Reverse Holder Inequality $\|fg\|_1\geq\| f\|_{\frac{1}{p}}\|g\|_{-\frac{1}{p-1}}$

Question

Let $p\in(1,\infty)$ and $(X,\mathcal{F},\mu)$ a measure space such that $\mu(X)\not=0$. Let $f,g:X\to\mathbb{R}$ be such that $g\not=0$ a.e., $\|fg\|_1<\infty$ and $\|g\|_{-\frac{1}{p-1}}<\infty$. Then prove:$$\|fg\|_1\geq\| f\|_{\frac{1}{p}}\|g\|_{-\frac{1}{p-1}}$$

I tried searching for "reverse holder inequality" and I found I thousand things that do not look similar to this.

I just wrote the definitions of the norms and couldn't do nothing because the only thing I thought that I could use was Holder inequality but that is in the reverse direction of the inequality above (laugh). I would aprecciate some hint to how I should proceed.

Answer 1

$$\int |f|^{1/p} =\int |fg|^{1/p} |g|^{-1/p}.$$ Apply Holder's inequality with indices $p$ and $q$ where $q =\frac p {p-1}$. You get $$\int |f|^{1/p} \leq (\int |fg|)^{1/p} (\int |g|^{-q/p})^{1/q}.$$ Substitute the value of $q$ in terms of $p$ and divide both sides by $(\int |g|^{-q/p})^{1/q}$. If you now raise both side to power $p$ you will get exactly the inequality you want.