21:00
• No, it is only experience. Let me tell you the thought process: The inequalitites in the problem look very similar to Cauchy-Schwarz inequality, Cauchy-Schwarz inequality looks very similar to the discriminant and actually there is a proof of Cauchy-Schwarz that is along these lines. Problem solving is always about imitation. Then we try building a polynomial with that discriminant. There are many ways to do this but keeping integrals multiplying is not convenient. One would like to use the linearity of the integrals. – Pp.. Feb 4 '15 at 21:06
• So, we put one integral in each coefficient and everything works out. – Pp.. Feb 4 '15 at 21:07
• A failed approach: Before thinking about polynomials I was trying something like using steps functions to get an approximated identity, i.e. to get a Dirac delta which gives $f(x)$ from some integral of $f$. But the given inequality is not linear, it has integrals multiplied together. So, it doesn't work. – Pp.. Feb 4 '15 at 21:10
• Thank you so much :) I will think about this. – mrprottolo Feb 4 '15 at 21:16