Proof.

1. Obviously:

   x∈kerT ⇔ ⟨ Tx,Tx  ⟩=0 ⇔ ⟨ T*Tx,x  ⟩=0    ⇔ ⟨ TT*x,x  ⟩=0 ⇔  ⟨ T*x,T*x  ⟩=0    ⇔ x∈kerT*.

   The second part holds because normalities of T and T−λ I are equivalent.
2. If Tx=λ x, Ty=µ y then from the previous statement T^* y =µy. If λ≠µ then the identity

   λ⟨ x,y  ⟩=⟨ Tx,y  ⟩ =⟨ x,T*y  ⟩=µ⟨ x,y  ⟩

   implies ⟨ x,y ⟩=0.
3. Let S=T^*T, then S is Hermitian (check!). Consequently, inequality

   ⎪⎪ ⎪⎪

   Sx

   ⎪⎪ ⎪⎪

   2=⟨ Sx,Sx  ⟩=⟨ S2x,x  ⟩≤

   ⎪⎪ ⎪⎪

   S2

   ⎪⎪ ⎪⎪

   ⎪⎪ ⎪⎪

   x

   ⎪⎪ ⎪⎪

   2

   implies ||S||²≤ ||S²||. But the opposite inequality follows from the Theorem 12, thus we have the equality ||S²||=||S||² and more generally by induction: ||S^2m||=||S||^2m for all m.

   Now we claim ||S||=||T||². From Theorem 12 and 18 we get ||S||=||T^*T||≤ ||T||². On the other hand if ||x||=1 then

   ⎪⎪ ⎪⎪

   T*T

   ⎪⎪ ⎪⎪

   ≥

   ⎪ ⎪

   ⟨ T*Tx,x  ⟩

   ⎪ ⎪

   =⟨ Tx,Tx  ⟩=

   ⎪⎪ ⎪⎪

   Tx

   ⎪⎪ ⎪⎪

   2

   implies the opposite inequality ||S||≥||T||². Only now we use normality of T to obtain (T^2m)^*T^2m=(T^*T)^2m and get the equality

   ⎪⎪ ⎪⎪

   T2m

   ⎪⎪ ⎪⎪

   2=

   ⎪⎪ ⎪⎪

   (T*T)2m

   ⎪⎪ ⎪⎪

   =

   ⎪⎪ ⎪⎪

   T*T

   ⎪⎪ ⎪⎪

   2m =

   ⎪⎪ ⎪⎪

   T

   ⎪⎪ ⎪⎪

   2m+1.

   Thus:

   r(T)=

   lim m→∞

   ⎪⎪ ⎪⎪

   T2m

   ⎪⎪ ⎪⎪

   1/2m=

   lim m→∞

   ⎪⎪ ⎪⎪

   T

   ⎪⎪ ⎪⎪

   2m+1/2m+1 =

   ⎪⎪ ⎪⎪

   T

   ⎪⎪ ⎪⎪

   .

   by the spectral radius formula (44).

□

Proof.

1. Let H₀ be the closed linear span of eigenvectors of T. Then T restricted to H₀ is a diagonal compact operator with the same set of eigenvalues λ_n as in H. Then λ_n→ 0 from Proposition 8 .
   Exercise 5  Use the proof of Proposition 8 to give a direct demonstration.
   Proof.[Solution] Or straightforwardly assume opposite: there exist an δ>0 and infinitely many eigenvalues λ_n such that | λ_n |>δ. By the previous Theorem there is an orthonormal sequence v_n of corresponding eigenvectors T v_n=λ_n v_n. Now the sequence (v_n) is bounded but its image T v_n=λ _n e_n has no convergent subsequence because for any k≠ l:

   ⎪⎪ ⎪⎪

   λ kvk−λ lel

   ⎪⎪ ⎪⎪

   =  (

   ⎪ ⎪

   λ k

   ⎪ ⎪

   2 +

   ⎪ ⎪

   λl

   ⎪ ⎪

   2)1/2≥

   √

   2

   δ ,

   i.e. T e_nk is not a Cauchy sequence, see Figure 16. □
2. Similarly if H₀=ker(T−λ I) is infinite dimensional, then restriction of T on H₀ is λ I—which is non-compact by Proposition 8. Alternatively consider the infinite orthonormal sequence (v_n), Tv_n=λ v_n as in Exercise 5.

□

Proof. Assume without lost of generality that T≠ 0. Let λ∈σ(T), without lost of generality (multiplying by a scalar) λ=1.

We claim that if 1 is not an eigenvalue then there exist δ>0 such that

⎪⎪ ⎪⎪

(I−T)x

⎪⎪ ⎪⎪

≥ δ

⎪⎪ ⎪⎪

x

⎪⎪ ⎪⎪

. (51)

Otherwise there exists a sequence of vectors (x_n) with unit norm such that (I−T)x_n→ 0. Then from the compactness of T for a subsequence (x_nk) there is y∈ H such that Tx_nk → y, then x_n→ y implying Ty=y and y≠ 0—i.e. y is eigenvector with eigenvalue 1.

Now we claim Im (I−T) is closed, i.e. y∈Im(I−T) implies y∈Im(I−T). Indeed, if (I−T)x_n → y, then there is a subsequence (x_nk) such that Tx_nk→ z implying x_nk→ y+z, then (I−T)(z+y)=y by continuity of I−T.

Finally I−T is injective, i.e ker(I−T)={0}, by (51). By the property 1, ker(I−T^*)={0} as well. But because always ker(I−T^*)=Im(I−T)^⊥ (by 2) we got surjectivity, i.e. Im(I−T)^⊥={0}, of I−T. Thus (I−T)⁻¹ exists and is bounded because (51) implies ||y||>δ ||(I−T)⁻¹y||. Thus 1∉σ(T).

The existence of eigenvalue λ such that | λ |=||T|| follows from combination of Lemma 13 and Theorem 3. □

Proof. Suppose T≠ 0. Then by the previous Theorem there exists an eigenvalue λ₁ such that | λ₁ |=||T|| with corresponding eigenvector e₁ of the unit norm. Let H₁=Lin(e₁)^⊥. If x∈ H₁ then

⟨ Tx,e1  ⟩=⟨ x,T*e1  ⟩=⟨ x,λ1 e1  ⟩=λ1⟨ x,e1  ⟩=0, (53)

thus Tx∈ H₁ and similarly T^* x ∈ H₁. Write T₁=T|_H1 which is again a normal compact operator with a norm does not exceeding ||T||. We could inductively repeat this procedure for T₁ obtaining sequence of eigenvalues λ₂, λ₃, …with eigenvectors e₂, e₃, …. If T_n=0 for a finite n then theorem is already proved. Otherwise we have an infinite sequence λ_n→ 0. Let

x=

n ∑ 1

⟨ x,ek  ⟩ek +yn  ⇒

⎪⎪ ⎪⎪

x

⎪⎪ ⎪⎪

2=

n ∑ 1

⎪ ⎪

⟨ x,ek  ⟩

⎪ ⎪

2 +

⎪⎪ ⎪⎪

yn

⎪⎪ ⎪⎪

2 ,    yn∈ Hn,

from Pythagoras’s theorem. Then ||y_n||≤ ||x|| and ||T y_n||≤ ||T_n||||y_n||≤ | λ_n |||x||→ 0 by Lemma 3. Thus

T x =

lim n→ ∞

⎛ ⎜ ⎜ ⎝

n ∑ 1

⟨ x,en  ⟩ Ten + Tyn

⎞ ⎟ ⎟ ⎠

=

∞ ∑ 1

λn⟨ x,en  ⟩ en

Conversely, if T x = ∑₁^∞λ_n⟨ x,e_n ⟩ e_n then

⟨ Tx,y  ⟩=

∞ ∑ 1

λn⟨ x,en  ⟩ ⟨ en,y  ⟩ =

∞ ∑ 1

⟨ x,en  ⟩ λn

⟨ y,en  ⟩

,

thus T^* y = ∑₁^∞λ_n⟨ y,e_n ⟩ e_n. Then we got the normality of T: T^*Tx=TT^*x= ∑₁^∞| λ_n |²⟨ y,e_n ⟩ e_n. Also T is compact because it is a uniform limit of the finite rank operators T_nx=∑₁ⁿ λ_n⟨ x,e_n ⟩e_n. □

Proof. Let (e_n) be the orthonormal sequence constructed in the proof of the previous Theorem. Then x is perpendicular to all e_n if and only if its in the kernel of T. Let (f_n) be any orthonormal basis of kerT. Then the union of (e_n) and (f_n) is the orthonormal basis (g_n) we have looked for. □

Proof. Operator T^*T is compact and Hermitian (hence normal). From the previous Corollary there is an orthonormal basis (e_k) such that T^*T x= ∑_n λ_n⟨ x,e_k ⟩e_k for some positive λ_n=||T e_n||². Let µ_n=||Te_n|| and f_n=Te_n/µ_n. Then f_n is an orthonormal sequence (check!) and

Tx=

∑ n

⟨ x,en  ⟩ Ten =

∑ n

⟨ x,en  ⟩ µn fn.

□

Proof. Sufficiency follows from 9.
Necessity: by the previous Corollary Tx =∑_n ⟨ x,e_n ⟩ µ_n f_n thus T is a uniform limit of operators T_m x=∑_n=1^m ⟨ x,e_n ⟩ µ_n f_n which are of finite rank. □