Equivalent norms

Two norms on one vector space are equivalent when constants $c,C>0$ satisfyfor every vector. Equivalent norms define the same convergent sequences, open sets, and continuous linear functionals.

Two norms $p,q$ on a vector space give the same topology if and only if $cp(x)\le q(x)\le Cp(x)$ for some positive constants. Indeed, continuity of the identity at zero gives $p(x)<\delta \Rightarrow q(x)<1$. For $x\ne 0$, apply this to $\delta x/(2p(x))$ to obtain $q(x)<2p(x)/\delta$. Reverse the roles of the norms for the lower bound. Conversely the two inequalities compare every translated open ball. This proof uses absolute homogeneity of a norm, not finite dimension.

Relative to a Schauder basis, take the supremum over successive finite intervals whose minima belong to a specified hereditary family $\mathcal{F}$, and signs ${\epsilon }_i\in \{-1,1\}$. If this supremum is bounded above by a multiple of the original norm and the family contains all singletons, it gives an equivalent norm. Composing two admitted signed interval projections costs at most a factor of two in this norm: split their nonempty intersections according to which interval supplies the later left endpoint. In each of the two resulting families the minima form a subset of one original list, so heredity preserves admissibility. This estimate lets bounded distortion of a Banach space give a uniform bound independent of the complexity of $\mathcal{F}$.

Bounded distortion means there is one finite bound on the ratio between the largest and smallest relative sizes of any equivalent norm after restriction to a further infinite-dimensional block subspace. The comparison is invariant under scalar multiplication of the new norm. A literal two-sided bound without an allowed rescaling would fail even for a scalar multiple of the original norm.

On $C(X)$ for compact Hausdorff $X$, any complete norm making every point evaluation functional continuous is equivalent to the uniform norm. If $f_n$ converges in the new norm to $f$ and uniformly to $g$, evaluation gives $f(x)=g(x)$ for every $x$, so the identity map between the two Banach spaces has closed graph. The closed graph theorem and open mapping theorem make the identity and its inverse bounded. Completeness of the proposed norm is essential to this argument.

Any two norms on a finite-dimensional real or complex vector space are equivalent. Relative to a fixed basis, continuity and positivity of a norm on the compact Euclidean unit sphere give uniform positive lower and finite upper bounds.