Suppose that $f \in L^{p}(\mu)$ for some $p > 0$ and $\mu(X) = 1$. Show that when $\log|f| \in L^1$, then $$\lim_{q\to 0_{+}}\|f\|_q = \exp\Big(\int_X \log |f| \, d\mu\Big).$$
I intend to use the L'Hospital in the following manner: $$\log \lim_{q\to 0_{+}}\|f\|_q = \lim_{q\to 0_{+}} \log\|f\|_q = \lim_{q\to 0_{+}} \frac{1}{q}\int |f|^q \, d \mu.$$ The next step is to use L'Hospital's law taking differentiatin inside the integration. The usual condition to gaurantee that is to bound the derivative of $\|f\|^q$ which is $q f'|f|^{q-1}$. I am having a hard time finding a dominating function for it.
As the comment pointed out, $f$ is not even necessarily differentiable. Then what is the right technique to tackle this problem?
Update: Okay, the right way to do this might be to first do it for a smooth function with compact support, which will settle all the problems.
We define $h(q):=|f(x)|^q = \exp(q \log|f(x)|)$ and find that $$h'(q) = |f(x)|^q \log|f(x)| $$ Using that $\log(x) \le c_\epsilon x^\epsilon$ for $x \ge 1$ we get for any $0 < p < q- \epsilon$ that $$|h'(p)| \leq c_\epsilon |f(x)|^q 1_{\{|f| \ge 1\}}(x) + |\log|f(x)|| 1_{|f| < 1}(x).$$ Thus $h'(p)$ is uniformly bounded in $0 < q < p- \epsilon$ by an integrable majorant. Because of $$\Big|\frac{h(q)-h(0)}{q} \Big| \leq \max_{0< q < p-\epsilon} |h'(q)|$$ we can apply the dominated convergence theorem in order to get that $$G(q):= \int |f(x)|^q \, \mathrm{d} \mu(x)$$ is right-differentiable in $p=0$ with $$G'(0) = \int \log|f| \, \mathrm{d} \mu(x).$$ Now let $H(q) := \frac{1}{q} \log G(q)$. Because $\log G(0) = 0$ we find that $$\lim_{q \downarrow 0} H(q) = \left. \frac{d}{dx} \log G(q) \right|_{q=0} = \frac{G'(0)}{G(0)} = G'(0).$$ This proves that $$\lim_{q \downarrow 0} \|f\|_q = \exp \Big \{ \int \log|f| \, \mathrm{d} \mu(x) \Big\}.$$