Equivalent norms

Equivalence of the Euclidean norm (blue) and the maximum norm (red) in two dimensions

In mathematics, equivalent norms are a pair of abstract distance concepts, so-called norms , which generate identical convergence concepts. A more detailed distinction is made between stronger norms (synonymously also called finer norms ) and weaker norms (synonymously also called coarser norms ) and two norms are called equivalent if they are both stronger and weaker than their counterpart.

definition

Given is a vector space over (in most cases or ) on which two norms and are defined. ${\ displaystyle X}$${\ displaystyle \ mathbb {K}}$${\ displaystyle \ mathbb {K} = \ mathbb {R}}$${\ displaystyle \ mathbb {K} = \ mathbb {C}}$ ${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle \ | \ cdot \ | _ {2}}$

Then means stronger or finer than if there is a positive number such that ${\ displaystyle \ | \ cdot \ | _ {2}}$ ${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle C}$

${\ displaystyle \ | x \ | _ {1} \ leq C \ cdot \ | x \ | _ {2} {\ text {for all}} x \ in X}$

is. Correspondingly, weaker or coarser than is also mentioned. ${\ displaystyle \ | \ cdot \ | _ {1}}$ ${\ displaystyle \ | \ cdot \ | _ {2}}$

The norms and are called equivalent if there are positive numbers such that ${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle c, C}$

${\ displaystyle c \ cdot \ | x \ | _ {2} \ leq \ | x \ | _ {1} \ leq C \ cdot \ | x \ | _ {2} {\ text {for all}} x \ in X}$

applies. Two norms are therefore equivalent if is stronger than and is stronger than . ${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$

Examples

Finite dimensional

Let it be given , provided with the maximum norm and the sum norm${\ displaystyle \ mathbb {R} ^ {n}}$

${\ displaystyle \ | x \ | _ {\ infty}: = \ max _ {1 \ leq i \ leq n} | x_ {i} | {\ text {and}} \ | x \ | _ {1}: = \ sum _ {i = 1} ^ {n} | x_ {i} |}$.

Then it is always because of${\ displaystyle | x_ {i} | \ leq \ max _ {1 \ leq i \ leq n} | x_ {i} |}$

${\ displaystyle \ sum _ {i = 1} ^ {n} | x_ {i} | \ leq n \ cdot \ max _ {1 \ leq i \ leq n} | x_ {i} |}$.

So is

${\ displaystyle \ | x \ | _ {1} \ leq n \ cdot \ | x \ | _ {\ infty}}$,

accordingly the maximum norm is stronger than the sum norm. It is always the other way around

${\ displaystyle \ max _ {1 \ leq i \ leq n} | x_ {i} | \ leq \ sum _ {i = 1} ^ {n} | x_ {i} | {\ text {, also}} \ | x \ | _ {\ infty} \ leq \ | x \ | _ {1}}$,

since the entry with the largest amount in a vector is never greater than the sum of the amounts of all entries in the vector. Thus the sum norm is stronger than the maximum norm. Overall then applies

${\ displaystyle \ | x \ | _ {\ infty} \ leq \ | x \ | _ {1} \ leq n \ cdot \ | x \ | _ {\ infty}}$,

The maximum norm and the sum norm im are therefore equivalent. In fact it can be shown that all norms are equivalent on arbitrary finite-dimensional vector spaces. ${\ displaystyle \ mathbb {R} ^ {n}}$

Infinitely dimensional

If one considers the vector space of the real-valued continuous functions on the closed interval from zero to one, two norms can be defined: ${\ displaystyle C ([0,1], \ mathbb {R})}$

• On the one hand the supremum norm , which is well-defined due to the boundedness of continuous functions on the compact interval .${\ displaystyle \ | f \ | _ {\ infty}: = \ sup _ {x \ in [0,1]} | f (x) |}$${\ displaystyle [0,1]}$
• On the other hand, continuous functions are always measurable in this context and are contained in Lp space because of their limited nature . This means that the L1 standard
${\ displaystyle \ | f \ | _ {L ^ {1}}: = \ int _ {[0,1]} | f (x) | \ mathrm {d} \ lambda (x)}$
define.

The integral can always be estimated upwards by the largest possible function value, so it applies here

${\ displaystyle \ int _ {[0,1]} | f (x) | \ mathrm {d} \ lambda (x) \ leq \ sup _ {x \ in [0,1]} | f (x) | }$

and thus

${\ displaystyle \ | f \ | _ {L ^ {1}} \ leq \ | f \ | _ {\ infty}}$.

The supreme norm is therefore stronger than the L1 norm.

However, the two standards are not equivalent: For example, the following applies to the functions defined by with and . So there can be no constant with for all functions in . ${\ displaystyle f_ {n} (x) = \ max (2n-2n ^ {2} x, 0)}$${\ displaystyle n \ in \ mathbb {N}}$${\ displaystyle \ | f_ {n} \ | _ {L ^ {1}} = 1}$${\ displaystyle \ | f_ {n} \ | _ {\ infty} = 2n}$${\ displaystyle C}$${\ displaystyle \ | f \ | _ {\ infty} \ leq C \ | f \ | _ {L ^ {1}}}$${\ displaystyle f}$${\ displaystyle C ([0,1], \ mathbb {R})}$

interpretation

If two norms and are given and is stronger than , then is the sphere ${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$

${\ displaystyle B_ {r} ^ {\ | \ cdot \ | _ {1}}: = \ {x \ in X \, | \, \ | x \ | _ {1} \ leq r \}}$

always included in the sphere . A convergence with respect to always automatically forces a convergence with respect to , since the standard spheres of always contain the standard spheres of after rescaling . Thus always "majorizes" . ${\ displaystyle C \ cdot B_ {r} ^ {\ | \ cdot \ | _ {2}}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$

The equivalence of the norms now means that both is stronger than and that is stronger than . According to the above argument, a sequence converges with respect to exactly when it converges with respect to . ${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$

properties

• If the norm is stronger than , then applies to the generated metrics${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$
${\ displaystyle d_ {1} (x, y) = \ | xy \ | _ {1} {\ text {and}} d_ {2} (x, y) = \ | xy \ | _ {2}}$,
that then is also stronger than .${\ displaystyle d_ {2}}$${\ displaystyle d_ {1}}$
• The same applies: Is stronger than , the topology generated by is finer or stronger than the topology generated by.${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$${\ displaystyle \ | \ cdot \ | _ {2}}$${\ displaystyle \ | \ cdot \ | _ {1}}$
• In finite-dimensional vector spaces, all norms are equivalent.