496 CHAPTER 18. TOPOLOGICAL VECTOR SPACES

Since Ank f → g weakly and Amk f → h weakly. Then 18.4.13 shows that∫ξ gdµ = lim

k→∞

∫ξ Ank f dµ =

∫ξ f dµ = lim

k→∞

∫ξ Amk f dµ =

∫ξ hdµ. (18.4.14)

Thus for these special ξ , it follows that∫ξ (g−h)dµ = 0. (18.4.15)

Next observe that for each fixed n, if nk→ ∞,

limk→∞

∥∥T ∗nAnk f −Ank f∥∥= 0 (18.4.16)

this follows like the arguments given above in Claim 1. Note that if L ∈ L (X ,X) andxn→ x weakly in X , then for φ ∈ X ′

⟨φ ,Lxn⟩= ⟨L∗φ ,xn⟩ → ⟨L∗φ ,x⟩= ⟨φ ,Lx⟩

and so Lxn → Lx weakly. Therefore, this simple observation along with the above strongconvergence 18.4.16 implies

T ∗ng = weak limk→∞

T ∗nAnk f = weak limk→∞

Ank f = g.

Similarly T ∗nh = h where Amk f → h weakly. It follows that An (g−h) = g−h so if g ̸= h,then g−h /∈M because

An (g−h)→ g−h ̸= 0.

It follows that since M is a closed subspace, there exists ξ ∈ Lp′ (Ω) such that∫ξ (g−h)dµ ̸= 0

but∫

ξ kdµ = 0 for all k ∈M, contradicting 18.4.15. This verifies Claim 2.Now

∥An f∥p =

(∫ ∣∣∣∣∣1n n−1

∑k=0

f(

T kω

)∣∣∣∣∣p

dµ

)1/p

≤ 1n

n−1

∑k=0

(∫ ∣∣∣ f (T kω

)∣∣∣p dµ

)1/p

=1n

n−1

∑k=0

∥∥∥T ∗k f∥∥∥

p≤ 1

n

n−1

∑k=0

K1/p ∥ f∥p = K1/p ∥ f∥p (18.4.17)

Hence, by the Eberlein Smulian theorem, Theorem 17.5.12, in case p > 1, there is a sub-sequence for which An f converges weakly in Lp (Ω). From the above, it follows that theoriginal sequence must converge. That is, An f converges weakly for each f ∈ Lp (Ω). LetA f denote this weak limit. Then it is clear that A is linear because this is true for eachAn. What of the claim about the estimate? From weak lower semicontinuity of the norm,Corollary 18.2.12,

∥A f∥p ≤ lim infn→∞∥An f∥ ≤ K1/p ∥ f∥p