If $\varphi, g$ are positive and non-zero in $L^p(\mathbb R^n)$ and $L^q(\mathbb R^n)$ respectively, then so is their product in $L^1(\mathbb R^n)$

Question

This appears somewhere in a proof of my functional analysis course and I do not see how can I justify it.

Let $\varphi \in L^p(\mathbb R^n)$ for all $p \in [1, + \infty]$, $\varphi > 0$ and $\varphi \neq 0$ in all the $L^p$'s. Let $g \in L^q(\mathbb R^n)$ for some $q \in [1, +\infty]$ also positive and non-zero in $L^q$. And from that, my teacher deduced that $$\int_\mathbb{R^n} \varphi(x) g(x) > 0.$$ Clearly, $\varphi(x) g(x) \ge 0$, but how can I bound this integral from below ? All obvious inequalities (Young, Holder, ..) are in the wrong direction.

Answer 1

Let $A$ be the set where $g$ is stricly positive. Since $\varphi$ as you said is strictly positive almost everywhere, there is a subset $B$ of $A$ with $\lambda(B) >0$ such that $g \cdot \varphi$ is strictly positive. Finally,

$$\int g \cdot\varphi d \lambda \geq \int_B g \cdot\varphi d \lambda >0.$$

Where you can check the last inequality in this link.

Answer 2

This is impossible. Counterexample $\phi = 1_{\lbrack 0,1 \rbrack}$. $\int \phi(x)^p dx = \int 1_{\lbrack 0,1 \rbrack}(x)^p dx = 1$ for all $p$. Picking $g =1_{\lbrack 2,3 \rbrack} $. Thn the integral you seek is 0.