Supremum norm ($\Vert f{\Vert }_{\infty }$)

The supremum norm of a bounded real- or complex-valued function on a set $X$ isFor a continuous function on a compact space, the extreme value theorem makes this supremum a maximum.

Let $K$ be a nonempty compact space and let $\alpha :K\to \mathbb{R}$ be continuous and nowhere zero. On $C(K)$, define $\Vert f{\Vert }_{\alpha }={\sup }_{x\in K}|\alpha (x)f(x)|$. Multiplication by $\alpha$ is an isometry onto $C(K)$ with its ordinary uniform norm, because division by the continuous nonvanishing $\alpha$ provides the inverse. Thus the weighted space is a Banach space. Equivalently, the extreme value theorem gives $0<m=\min |\alpha |\le \max |\alpha |=M<\infty$, and $m\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_{\alpha }\le M\Vert f{\Vert }_{\infty }$. A weighted Cauchy sequence is uniformly Cauchy, has a continuous uniform limit, and the upper bound gives convergence also in the weighted norm.

On a nondegenerate interval, point evaluation of the derivative is unbounded for the supremum norm on smooth functions. The functions $f_N(x)=\sin (N(x-x_0))$ have supremum norm at most one but $f_N^{\prime }(x_0)=N$. Adding the supremum norm of the derivative makes this evaluation a bounded linear functional.

For a function with bounded derivatives through order $k$, the derivative supremum norm is $|u{|}_{k;U}={\sum }_{j=0}^k{\sup }_U|D^{j}u|$, with any equivalent finite-dimensional norm for each derivative tensor. This is a classical $C^k$ norm; it is different from the Lp norm with exponent $k$.