Definition of $L_\infty$ norm

Question

Where does the definition of the $L_\infty$ norm come from?

$$\|x\|_\infty=\max \{|x_1|,\dots,|x_k|\}$$

Answer 1

General Approach

As shown in this answer, $$ \lim_{p\to\infty}\left(\int_A|f(x)|^p\,\mathrm{d}x\right)^{1/p}=\sup_{x\in A}|f(x)|\tag{1} $$ Using a discete measure, $(1)$ shows that $$ \lim_{p\to\infty}\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}=\max_{1\le k\le n}|x_k|\tag{2} $$ Since $$ \|x_k\|_p=\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}\tag{3} $$ if we take the limit of $(3)$ and compare with $(2)$, we get $$ \|x_k\|_\infty=\max_{1\le k\le n}|x_k|\tag{4} $$

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Discrete Approach

The discrete $\ell^p$ norm is defined as $$ \|x_k\|_p=\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}\tag{5} $$ suppose that $m=\max\limits_{1\le k\le n}|x_k|$, and that $q$ of the $|x_k|$ equal $m$ (that is, $1\le q\le n$). For the $|x_k|\lt m$, we have $\lim\limits_{p\to\infty}\left|\frac{x_k}{m}\right|^p=0$. Therefore, $$ \begin{align} \lim_{p\to\infty}\|x_k\|_p &=\lim_{p\to\infty}\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}\\ &=m\lim_{p\to\infty}\left(\sum_{k=1}^n\left|\frac{x_k}{m}\right|^p\right)^{1/p}\\[4pt] &=m\lim_{p\to\infty}q^{1/p}\\[12pt] &=m\tag{6} \end{align} $$ Taking the limit of $(6)$, we get that $$ \|x_k\|_\infty=\max_{1\le k\le n}|x_k|\tag{7} $$

Answer 2

Here's the way I like to think about it (which is not too rigorous but can be made rigorous). We have, more generally, the $L^p$ norms ($1\le p < \infty$):

$$(\|x\|_p)^p:=\sum_{i=1}^n |x_i|^p.$$

The Euclidean norm is a special case of this (take $p = 2$); the taxicab norm is also a special case (take $p=1$). Suppose $|x_i|\ge |x_j|$ for all $1\le j\le n$. What happens if $p$ gets really large? Well we would see that the $|x_i|$ term would dominate the sum and asymptotically, all of the others would be inconsequential (due to the function being very convex). So what we could say is that for large $p$,

$$(\|x\|_p)^p \approx |x_i|^p.$$

Taking a $p$th root of both sides gives us

$$\|x\|_p \approx |x_i|.$$

However note that we said that $|x_i|$ was the largest of the components of our vector. Thus, informally, we would say that $\|x\|_{\infty} = |x_i| = \max \{|x_1|,\ldots,|x_n|\}$.

Answer 3

The $p$ norm of a vector is defined as such:

$\|x\|_p = (\sum_{i=1}^{n}|x_i|^p)^\frac{1}{p}$.

Notice that when $p=2$ this is the simple euclidean norm.

You asked about the infinity norm.

When $p$ tends to infinity, we can see that:

$$\lim_ {p \to \infty} \|x\|_p = \lim_ {p \to \infty} (\sum_{i=1}^{n}|x_i|^p)^\frac{1}{p}$$

Convince yourself that if $a>b>0$ then:

$\lim_{p \to \infty} a^p+b^p=\lim_{p \to \infty} a^p$

Combine the two statements to reach the desired results.

Answer 4

The following adds nothing to the above proofs other than a slightly geometric flavour. It is too long for a comment, so I am including it as an answer.

I am taking the space to be $\mathbb{C}^n$.

It is not hard to show (see https://math.stackexchange.com/a/424335/27978 for example) that $p \mapsto \|x\|_p$ is non-increasing, $\lim_{p \to \infty} \|x\|_p = \|x\|_\infty$, and $\|x\|_\infty \le \|x\|_p \le \|x\|_1 \le \sqrt{n} \|x\|_\infty$, hence the concepts of interior and boundedness are independent of $p$.

The above shows that $B_\infty(0,1) = \cup_{ p \ge 1 } B_p(0,1)$.

Since each $\|\cdot \|_p$ is a norm, we see that each $B_p(0,1)$ is bounded (uniformly by comment above), balanced, convex and contains $0$ in its interior. Hence $\cup_{ p \ge 1 } B_p(0,1)$ is bounded, balanced, convex and contains $0$ in its interior, and so is the unit ball of some norm, in this case, $\|\cdot \|_\infty$.

The point here is that there is a natural motivation in terms of the unit balls which 'converge' (in the above sense) to the $\|\cdot \|_\infty$ unit ball.