The corollary is given below:
And the question is:
Show that if $f(x) = \ln{1/x}$ for $x \in (0,1],$ then $f$ belongs to $L^{p}(0,1]$ for all $1 \leq p < \infty$ but not to $L^{\infty}(0,1]$.
My questions are:
1-I was given a hint that it is enough to prove that the question is correct for $p=1,$ I was assuming that this hint is based on the corollary I mentioned above (am I correct?) but the corollary is including the case when $p_{2}$ is $\infty$ which is excluded from the problem given above, does not this contradict the problem?
2-Also, I am not convinced with that if $f \in L^{1}$ then it is in all $L^P$ with p > 1. because by set theory, if we have $A \subset B$ and we know that $x \in B,$ this does not imply that $x \in A.$ may be $x \in B \setminus A.$ Could anyone clarify this point for me please?
It says that, in a finite measure space, for all $p\in[1,\infty]$, if $f\in L^{p}(E)$ then $f\in L^{1}(E)$, not in the reverse direction as your second question concerns.
To show that $\log(1/x)$, $x\in(0,1]$ does not belong to $L^{\infty}(0,1]$, note that for each $M>0$, and $x<1/(e^{1+M})$, so that $x\in(0,1]$, then $1/x>e^{1+M}$ and hence $\log(1/x)>\log(e^{1+M})=1+M>M$, this shows that $\log(1/x)\rightarrow\infty$ as $x\rightarrow 0^{+}$, so that it is not essentially bounded.
To show that $\log(1/x)\in L^{1}(0,1]$, use the inequality that $\log u\leq u^{1/2}$ for large $u>0$, then $\log(1/x)\leq 1/\sqrt{x}$ for small $x\in(0,1]$, but we know that $\displaystyle\int_{0}^{1}1/\sqrt{x}dx=2<\infty$.