Topics In Analysis

32.3 An Important Formula

It is not necessary to have p > 2 in order to do the sort of thing just described. First is an approximation theorem which says that a given functionin L^p

can be approximated by step functions.

Note that it would make no difference in terms of the conclusion of this lemma if you defined

m l ∑k ( k ) Φ k(t) ≡ Φ tj−1 X(tkj−1,tkj](t) j=1

because the modified function equals the one given above off a countable subset of

, the union of the mesh points.

Proof: For t ∈ ℝ let γ_n

≡ k∕2ⁿ,δ_n ≡∕2ⁿ, where

n n t ∈ (k∕2 ,(k+ 1)∕2 ],

and 2⁻ⁿ < T∕4. Also suppose Φ is defined to equal 0 on

^C × Ω. There exists a set of measure zero N such that for ωN,t → is in L^p. Therefore by continuity of translation, as n →∞ it follows that for ωN, and t ∈,

∫ p ||Φ(γn(t)+ s)− Φ(t+ s)||E ds → 0 ℝ

The above is dominated by

Consider

∫ 2T (∫ ) ||Φ(γn (t)+ s)− Φ (t +s)||pE ds dt − 2T ℝ

By the dominated convergence theorem, this converges to 0 as n →∞. Now Fubini. This yields

∫ ∫ 2T p ℝ − 2T ||Φ (γn (t)+ s)− Φ (t+ s)||E dtds

Change the variables on the inside.

∫ ∫ 2T+s p ℝ − 2T +s||Φ (γn (t − s)+ s)− Φ (t)||E dtds

Now by definition, Φ

vanishes if t, thus the above reduces to

Also by definition, γ_n + s is within 2⁻ⁿ of t and so the integrand in the integral on the right equals 0 unless t ∈ ⊆. Thus the above reduces to

∫ ∫ 2T p ||Φ(γn(t− s)+ s)− Φ (t)||E dtds. ℝ −2T

This converges to 0 as n →∞ as was shown above. Therefore,

∫ ∫ T T p 0 0 ||Φ (γn (t− s)+ s)− Φ (t)||E dtds

also converges to 0 as n →∞. The only problem is that γ_n

+ s ≥ t − 2⁻ⁿ and so γ_n + s could be less than 0 for t ∈. Since this is an interval whose measure converges to 0 it follows

∫ T∫ T |||| ( + ) ||||p ||Φ (γn(t− s)+ s) − Φ (t)||E dtds 0 0

converges to 0 as n →∞. Let

∫ T|| ( ) ||p mn (s) = ||||Φ (γn (t− s)+ s)+ − Φ(t)|||| dt 0 E

Then letting μ denote Lebesgue measure,

1∫ T μ([mn (s) > λ]) ≤ λ mn (s)ds. 0

It follows there exists a subsequence n_k such that

([ 1]) μ mnk (s) > -- < 2−k k

Hence by the Borel Cantelli lemma, there exists a set of measure zero N such that for s

N,

mnk (s) ≤ 1∕k

for all k sufficiently large. Pick such an s. Then consider t → Φ

. For n_k, t →⁺ has jumps at points of the form 0, s + l2^−n_k where l is an integer. Thus P_nk consists of points of which are of this form and these partitions are nested. Define Φ_k^l ≡ 0, Φ_k^l ≡ Φ. Now suppose N₁ is a set of measure zero. Can s be chosen such that all jumps for all partitions occur off N₁? Let be an interval contained in . Let S_j be the points of which are translations of the measure zero set N₁ by t_j^l for some j. Thus S_j has measure 0. Now pick s ∈ ∖∪_jS_j. To get the other sequence of step functions, the right step functions, just use a similar argument with δ_n in place of γ_n. Just apply the argument to a subsequence of n_k so that the same s can hold for both. ■

Proof: By Lemma 32.3.1, there exists a sequence of uniform partitions

_k=0^m_n = P_n,P_n ⊆P_n+1, of such that the step functions

converge to X in K and in L².

Proof: It follows from the following computations

∫ t X (t)− X (s) = Y (u)du s

2 2 2 − |X (t)− X (s)| = − |X (t)| + 2(X (t),X (s))− |X (s)|

Hence

∫ 2 2 t 2 |X (t)| = |X (s)| +2 s 〈Y (u),X (t)〉du − |X (t)− X (s)| ■

Proof: From the above formula applied to the k^th partition of

described above,

m∑−1 |X (tm)|2 − |X0 |2 = |X (tj+1)|2 − |X (tj)|2 j=0

Thus, discarding the negative terms and denoting by P_k the k^th of these partitions,

∫ 2 2 T r stju∈pPk|X (tj)|H ≤ |X0| + 2 0 |〈Y (u) ,X k (u)〉|du

∫ T ≤ |X0 |2 +2 ∥Y (u)∥V ′ ∥Xrk (u)∥V du 0

( )1∕p′ ( )1 ∕p 2 ∫ T p′ ∫ T r p ≤ |X0| + 2 0 ∥Y (u)∥V ′ du 0 ∥X k (u)∥V du ≤ C (∥Y ∥K′ ,∥X ∥K)

because these partitions are chosen such that

( ∫ )1∕p ( ∫ )1∕p lim T∥Xr (u)∥p = T∥X (u)∥p k→∞ 0 k V 0 V

and so these are bounded. This has shown that for the dense subset of

, D ≡∪_kP_k,

sup|X (t)| < C (∥Y∥K′ ,∥X ∥K) t∈D

Now let

_k=1^∞ be linearly independent vectors of V whose span is dense in V . This is possible because V is separable. Then let _j=1^∞ be an orthonormal basis for H such that e_k ∈ span and each g_k ∈ span. This is done with the Gram Schmidt process. Then it follows that span is dense in V . I claim

∞∑ |y|2H = |〈y,ej〉|2. j=1

This is certainly true if y ∈ H because

〈y,e〉 = (y,e ) j j H

If y

H, then the series must diverge since otherwise, you could consider the infinite sum

∑∞ 〈y,ej〉 ej ∈ H j=1

because

||∑q ||2 ∑q || 〈y,ej〉ej|| = |〈y,ej〉|2 → 0 as p,q → ∞. ||j=p || j=p