# showing that $(\int f )(\int g) \geq 1$

Let $\mu(X) =1$.
Let $f,g \in L^1(X)$ be two positive functions satisfying $f(x) g(x)>1$ for almost all $x$, Then $$\left(\int f ~dx\right) \left(\int g~dx\right) \geq 1.$$

Show also that if $f,g\in L^2(X)$ with $\int f ~dx= 0$, then $$\left(\int fg~dx\right)^2 \leq \left[ \int g^2 ~dx - \left(\int g~dx\right)^2 \right] \int f^2~dx.$$

I think I have to use Holder's inequality for both questions:

For the first question, since $\mu(X) =1$, $1\lt \int fg~dx$. How do I apply Holder's inequality.

For the first, try applying the inequality to $\int \sqrt{f g}$, and obtain a lower bound for $\int \sqrt{f g}$.

For the second, let $\overline{g} = \int g$, and apply the inequality to $\int f (g-\overline{g})$.

• what do you mean by "apply the inequality?" – Daniel May 8 '12 at 2:37
• Sorry, I forgot the square root. Apply the inequality to $\int \sqrt{fg} = \int \sqrt{f} \sqrt{g}$. – copper.hat May 8 '12 at 2:45
• To apply Holder's inequality don't I need $f\in L^p$ and $g\in L^q$ where $p$ and $q$ are conjugates? The problem I have is that both $f,g\in L^1$. – Daniel May 8 '12 at 2:48
• Since $f \in L^1$, it follows that $\sqrt{f} \in L^2$. Similarly for $g$. The conjugate of 2 is 2. – copper.hat May 8 '12 at 3:04
• Is there a general theory for that, because I'm not aware of it. Using that I get $$1\le \left[\int f \int g\right]^{1/2}.$$ right? – Daniel May 8 '12 at 3:13

Hint:

For the first inequality use Hölder for $\sqrt{gf}$.

• what will be the conjugate exponents? – Daniel May 8 '12 at 2:56
• @Daniel: 2 and 2. In other words, the Cauchy-Schwarz inequality. – Nate Eldredge May 8 '12 at 3:29