measure theory - "Scaled $L^p$ norm" and geometric mean

The $L^p$ norm in $\mathbb{R}^n$ is

\begin{align}
\|x\|_p = \left(\sum_{j=1}^{n} |x_j|^p\right)^{1/p}.
\end{align}
Playing around with WolframAlpha, I noticed that, if we define the "scaled" $L^p$ norm in $\mathbb{R}^n$ to be

\begin{align}
\| x \|_p = \left(\frac{1}{n}\sum_{j=1}^{n} |x_j|^p\right)^{1/p}
\end{align}
then
\begin{align}

\lim_{p \to 0} \|x\|_p &= \left( \prod_{j=1}^{n} |x_j| \right)^{1/n},
\end{align}
which is the geometric mean of the coordinates' absolute values. This is interesting maybe because the $L^p$ norm doesn't have a nice limit at zero.

My questions:

1. How do I prove this?


2. Is this definition of "scaled $L^p$ norm" interesting, or known by another name, or used anywhere?


3. Is there any interesting reason to define the $L^0$ norm as the geometric mean, as above?



4. Further reading?

Thanks!

Answer

I'll prove the general case in which $\mu$ is a positive measure on a space $X$ and $\mu(X) = 1$. Assuming that $\|f\|_q < \infty$ for at least one $0 < q < 1$, we have
$$
\lim_{p \to 0} \|f\|_{p} = \exp\left(\int_X \log|f| \,d\mu\right).
$$

Your particular case follows by setting $X = \{1, \ldots, n\}$ and $\mu(i) = 1/n$.

---

By definition
$$
\|f\|_p = \left\{\int_X |f|^p \,d\mu\right\}^{1/p}.
$$

---

Lemma 1: If $0 < r < s < 1$, then $\|f\|_r \le \|f\|_s$.

Proof: $\varphi(x) = x^{s/r}$ is a convex function. Apply Jensen's inequality to $\int_X |f|^r \,d\mu$ to get
$$
\left\{\int_X |f|^r \,d\mu\right\}^{s/r} \le \int_X |f|^s \,d\mu.
$$

Hence $\|f\|_r \le \|f\|_s$.

---

Lemma 2: If $0 < p < 1$, then $\int_X \log|f| \,d\mu \le \log \|f\|_p$.

Proof:
$\log$ is a concave function. Apply Jensen's inequality to $\int_X |f|^p \,d\mu$ to get the desired inequality.

---

From lemmas 1 and 2, it follows that $\log\|f\|_{1/n}$ is decreasing and bounded from below. Therefore, it converges as $n \to \infty$.

To find the limit, apply the inequality $\log a \le n(a^{1/n} - 1)$ with $a = \left\{\int_X |f|^{1/n}\,d\mu\right\}^{n} $ to get
$$
\log \|f\|_{1/n} \le \int_X \frac{|f|^{1/n} - 1}{1/n} \,d\mu. \tag{1}
$$

Use L'Hôpital's rule to obtain
$$
\lim_{x \to 0} \frac{a^x - 1}{x} = \log a.
$$

Take the limit of (1) as $n \to \infty$. Since $\dfrac{|f|^{1/n} - 1}{1/n}$ is dominated by the integrable function $\dfrac{|f|^{q} - 1}{q}$ for large enough $n$ and the value of $q$ stated in the assumptions, apply the dominated convergence theorem to get
$$
\lim_{n \to \infty} \log \|f\|_{1/n} \le \int_X \log|f| \,d\mu.
$$

Apply the squeeze theorem with lemma 2 to obtain
$$
\lim_{n \to \infty} \log \|f\|_{1/n} = \int_X \log|f| \,d\mu.
$$

Since $\log$ is continuous, we conclude
$$
\lim_{n \to \infty} \|f\|_{1/n} = \exp\left(\int_X \log|f| \,d\mu\right).
$$

The above argument applies to any sequence $s_n$ that converges to $0$, not just $1/n$. The general result stated at the beginning now follows. This is a standard argument in measure theory. It is usually used to apply the dominated convergence theorem to general limits, not just limits of countable sequences.

---

To answer your other questions, the "scaled norm" follows from the general case as I explained at the beginning of my answer. I've never seen the geometric mean called $L^0$. As for further readings, check out Rudin's Real and Complex Analysis or Folland's Real Analysis. The above is an exercise in one of them (I think the former).