Factoring through Hilbert spaces

Factorization Theory

7.3 Factoring through Hilbert spaces

In the first section of this chapter we saw that ifX has type 2 and 1≤p <2 then any operatorT :X→L_pfactors through a Hilbert space. In this section we are giving a characterization for an operator between Banach spaces to factor through a Hilbert space.

Definition 7.3.1.Suppose thatX andY are Banach spaces. We say that an operatorTfromXtoY factors through a Hilbert spaceif there exist a Hilbert spaceH and operatorsS:X −→H andR:H −→Y verifyingT =RS.

7.3 Factoring through Hilbert spaces 181 We will begin by making some remarks that will lead us to the necessary condition we are seeking. We will only consider real scalars, although at the appropriate moment we will discuss the alterations necessary to handle com- plex scalars. Throughout this section H will denote a generic Hilbert space with a scalar product · .

Suppose we havenarbitrary vectorsx₁, . . . , x_ninH. Given a real orthog- onal matrix A = (aij)1≤i,j≤n, let us consider the new vectors in H defined fromA,

zi= n j=1

aijxj, 1≤i≤n. (7.14) Then,

n i=1

zi²= n i=1

n j=1

aijxj²

= n i=1

n j=1

aijxj, n k=1

aikxk

= n i=1

n j=1

n k=1

a_ija_ikx_j, x_k

= n j=1

x_j, x_j

= n j=1

x_j².

Any real n×n matrixA = (a_ij) defines a linear operator (that will be denoted in the same way)A:ⁿ₂ −→ⁿ₂ via

A

⎛

⎜⎜

⎜⎝ s₁ s₂ ... s_n

⎞

⎟⎟

⎟⎠=

⎛

⎜⎜

⎜⎝

a₁₁ a₁₂ . . . a_1n a₂₁ a₂₂ . . . a_2n ... ... . .. ... a_n1 a_n2. . . a_nn

⎞

⎟⎟

⎟⎠

⎛

⎜⎜

⎜⎝ s₁ s₂ ... s_n

⎞

⎟⎟

⎟⎠.

The matrix (a_ij)_1≤i,j≤n is orthogonal if and only if the operator A is an isometry. If (a_ij)ⁿ_i,j=1 is not orthogonal but A ≤ 1, it is an exercise of linear algebra to prove that (a_ij) can be written as a convex combination of orthogonal matrices. In fact, it is always possible to find orthonormal basis (e_j)ⁿ_j=1 and (f_j)ⁿ_j=1 in ⁿ₂ so that Ae_j = λ_jf_j with λ_j ≥ 0: Just find an orthonormal basis of eigenvectors (e_j)ⁿ_j=1 forAA whereA is the transpose.

Then A =DU where Df_j =λ_jf_j and U e_j =f_j. U is orthogonal and since 0≤λ_j ≤1 we can writeDas a convex combination of the orthogonal matrices Vfj=jfj wherej =±1.

Thus, ifx₁, . . . , x_n, z₁, . . . , z_nare arbitrary vectors inHsatisfying equation (7.14), where(a_jk)ⁿ_j,k=1ⁿ₂→ⁿ₂ ≤1, we will have

n i=1

z_i²≤ n j=1

x_j².

This can easily be extended to the case of differing numbers ofx_j’s andz_i’s by adding zeros to one of the two collections of vectors.

Theorem 7.3.2.LetT be an operator from a Banach spaceX into a Banach space Y. Suppose that there exist operators S :X −→ H and R :H −→Y verifying T = RS. If (xj)^m_j=1 and (zi)ⁿ_i=1 are vectors in X related by the equation

zi= m j=1

aijxj, 1≤i≤n, (7.15) where(a_ij)is a realn×m matrix such thatA^m₂→ⁿ₂ ≤1, then

ⁿ

i=1

T z_i²1/2

≤ S R^m

j=1

x_j²1/2

.

Proof. The proof easily follows from the comments we made. Indeed, given x₁, . . . , x_mandz₁, . . . , z_ninXsatisfying (7.15), since the collections of vectors (Sx_j)^m_j=1 and (Sz_i)ⁿ_i=1 lie insideH we have

n i=1

T z_i²= n i=1

RSz_i²

≤ R² n i=1

Sz_i²

≤ R² m j=1

Sxj²

≤ R²S² m j=1

x_j².

In light of the previous theorem we want to give an alternative formulation of the property that (x_j)^m_j=1 and (z_i)ⁿ_i=1 are vectors in X related by the equation

z_i= m j=1

a_ijx_j, 1≤i≤n,

whereA= (a_ij) is a realn×mmatrix such thatA^m₂→ⁿ₂ ≤1.

7.3 Factoring through Hilbert spaces 183 Proposition 7.3.3.Givenn, m∈Nand any two sets of vectors (x_j)^m_j=1 and (z_i)ⁿ_i=1 in a Banach spaceX, the following are equivalent:

(a)There is a realn×m matrixA= (a_ij)so that A^m₂→ⁿ₂ ≤1 and z_i=

m j=1

a_ijx_j, 1≤i≤n;

(b) m j=1

|x^∗(zj)|²≤ n i=1

|x^∗(xi)|², for all x^∗∈X^∗.

Proof. Assume that (a) holds. Then, sinceA^m₂→ⁿ₂ ≤1, it follows that n

i=1

|x^∗(zi)|²= n i=1

x^∗m

j=1

aijxj²= n i=1

m j=1

aijx^∗(xj)²≤ m j=1

|x^∗(xj)|².

For the reverse implication, (b)⇒(a), consider the linear operators α:X^∗−→^m₂ , x^∗→(x^∗(x_j))^m_j=1

and

β :X^∗ −→ⁿ₂, x^∗→(x^∗(z_i))ⁿ_i=1. The hypothesis says that βx^∗_m

2 ≤ αx^∗_n

2 for all x^∗ ∈X^∗. Thus we can define an operatorA0:α(X^∗)→β(X^∗) withA0 ≤1 andβ=A0◦α.Then A₀ can be extended to an operatorA:^m₂ →ⁿ₂ withA ≤1. Let (a_ij) be the matrix associated withA.

x^∗(z_i) = m j=1

a_ijx^∗(x_j) for all x^∗∈X^∗, which implies

zi= m j=1

aijxj, i= 1, . . . , n.

The main result of this section is the following criterion:

Theorem 7.3.4.LetX andY be Banach spaces. SupposeEis a closed linear subspace ofX andT :E→Y is an operator. In order that there exist a Hilbert spaceH and operatorsR:X →H, S :H →Y with RS ≤C such that T =RS|Eit is necessary and sufficient that for all sets of vectors(xj)^m_j=1⊂X and(z_i)ⁿ_i=1⊂E such that

n i=1

|x^∗(z_i)|²≤ m j=1

|x^∗(x_j)|², x^∗∈X^∗,

we have

ⁿ

i=1

T z_i²1/2

≤C ^m

j=1

x_j²1/2

.

In the proof of this result and other ones in the next chapter we will make use of the following lemma. IfAis a subset of real vector space we define

cone (A) = ⁿ

j=1

αjaj: a1, . . . , an ∈A, α1, . . . , αn≥0, n= 1,2, . . .

. Lemma 7.3.5.LetV be a real vector space. GivenA,Btwo subsets ofV such that V =cone(B)−cone(A), and two functions φ:A →R,ψ:B →R, the following are equivalent:

(i)There is a linear functionalL onV verifying φ(a)≤ L(a), a∈ A and

ψ(b)≥ L(b), b∈ B.

(ii) If (α_i)^m_i=1, (β_j)ⁿ_j=1 are two finite sequences of nonnegative scalars such

that m

i=1

α_ia_i= n j=1

β_jb_j for some(a_i)^m_i=1⊂ A,(b_j)ⁿ_j=1⊂ B, then

m i=1

α_iφ(a_i)≤ n j=1

β_jψ(b_j).

Proof. The implication (i)⇒(ii) is immediate.

(ii)⇒(i) Let us define the mapp:V →[−∞,∞) as follows:

p(v) = inf ⁿ

j=1

βjψ(bj)− m i=1

αiφ(ai)

,

the infimum being taken over all possible representations in the form v = n

j=1β_jb_j −m

i=1α_ia_i, where α₁, . . . , α_m, β₁, . . . , β_n ≥ 0, a₁, . . . , a_m ∈ A, andb₁, . . . , b_m∈ B.

pis well-defined since V= cone (B)−cone (A). Besides, one easily checks that p is positive-homogeneous and satisfies p(v₁+v₂) ≤ p(v₁) +p(v₂) for any v₁, v₂ in V. In order to prove thatpis a sublinear functional we need to show that p(v)> −∞for every v ∈ V. This will follow ifp(0) = 0. Indeed, p(v) +p(−v)≥p(0), so neither p(v) norp(−v) could be−∞ifp(0) = 0.

7.3 Factoring through Hilbert spaces 185 For each representation of 0 in the form 0 =n

j=1β_jb_j−m

i=1α_ia_i, by the hypothesis it follows that n

j=1β_jψ(b_j) ≥ m

i=1α_iφ(a_i). Therefore, by the definition,p(0)≥0 hencep(0) = 0.

As an consequence of the Hahn-Banach theorem, there is a linear func- tional L onV such that L(v)≤p(v) for everyv ∈ V and so φ(a)≤ L(a) for alla∈ AandL(b)≤ψ(b) for allb∈ B.

Proof of Theorem 7.3.4. We need only show that the condition is sufficient.

Let F(X^∗) denote the set of all functions from X^∗ to R, and consider the natural mapX→ F(X^∗),x→x, whereˆ

ˆ

x(x^∗) =x^∗(x), x^∗∈X^∗.

Let V be the linear subspace of F(X^∗) of all finite linear combinations of functions of the form ˆxˆz, withx,zin X. That is,

V = ^N

k=1

λ_kxˆ_kzˆ_k: (λ_k)^N_k=1inR, (x_k)^N_k=1and (z_k)^N_k=1 inX,andN∈N . Clearly, the set{xˆ² : x∈X}spansV since each product ˆxˆzwithxandzin X can be written in the form

ˆ xˆz= 1

4

(ˆx+ ˆz)²−(ˆx−z)ˆ ² .

We want to construct a linear functionalLonV with the following prop- erties:

0≤ L(ˆx²)≤C²x², x∈X (7.16) and

T x²≤ L(ˆx²), x∈E. (7.17) To this end, let us apply Lemma 7.3.5 in the caseA=B={xˆ²: x∈X}by putting

φ(ˆx²) =

0 if x∈X\E T x² if x∈E and

ψ(ˆx²)²=C²x². Suppose that

n i=1

β_i²zˆ_i²= m j=1

α²_jxˆ²_j

for some (ˆx_j)^m_j=1, (ˆz_i)ⁿ_i=1vectors inX, and some nonnegative scalars (α²_j)^m_j=1, (β_j²)ⁿ_j=1. Let us supposez₁, . . . , z_l∈E andz_l+1, . . . , z_n∈X\E.Then

l i=1

β²_izˆ_i²≤ m j=1

α²_jxˆ²_j, hence

l i=1

T(β_jz_i)²≤C² m j=1

α_jx_j².

Thus n

i=1

β_i²φ(ˆz²_i)≤ m j=1

α²_jψ(ˆx²_j).

Lemma 7.3.5 yields a linear functionalL onV with φ(ˆx²)≤ L(ˆx²)≤ψ(ˆx²), x∈X.

L, in turn, induces a symmetric bilinear form · onX given by x, z=L(ˆxˆz),

so the mapX −→[0,∞),x→1

x, x=1

L(ˆx²) defines a seminorm onX. Thus,X (modulo the subspace{x; x, x= 0}) endowed with the (now) inner product, is an inner product space, andx0=1

x, xa norm on X. LetH be the completion ofX₀ under this norm.H is a Hilbert space.

TakeSto be the induced operatorS:X →Hmappingxto its equivalence class in X₀.Then we have

Sx ≤Cx, x∈X.

S has norm one and dense range. By construction, ifx∈E we have T x ≤ Sx,

therefore we can find an operator R0 : S(E)→ Y with R0 ≤ 1 and T = R₀S|E.ComposeR₀with the orthogonal projection ofH ontoS(E) to create R.

The proof for complex scalars.In the case whenX andY are complex Banach spaces we proceed as first by “forgetting” their complex structure and treating them as real spaces. The argument creates a real symmetric bilinear form · onX which is continuous for the original norm. We can then define a complex inner product by “recalling” the complex structure ofX and setting

(x, z) = 1 2π

2π 0

e^iθx, e^iθz −iie^iθx, e^iθzdθ.

We leave it to the reader to check that this induces a complex inner product and that using this to defineH gives the same conclusion.