# If $1 \leq |f| |g|$ $\Rightarrow$ $1 \le \Vert g\Vert_{1} \Vert f \Vert_1$

Let $(X,A,\mu)$ be a measure space such that $\mu(X) = 1$ and let $f,g : X → [0,\infty]$ be measurable functions such that $fg ≥ 1$. Prove that

$$1 \le \Vert g\Vert_{1} \Vert f\Vert_1$$

Where $\Vert f\Vert_1 = \int_X|f| d\mu.$

Using Hölder's inequality we get that

$$1 \leq \Vert g \Vert_1 \Vert f \Vert_\infty, \quad\mbox{and} \quad 1 \leq \Vert g \Vert_\infty \Vert f \Vert_1,$$ but I don't know if this implies something good.

Does anyone know how to solve this question?

• Did you try to use the definition alone, without Holder's inequality?
– Guy
Nov 11, 2017 at 20:56
• Do you mean $1 \leq fg \Rightarrow1 \leq \int_X f g d\mu$? I'm missing something obvious here? Nov 11, 2017 at 20:58
• @Guy Does it implies that $\int_X f g d \mu \leq \int g d\mu \int f d\mu$ for some reason? Nov 11, 2017 at 21:01
• Essentially a duplicate of math.stackexchange.com/questions/157439/…. Nov 11, 2017 at 21:12

We have $\sqrt{|f||g|}\ge1$, then by Holder:
$$1\le\int_X \sqrt{|f||g|}\le\bigg(\int_X |f|\bigg)^{1/2}\bigg(\int_X |g|\bigg)^{1/2}.$$
Then by squaring we get: $1\le\int_X|f|\int_X|g|.$
Since $f$ and $g$ map into $[0,\infty]$ with $fg\geq 1$, it follows that $f,g$ are positive everywhere, and so $||f||_1,||g||_1>0$. The inequality is therefore trivially satisfied when $||f||_1=\infty$, so assume that $f$ is integrable.
Rewriting $fg\geq 1$ as $g\geq \frac{1}{f}$ and using the convexity of $t\mapsto \frac{1}{t}$ on $(0,\infty)$, we obtain $$||g||_1=\int_Xg\geq \int_X\frac{1}{f}\geq \frac{1}{\int_Xf}=\frac{1}{||f||_1}$$ by Jensen's inequality.