= Dual norm
{title2=$\|p\|_* =\sup_{\|v\|\le1}|\langle p,v\rangle|$}
{wiki}
A <dual norm> measures how strongly a <bounded linear functional> can respond to a unit-sized input. If $p$ is a <bounded linear functional> on a <normed vector space> $X$, write $\langle p,x\rangle=p(x)$ and define
$$
\|p\|_* =\sup_{\|x\|\le1}|p(x)|.
$$
It is the <operator norm> of $p$ when its scalar output is measured by <absolute value>. On $\mathbb R^n$, represent $p(x)=a^Tx$ using a coefficient <vector> $a$. The <unit ball> is symmetric, so maximizing $|a^Tx|$ gives the same value as maximizing $a^Tx$: negate an input whenever its output is negative.<Convex Optimization by Stephen Boyd and Lieven Vandenberghe>[§A.1.6, printed p. 636: dual norm and scalar operator-norm interpretation]{ref}
For an original two-coordinate example, fix $a=(2,-1)^T$, so $p(x)=2x_1-x_2$. The input <norm> changes the allowable <unit ball> and hence the <dual norm>:
$$
\begin{aligned}
\|x\|_1\le1&:\quad\|p\|_*=2,
&&x_*=(1,0)^T,\\
\|x\|_2\le1&:\quad\|p\|_*=\sqrt5,
&&x_*=(2,-1)^T/\sqrt5,\\
\|x\|_\infty\le1&:\quad\|p\|_*=3,
&&x_*=(1,-1)^T.
\end{aligned}
$$
Under the <L1 norm>, $|2x_1-x_2|\le2(|x_1|+|x_2|)\le2$, and $(1,0)^T$ attains the bound. Under the <Euclidean norm>, the <Cauchy-Schwarz inequality> gives $|a^Tx|\le\|a\|_2\|x\|_2\le\sqrt5$, with equality at the displayed normalized <vector>. Under the <maximum norm>, each coordinate can reach unit <absolute value>, so $|2x_1-x_2|\le2+1=3$; $(1,-1)^T$ attains it. Thus the <dual norms> of these three <vector norms> are respectively the <maximum norm>, the <Euclidean norm>, and the <L1 norm>. For $1\le p\le\infty$, <Lp norms> on a finite coordinate space pair with their <Holder conjugate exponents>.<Convex Optimization by Stephen Boyd and Lieven Vandenberghe>[§A.1.6, printed p. 637: Euclidean self-duality and one-, infinity- and p-norm duality; the numerical vector and maximizing points here are independent examples]{ref}
\Image[/dual-norm-unit-balls-supporting-lines.png]
{title=The same <linear functional> has three <dual norms> on three input <unit balls>}
{description=Original geometry for $p(x)=2x_1-x_2$. The orange <supporting line> $2x_1-x_2=\|p\|_*$ touches each <unit ball> at the marked maximizing point. Blue regions use the <L1 norm>, <Euclidean norm> and <maximum norm>, respectively.}
This geometry explains the usefulness of a <dual norm>: it translates a bound on inputs into the sharp bound on a linear output. The <support function> of the <unit ball> gives its value in the coefficient direction $a$. Changing the input budget changes the set of possible inputs; the three different answers do not conflict. In finite dimensions a closed <unit ball> is <compact>, so a <continuous linear functional> achieves its maximum <absolute value>. In an infinite-dimensional <normed vector space>, the <supremum> remains the definition and need not be achieved.
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