Evaluation of $\lim_{q \to 0} (\int_0^1 |f|^q)^{1/q} $ for $f \in L^1([0,1])$.

Question

I'm given that $f \in L^1$ and $f(x) \neq 0$ a.e. and trying to prove that for q > 0

$$\lim_{q \to 0} (\int_0^1 |f|^q)^{1/q} = e^{\int_0^1 \ln|f|}$$

My work:

I took the log of both sides to reduce the problem to

$$\lim_{q \to 0} \frac{1}{q} \ln(\int_0^1 |f|^q) = \int_0^1 \ln|f|$$

Since $\ln$ is a concave function I used Jensen's inequality to show that

$$\frac{1}{q} \ln(\int_0^1 |f|^q) \geqslant \int_0^1 \frac{1}{q}\ln|f|^q = \int_0^1 \ln|f| $$

So if the limit exists, I know

$$\lim_{q \to 0} \frac{1}{q} \ln(\int_0^1 |f|^q) \geqslant \int_0^1 \ln|f|$$

Now I can't make any more progress to reverse the inequality. Maybe I need a completely different approach. Also what about the necessity of $f \in L^q$? Please help.

Answer 1

Following your idea, let's denote $F(q)=\displaystyle\int_{0}^{1}|f(x)|^{q}dx$. Note that $|f(x)|^{q}\leq 1+|f(x)|$ for all sufficiently small $q$. Adn we have $\displaystyle\int_{0}^{1}(1+|f(x)|)dx<\infty$ so Lebesgue Dominated Convergence Theorem is applicable here, we have $\displaystyle\int_{0}^{1}|f(x)|^{q}dx\rightarrow\displaystyle\int_{0}^{1}1dx=1$, so L'Hopital is taken place here. We can use Lebesgue to deduce once again that $F'(q)=\displaystyle\int_{0}^{1}|f(x)|^{q}\ln|f(x)|dx$. Now you see the later integral approaches $\displaystyle\int_{0}^{1}\ln|f(x)|dx$ as $q\downarrow 0$ by Lebesgue again.

Answer 2

Following Paul's hint, realize that

$$\lim\limits_{q \rightarrow 0} \frac{1}{q} \ln \int_0^1 |f|^q$$

is in the $\frac{0}{0}$ indeterminant form. Then we can take the derivative with respect to $q$ of the numerator $$\ln \int_0^1 |f|^q$$ and denominator $q$ to yield

$$\lim\limits_{q \rightarrow 0} \frac{1}{\int_0^1 |f|^q} \cdot \int_0^1 \ln |f| \cdot |f|^q$$

Now we are no longer in an indeterminant form, and since this is a continuous function of $q$, just set $q=0$ to yield the desired result.