Suppose that V1 and V2 are vector spaces, both real or both complex, and that k · k1 and k · k2 are norms on V1 and V2, respectively. Let Y be a vector subspace of V1, and assume that we have a linear mapping T : Y → V2. Under what conditions can we find an extension of T to a linear mapping from V1 to V2 which has the same operator norm as the original mapping, or perhaps an operator norm which is not too much larger? Here the operator norms use k · k1 on the domain and k · k2 on the range.
In general an extension with the same norm does not exist, and it is not so easy to find an extension whose norm is not too much larger than the original mapping. Let us mention a few basic points related to this, however.
Suppose that P is a projection from V1 onto Y , and let k be the operator norm of P (relative to k · k1 in both the domain and range). One way to get an extension of T is to simply take the composition T ◦ P . The operator norm of T ◦ P (using k · k1 on the domain and k · k2 on the range) is bounded by k times the operator norm of T , as in (4.30) in Section 4.4.
This works quite well when the norm k · k1 on V1 comes from an inner product, so that there is a projection of norm 1 onto any subspace of V1, namely, the orthogonal projection.
Note that one might take V2 = Y , k · k2 = k · k1 on Y , and T to be the
8.3. MAPPINGS BETWEEN VECTOR SPACES 71 identity on Y . Extensions of T to mappings from V1 to V2 = Y are then exactly projections of V1 onto Y . Thus the original question becomes one about norms of projections onto Y .
In another vein, if V2 is 1-dimensional, then the question reduces to that of extending linear functionals, and this is possible while keeping the same norm because of Theorem 4.17.
This can be carried over to the case where V2 is Rn or Cn for some n and one employs the norm k · k∞ from Section 4.1. A linear mapping from a vector space V1 into Rn or Cn (according to whether V1 is real or complex) can be described in terms of n linear functionals on V1, corresponding to the n coordinates for Rn or Cn. When we employ the norm k · k∞ on Rn or Cn, then the norm of a mapping from V1 to Rn or Cn is the same as the maximum of the norms of these n linear functionals on V1 (using the norm that we have on V1). In this way the extension problem reduces to its counterpart for linear functionals.
Here is a “dual” version of the question at the beginning of the section.
Suppose that Z1 and Z2 are vector spaces, both real or both complex, and equipped with norms. Assume also that we are given a subspace Y of Z2, which we can use to define the quotient Z2/Y . Let q : Z2 → Z2/Y be the corresponding quotient mapping. If A is a linear mapping from Z1 to Z2/Y , under what conditions can we “lift” A to a linear mapping A : Zb 1 → Z2 such that A = q ◦A and the norm ofb A is the same as the norm of A, or not toob much larger? For the norm of A, we use the quotient norm on Z2/Y coming from the given norm on Z2.
There are natural duals of the remarks above for the previous question.
If U is a subspace of Z2 which is complementary to Y , then the restriction of q to U is one-to-one and maps U onto Z2/Y . Thus we can obtain liftings A of A which take values in A. There is then the question of the norm of theb
lifting, which can be analyzed in terms of projections, through the remarks in Section 8.2. If the norm on Z2 comes from an inner product, then one can take U to be the orthogonal complement of Y , and the lifting to U has the same norm as the original mapping A.
If Z1 has dimension 1, then it is easy to do the lifting, while keeping the norm fixed. If instead Z1 is Rn or Cn, equipped with the norm k · k1 from Section 4.1 (so that k · k1 is not now just a generic name for a norm, but rather a very specific one), then the argument for 1-dimensional domains can be carried over in a simple manner. More precisely, let ej, 1 ≤ j ≤ n, be the standard basis vectors for Rn or Cn (according to whether one is
72 CHAPTER 8. SUBSPACES AND QUOTIENT SPACES working with real or complex vector spaces). The idea is to first choose A(eb j) for each j so that q(A(eb j)) = A(ej) and the norm of A(eb j) is equal to the (quotient) norm of A(ej) for each j. Once this is done, A is determinedb on the whole domain by linearity. Because of the specific norm that we have on the domain, the norm of A as a linear mapping will be the same as theb norm of A.
Chapter 9
Variation seminorms
9.1 Basic definitions
Let Z denote the set of integers, and let Zn denote the set of n-tuples of positive integers (for a given positive integer n).
If x, y are elements of Zn, then we shall say that x and y are neighbors if there is an integer i, 1 ≤ i ≤ n, such that yi− xi = ±1 and yj = xj for all j 6= i, 1 ≤ j ≤ n, where xland yl denotes the components of x and y. This is clearly symmetric in x and y. We shall denote by N(x) the set of neighbors of x. Note that x is not considered to be a neighbor of itself.
If U is a subset of Zn, then we define Int U, ∂U ⊆ U by Int U = {x ∈ U : N(x) ⊆ U}
(9.1) and
∂U = U\∂U.
(9.2)
This notation is somewhat at odds with that in topology, but these definitions will play an analogous role.
In this chapter we shall make the standing assumption that U ⊆ Zn is finite and Int U 6= ∅.
(9.3)
Let f be a function on U. By “function” we mean one that is real-valued or complex-real-valued, so that we have the real vector space of real-real-valued functions on U, or the complex vector space of complex-valued functions on U. Sometimes it will be helpful to specify one or the other, but often it will not matter.
73
74 CHAPTER 9. VARIATION SEMINORMS For each real number p ≥ 1, define the p-variation Vp(f ) of f (on U) by
Vp(f ) = X
x∈Int U
X
y∈N (x)
|f (x) − f (y)|p
1/p
. (9.4)
This is a seminorm on the vector space of functions on U (real or complex), which means that it satisfies the same properties as a norm except that Vp(f ) might be 0 even if f is not the zero function. In particular, Vp(f ) = 0 whenever f is a constant function. The fact that Vp(·) satisfies the triangle inequality
Vp(f + g) ≤ Vp(f ) + Vp(g) (9.5)
for all functions f and g on U is a consequence of Minkowski’s inequality for sums (using the same choice of p).
As in the case of norms, we can apply the triangle inequality twice to obtain that
Vp(f ) ≤ Vp(g) + Vp(f − g) and Vp(g) ≤ Vp(f ) + Vp(f − g), (9.6)
and hence that
|Vp(f ) − Vp(g)| ≤ Vp(f − g) (9.7)
for all functions f and g on U. One can use this to show that Vp(·) is continuous on the vector space of functions on U, because Vp(·) is bounded by a constant multiple of a standard norm on functions.
Observe that elements of U which are not neighbors of elements of Int U are not used in (9.4), so that the values of f at such points does not affect Vp(f ). One might as well take out these points from U.
It will sometimes be of interest to restrict our attention to functions f on U which are equal to 0 on ∂U. In this case Vp(f ) can be rewritten as
Vp(f ) =
X
(x,y)∈D(Int U )
|f (x) − f (y)|p
1/p
, (9.8)
where
D(Int U) = {(x, y) ∈ Int U × Int U : x, y are neighbors}.
(9.9)
On the vector space of functions on U that are 0 on ∂U, Vp(f ) is a norm, i.e., Vp(f ) = 0 implies that f is the zero function on U. (Exercise.)