Let $f \in L^p(\mathbb{R})$ be such that $f(x)>0$ almost everywhere on $\mathbb{R}$. Suppose further that $\inf f(x) = 0$. Does it follow that $\int_{\mathbb{R}} h(x) dx = \infty$, where $h(x):=\frac{1}{f(x)}$?
For some context: This problem arrises when I am trying to show that the multiplication operator $M_f: L^p(\mathbb{R}) \to L^p(\mathbb{R})$ is Fredholm, where $p< \infty$. I would like to prove that if $M_f$ is Fredholm then $f$ must be bounded away from zero. The case when $p= \infty$ was relatively easy, but somehow this case seems a lot harder.
Thanks!
This argument seems a bit too elementary but I think it should be fine. Consider the closed interval $[n,n+1]$ for some $n \in \mathbb{N}$. Then, by Jensen's inequality and the fact that $\frac1x$ is convex, we have \begin{align} \frac{1}{\int_{n}^{n+1}f(x) \, \mathrm{d}x} \leq \int_n^{n+1}\frac{1}{f(x)}\, \mathrm{d}x \, . \end{align} The left hand side should converge to $+\infty$ since again by Jensen's we have \begin{align} \frac{1}{\int_{n}^{n+1}f(x) \, \mathrm{d}x} \geq \frac{1}{\left(\int_{n}^{n+1}(f(x))^p \, \mathrm{d}x\right)^{1/p}} \, , \end{align} and since $f \in L^p$.
Regarding your condition that $\inf f=0$, I think the more appropriate condition should be $\mathrm{ess}\inf f=0$ and even this condition is redundant as it is satisfied for any nonnegative $f\in L^p$. Assume it is not and you should be able to derive a contradiction.