Norm ($\Vert \cdot \Vert$)

A norm assigns a nonnegative length to each vector, is zero only at the zero vector, is absolutely homogeneous, and satisfies the triangle inequality.

A quasi-norm on a vector space is nonnegative, vanishes only at zero, is absolutely homogeneous, and satisfies a triangle inequality with one constant factor $C\ge 1$. For $0<p<1$, $(\int |f{|}^p{)}^{1/p}$ on an Lp space is a quasi-norm: $\Vert f+g{\Vert }_p^p\le \Vert f{\Vert }_p^p+\Vert g{\Vert }_p^p$ gives a quasi-triangle inequality. Measure-preserving composition preserves this quantity even though the Minkowski inequality is unavailable below exponent one.

The block mixed norm sums the ${\ell }^q$ norm of each block, for $q\ge 1$. Its dual norm is $\Vert p{\Vert }_{q^{\prime },\infty }={\max }_i\Vert p_i{\Vert }_{q^{\prime }}$, where $q^{\prime }=q/(q-1)$ for $q>1$ and $q^{\prime }=\infty$ for $q=1$. Apply Holder inequality blockwise and choose a norming block to attain equality. Absolute values are essential for signed components.

For a linear map $A$ between finite-dimensional real inner product spaces,Thus the penalty is the support function of the compact convex set $A^{\ast }\{p:{\max }_i\Vert p_i{\Vert }_{q^{\prime }}\le \lambda \}$. The Moreau decomposition makes its proximal operator the identity minus the Euclidean projection onto a convex set.

Definiteness says that a norm vanishes only at the zero vector. The box norm on a nonempty finite Cartesian product is definite because its defining sum of squares includes equal-column terms, each of which is the square of an average of squares. Omitting repeated coordinates would destroy this proof and can produce a different quantity.

Absolute homogeneity is the norm axiom $\Vert av\Vert =|a|\Vert v\Vert$ for every scalar $a$ and vector $v$. It distinguishes a norm from a merely homogeneous expression of a different degree; for example the fourth root in the box norm converts a degree-four expression into a degree-one norm.

Suppose $B$ contains zero, is convex, and meets every line $\mathbb{R}v$, $v\ne 0$, in the finite symmetric open segment $\{tv:|t|<\lambda (v)\}$ with $\lambda (v)>0$. Then its Minkowski functional is a norm, with $p_B(v)=1/\lambda (v)$ and $B=\{x:p_B(x)<1\}$. Absolute homogeneity follows by rescaling the segment. If $s>p_B(x)$ and $t>p_B(y)$, convexity puts $(x+y)/(s+t)$ in $B$, giving $p_B(x+y)<s+t$; passing to the infima proves the triangle inequality. Conversely the open unit ball uniquely determines any norm by the displayed formula. No prior topology on the vector space is needed for these radial hypotheses.

On a uniform $d$-dimensional spatial grid, the discrete L2 norm weights the squared Euclidean norm by the cell volume. It approximates the continuous L2 norm and is induced by the discrete L2 inner product. A contraction bound in this norm with a mesh-independent constant proves stability of a numerical method for the spatially discretized evolution.