I'm currently having some trouble solving the following question.
Q. Suppose $X,M, \mu $ is a measure space for which $\mu(A) > 0 \Longrightarrow \mu(A) \geq 1$
Prove that if $1 \leq p < q \leq \infty $, then $ L^p \subset L^q$, and $\|f\|_{\infty } \leq \|f\|_{q} \leq \|f\|_{p} \leq \|f\|_{1}$
I'm aware that if $\mu(X) < \infty$, then $L^q \subset L^p$, but since this question does not assume that the measure is finite, I could not use the aforementioned property.
Would appreciate any hints/solutions to the question!
I will prove the last inequality and leave the rest to you. Suppose $\|f\|_1 \leq 1$. Then $1 \geq \int_{|f| >1} |f|> \mu \{x:|f(x)| >1\}\geq 1$ if $\mu \{x:|f(x)| >1\}>0$ (which is a contradiction). Hence, $|f| \leq 1$ a.e.. Now $\int |f|^{p}d\mu \leq \int |f| d\mu$ because $|f|^{p-1} \leq 1$. Hence $\int |f|^{p}d\mu \leq1$ and $\|f\|_p \leq 1$. We have proved that $\|f\|_1 \leq 1$ implies that $\int |f|^{p}d\mu \leq1$. This implies (by scaling) that $(\int |f|^{p}d\mu)^{1/p} \leq \|f\|_1 $.
Note: by an argument similar to the one above $\|f\|_r \leq 1$ for some $r\geq 1$ implies that $|f| \leq 1$ a.e. This makes the first inequality very easy. Let me know if you need help with $\|f\|_q \leq \|f\|_q$. ,