19.7. COMPACT OPERATORS 541

and if it is ever the case that λ n = 0, it follows from the above argument that the conclusionof the theorem is obtained.

I claim limn→∞ λ n = 0. If this were not so, then for some ε > 0, 0 < ε = limn→∞ |λ n|but then

∥Aun−Aum∥2 = ∥λ nun−λ mum∥2

= |λ n|2 + |λ m|2 ≥ 2ε2

and so there would not exist a convergent subsequence of {Auk}∞

k=1 contrary to the assump-tion that A is compact. This verifies the claim that limn→∞ λ n = 0.

It remains to verify that span({ui}) is dense in A(H). If w ∈ span({ui})⊥ then w ∈ Hnfor all n and so for all n,

∥Aw∥ ≤ ∥An∥∥w∥ ≤ |λ n|∥w∥ .Therefore, Aw = 0. Now every vector from H can be written as a sum of one from

span({ui})⊥ = span({ui})⊥

and one from span({ui}). Therefore, if x ∈ H, x = y+w where y ∈ span({ui}) and w ∈span({ui})

⊥and Aw = 0. Also, since y ∈ span({ui}), there exist constants, ck and n such

that ∥∥∥∥∥y−n

∑k=1

ckuk

∥∥∥∥∥< ε, (w,uk) = 0 for all uk.

Therefore, from Corollary 19.5.3,∥∥∥∥∥y−n

∑k=1

(y,uk)uk

∥∥∥∥∥=∥∥∥∥∥y−

n

∑k=1

(x,uk)uk

∥∥∥∥∥< ε.

Therefore,

∥A∥ε >

∥∥∥∥∥A

(y−

n

∑k=1

(x,uk)uk

)∥∥∥∥∥=∥∥∥∥∥Ax−

n

∑k=1

(x,uk)λ kuk

∥∥∥∥∥ .Since ε is arbitrary, this shows span({ui}) is dense in A(H) and also implies 19.7.31.

Define v⊗u ∈L (H,H) by

v⊗u(x) = (x,u)v,

then 19.7.31 is of the form

A =∞

∑k=1

λ kuk⊗uk

This is the content of the following corollary.

Corollary 19.7.3 The main conclusion of the above theorem can be written as

A =∞

∑k=1

λ kuk⊗uk

where the convergence of the partial sums takes place in the operator norm.