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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!

Ayman Hourieh
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usul
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2 Answers2

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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).

Ayman Hourieh
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  • By the way, if you're only familiar with $L^p$ norms and Jensen's inequality in finite spaces, you can follow the same proof by replacing integrals with sums. Try it! – Ayman Hourieh Jan 19 '13 at 22:40
  • Sorry for the delay, but thanks for your answer, your proof was very clear and easy to follow! – usul Jan 25 '13 at 15:14
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    @ Ayman Hourieh ,in inequality $(1)$ , you are taking the limit as $ n \to \infty $ & using D.C.T , you are passing the limit within the integral in the Right hand Side. But it is not clear to me that; to pass the limit you must have $|\frac {|f|^{1/n}-1}{1/n}| \le g(x)$ for some $g(x) \in L^{1}$. So, here what is the dominating function?? – user92360 Apr 21 '14 at 09:28
  • @user92360 I edited my answer to clarify this. In [your question](http://math.stackexchange.com/q/761654/4583), you have the assumption that $\|f\|_r < \infty$ for some $r$. This gives you the dominating function. – Ayman Hourieh Apr 21 '14 at 11:10
  • @ Ayman Hourieh , as you say, the numerator can be dominated ( for large enough $n$ , $\frac{1}{n} \le q$ , but this argument yields reverse inequality for the denominator... i.e. $\frac{1}{1/n} \ge \frac{1}{q}$ . Isn't it?? Please check my argument & rectify if I am wrong! – user92360 Apr 21 '14 at 14:46
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    @user92360 Study the function $x \mapsto \dfrac{|f|^x - 1}{x}$ as a whole. It is decreasing as $x \to 0$. – Ayman Hourieh Apr 21 '14 at 17:54
  • The recent http://math.stackexchange.com/questions/1534870/limit-of-l-p-norm-as-p-rightarrow-0 led me to your contribution here. I have a remark and a question. As an alternative one can obtain the result by showing it first for simple functions which is easy. The question is about what can be said if $X=\mathbb{R}^{n}$. In case $n=1$ one is led to \begin{equation*} \frac{1}{l}\int_{0}^{l}dx\ln |f(x)|=\int_{0}^{1}du\ln |f(\frac{u}{l})| \overset{l\rightarrow \infty }{\rightarrow }\ln |f(0)| \end{equation*} provided $|f(0)|$ exists and is positive together with a continuity requirement. – Urgje Nov 18 '15 at 13:26
  • @AymanHourieh When you wrote "Since $\dfrac{|f|^{1/n} - 1}{1/n}$ is dominated by the integrable function $\dfrac{|f|^{q} - 1}{q}$ for large enough $n$", I think you meant " $\dfrac{|f|^{1/n} - 1}{1/n}$ is dominated by the integrable function $\left\vert\dfrac{|f|^{q} - 1}{q}\right\vert$ ... ". Otherwise, the statement would false for any $f$ such that $\vert f \vert <1$. – Ramiro Nov 18 '15 at 14:37
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    @AymanHourieh , You are correct when you write that the function $s \mapsto \frac{a^s -1}{s}$ is increasing for $s\in(0,q]$ (decreasing as $s \to 0$). HOWEVER, if $a<1$, the function has negative values and then, in this case, $s \mapsto \left\vert \frac{a^s -1}{s}\right\vert$ is decreasing for $s\in(0,q]$ (incrasing as $s \to 0$). So, if $\vert f \vert <1$, your argument to apply the Dominated Convergence Theorem fails. – Ramiro Nov 19 '15 at 17:04
  • @AymanHourieh , Supose $0 – Ramiro Nov 19 '15 at 17:07
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    @AymanHourieh , A nice suggestion (from @dreammonger) to fix your proof is to use the Dominated Convergence Theorem for $\frac{\vert f\vert^{1/n}-1}{1/n}$ on the set $\{x : \vert f(x)\vert \geqslant 1\}$ and the Monotone Convergence Theorem for $\frac{1-\vert f\vert^{1/n}}{1/n}$ on the set $\{x : \vert f(x)\vert < 1\}$. – Ramiro Nov 19 '15 at 17:16
  • @Ramiro Notice that function $x\mapsto (a^x-1)/x$ is monotone increasing whenever $a\geq 0$, thus $\frac{|f|^{1/n}-1}{1/n}\leq |f|-1$ for all $n$. The sequence $\frac{|f|^{1/n}-1}{1/n}$ is monotone decreasing and we can thus apply monotone converge theorem to the sequence $|f|-1-\frac{|f|^{1/n}-1}{1/n}$ which is a non-negative monotone increasing sequence. – Xiang Yu Jan 04 '16 at 14:57
  • @XiangYu Your argument is another way to correct the answer. – Ramiro Jan 05 '16 at 15:12
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    @XiangYu How is the function $(a^x - 1) / x $ monotone increasing whenever $a \geq 0$? I've taken the derivative and it doesn't always seem positive. You need $(x \log(a) - 1)$ to be positive if I'm not mistaken. – AnlamK Sep 23 '17 at 17:18
  • @XiangYu How about something like if $a$ and $\log(a)$ are bounded, then for small enough $x \leq 1/log(a)$, the derivative is always negative - so as x increases, function's value decreases - as $x$ decreases, function's value increases. So as $x$ and $1/n$ approach zero, the series of functions $f_n$ become monotone *increasing*. And here use MCT? Rechecking the author's argument, I think he already established the boundedness of $log(a)$ or $\log||f_{1/n}||$ ? – AnlamK Sep 24 '17 at 17:00
  • @AnlamK Sorry, I made a mistake. It is true that $x\mapsto (a^x-1)/x$ is monotone increasing for all $a\geq 0$, because $x\mapsto a^x$ is a convex function (note that its second derivative is $a^x(\log a)^2$) for all $a\geq 0$. – Xiang Yu Sep 24 '17 at 17:18
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    @AnlamK We can also show this by computing derivatives. Let $f(x)=(a^x-1)/x$, then $$f'(x)=\frac{a^x(\log (a)x-1)+1}{x^2}.$$ It suffices to show that $a^x(\log(a)x-1)+1$ is always non-negative. Let $g(x)=a^x(\log(a) x-1)+1$, then $g'(x)=a^x(\log (a))^2x$. Thus $g(x)$ attains its minimum at $x=0$ which is $0$, so $g(x)\geq 0$. – Xiang Yu Sep 24 '17 at 17:23
  • @XianYu Thank you for the clarifying comments - I realize my mistake. In the expression $a^x (x*\log a - 1) + 1$, I was just looking at $(x*\log a - 1)$ and ignoring the rest of the terms. That was my mistake. Thanks again. – AnlamK Sep 24 '17 at 17:33
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Scaled $L^p$ norm (or rather $\ell^p$, since you work with vectors) is known as Generalized mean. A bunch of interesting inequalities involving the means are found in the book Inequalities by Hardy, Littlewood, and Pólya.


The integral geometric mean $\exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log f(\theta)\,d\theta \right)$ comes up in complex analysis, especially as it relates to operator theory and involves the name of Gabor Szegő. See the terse Wikipedia article on Szegő limit theorems and the not-at-all-terse book by Barry Simon Szego's Theorem and Its Descendants

In a visit back to his native Budapest, Pólya mentioned this conjecture to Szegő, then an undergraduate, and he proved the theorem below, published in 1915... At the time, Szegő was nineteen, and when the paper was published, he was serving in the Austrian Army in World War I

The book Banach spaces of analytic functions by Kenneth Hoffman presents this topic from the viewpoint of complex analysis without much operator theory. The quantity $\exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log f(\theta)\,d\theta \right)$ turns out to be equal to $\inf_{p}\int|1-p|^2 f(\theta)\,d\theta$ where $p$ runs over all polynomials vanishing at $0$. In particular, this gives a criterion for the density of polynomials in weighted $L^2$ spaces.


In a rather different direction, the integral geometric mean comes up in number theory. If $p$ a complex polynomial, the quantity $\exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log |p(\theta)|\,d\theta \right)$ is called the Mahler measure of $p$, denoted $M(p)$. Lehmer's conjecture asserts that there is a gap $(1,\mu)$ in the possible values of $M(p)$: that is, either $M(p)=1$ or $M(p)\ge \mu>1$. Conjecturally, $\mu$ is attained by the polynomial $$p(z)= z^{10}-z^9+z^7-z^6+z^5-z^4+z^3-z+1$$ But even the existence of such $\mu$ remains unknown, let alone its precise value. The Wikipedia article has a good list of references.


That said, please do not call the integral geometric mean "the $L^0$ norm". This term is ambiguous and misleading enough as it is.