18.4. MEAN ERGODIC THEOREM 495

and so∥T ∗ms∥ ≤ K1/p ∥s∥

and so the density of the simple functions implies that ∥T ∗m∥ ≤ K1/p.Next let

M ≡{

g ∈ Lp (Ω) : ||Ang||p→ 0}

It follows from 18.4.10 that M is a closed subspace of Lp (Ω) containing (I−T ∗)(Lp (Ω)).This is shown next.

Claim 1: M is a closed subspace which contains (I−T ∗m)(Lp (Ω)).First it is shown that this is true if m = 1 and then it will be observed that the same

argument would work for any positive integer m.

An ( f −T ∗ f ) ≡ 1n

n−1

∑k=0

T ∗k f −T ∗k+1 f

=1n

n−1

∑k=0

T ∗k f − 1n

n

∑k=1

T ∗k f =1n( f −T ∗n f )

Hence

∥An ( f −T ∗ f )∥p ≤1n

(∥ f∥p +∥T

∗n f∥p

)≤ 1

n

(∥ f∥p +K1/p ∥ f∥p

)and this clearly converges to 0. In fact, the same argument shows that M contains the set(I−T ∗m)(Lp (Ω)) for any m. Now suppose gn ∈M and gn→ g. Does it follow that g ∈Malso? Note that T ∗m is clearly linear. Thus

∥T ∗mg∥ ≤ ∥T ∗mg−T ∗mgn∥+∥T ∗mgn∥ ≤ K1/p ∥g−gn∥+∥T ∗mgn∥

Now pick n large enough that ∥gn−g∥< ε/(2K1/p

)so that

∥T ∗mg∥ ≤ ε

2+∥T ∗mgn∥

Then for all m large enough, the right side of the above is less than ε and this shows thatg ∈M. Note that M is also a subspace and so it is a closed subspace.

Claim 2: If Ank f → g weakly and Amk f → h weakly, then g = h.It is first shown that if ξ ∈ Lp′ (Ω) and

∫ξ gdµ = 0 for all g ∈M, then

∫ξ (g−h)dµ =

0.If ξ ∈ Lp′ (Ω) is such that

∫ξ gdµ = 0 for all g∈M, then since M ⊇ (I−T ∗n)(Lp (Ω)),

it follows that for all k ∈ Lp (Ω),∫ξ kdµ =

∫(ξ T ∗nk+ξ (I−T ∗n)k)dµ =

∫ξ T ∗nkdµ

and so from the definition of An as an average, for such ξ ,∫ξ kdµ =

∫ξ Ankdµ. (18.4.13)