$f, g$ measurable, $\lVert g\rVert_1=1$ and $0 \le a \le f \le b$ so $\int_{\mathbb{R^n}} \frac{g}{f}dx \ge \frac{1}{\int_{\mathbb{R^n}} fg dx}$

Question

Let $g \ge 0$ be measurable in $\mathbb{R^n}$ s.t. $\lVert g\rVert_1=1$ and $f$ measurable s.t. $a \le f \le b$, for $a, b \gt0$, show that: $$\int_{\mathbb{R^n}} \frac{g}{f}dx \ge \frac{1}{\int_{\mathbb{R^n}} fg dx}$$

My idea was:

$$1 = \lVert g\rVert_1 =\int_{\mathbb{R^n}} |g|dx = \int_{\mathbb{R^n}} \sqrt{fg}\sqrt\frac{g}{f}dx \le \biggl( \int_{\mathbb{R^n}} fg dx \biggr)^{\frac{1}{2}}\biggl(\int_{\mathbb{R^n}} \frac{g}{f}dx\biggr)^{\frac{1}{2}} $$

but, to use Hölder's inequality I have to be sure that $f \in L^1(\mathbb{R^n})$, how to see that (if my reasoning was right)?

Answer 1

You cannot be sure that $f \in L^1(\mathbb{R}^n)$ because it is not True. For example, if $f = a$, you don't have $f \in L^1(\mathbb{R}^n)$.

However, you can "make" $f$ be in $L^1(\mathbb{R}^n)$ by considering $f\mathbb{1}_{||x||<n}$ for a natural number n. Appling the reasoning you used, you get $1 \leq (\int_{||x|| < n} fg)^{\frac{1}{2}} (\int_{||x|| < n} \frac{g}{f})^{\frac{1}{2}}$.

You can conclude by letting $n \to +\infty$, as $\frac{g}{a}$ and $gb$ are dominating functions for the dominated convergence theorem.