Problem 6.1.8 - Suppose $\mu(X) = 1$ and $f\in L^p$ for some $p > 0$, so that $f\in L^q$ for $0 < q < p$.
a.) $\log\|f\|_{q}\geq \int \log |f|$
b.) $\left(\int |f|^{q} - 1\right)/q \geq \log \|f\|_{q}$, and $\left( \int |f|^q - 1\right)/q \rightarrow \int \log |f|$ as $q\rightarrow 0$
c.) $\lim\limits_{q\rightarrow 0}\|f\|_{q} = \exp\left(\int \log |f|\right)$
Attempted proof a.) We know that, $$\|f\|_{q} = \left[\int|f|^q d\mu\right]^{1/q} \geq \int |f| d\mu$$ where the last inequality is from Holder's inequality. Now taking the $\log$ on both sides we have $$\log\|f\|_{q} = \frac{1}{q}\log\int |f|^q d\mu \geq \frac{1}{q}\int \log |f|^q d\mu = \int \log |f|d\mu$$ where the last inequality follows from Jensen's inequality.
Attempted proof b.) We know that $\log x \leq x - 1$ for every $x > 0$, also we have that $$\log\|f\|_{q} = \frac{1}{q}\log\left(\int |f|^q d\mu\right) \leq \frac{1}{q}\left(\int |f|^q - 1 d\mu\right)$$ Now suppose $q\rightarrow 0$ then for $\frac{|f|^q - 1}{q}\rightarrow \log|f|$ so by the monotone convergence theorem the result follows.
Attempted proof c.) If we exponentiate the result from part a.) we have $$\liminf_{q\rightarrow 0}\|f\|_{q} \geq \exp(\int \log |f|)$$ Then exponentiating the result from part b.) we get $$\limsup_{q\rightarrow 0}\|f\|_{q} \leq \exp(\int \log |f|)$$ thus we get the desired result $$\lim\limits_{q\rightarrow 0}\|f\|_{q} = \exp(\int \log |f|)$$
Not sure if this is right will edit more as I progress. Any suggestions or comments is greatly appreciated.
Part $(a)$ follows immediately from Jensen's inequality.
For $(b)$ recall that $\log x \le x - 1$ for every $x > 0$, then $$\log\|f\|_q = \frac 1q\log\Big(\int_X|f|^q\Big)\le \frac 1q\Big(\int_X|f|^q - 1\Big).$$ Moreover it is known that $\frac{|f|^q - 1}{q} \to \log|f|$ as $q \to 0$, so that by the MCT the convergence in $(b)$ is proved.
For $(c)$: exponentiating the result in $(a)$ gives $\liminf_{q\to 0}\|f\|_q \ge e^{\int\log|f|}$ while exponentiating the one in $(b)$ we obtain $\limsup_{q \to 0}\|f\|_q\le e^{\int\log|f|}$.