1  Clearly the norm is non-negative. If ||f||=0, then f is constantly zero (since then f(0)=0 and f′(t)=0 everywhere). Also

||λ f || = |λ f(0)|+ sup|λ f′(t)|    = |λ| |f(0)|+|λ|sup|  f′(t)| =|λ| ||f||,

and

||f+g|| = |f(0)+g(0)| +sup|f′(t)+g′(t)|    ≤ |f(0)|+|g(0)|+sup|f′(t)|+sup|g′(t)|    = ||f||+||g||.

 2  Clearly the sum is absolutely convergent (since ∑1/n² < ∞), and checking the conditions:

⟨ y,x ⟩ = ⟨ x,y⟩,   ⟨λ x, y⟩ = λ ⟨ x,y ⟩,   ⟨ x+y, z ⟩=⟨ x,z⟩ + ⟨ y,z⟩,   and

⟨ x,x ⟩ > 0 except when x=0, when ⟨ x,x⟩ =0 ,

is fairly straightforward algebra.

Also ||(x_n)|| = ⟨(x_n),(x_n)⟩^1/2 = (∑_n=1^∞|x_n|²/n²)^1/2.

 3  With the usual inner product ⟨ f,g⟩=∫f ḡ d x, we have that |⟨ f,g⟩ | ≤ ||f|| ||g||, where g is the function g(x)=x. Now ||g||=1/√3 and this is the smallest possible constant C, since we do have ⟨ g,g⟩ =||g|| ||g||.

 4  I’ll omit the proof that this is a norm (but if it gives any trouble, ask me). The proof of completeness is a bit like the l₂ proof, only simpler. Suppose that (x⁽ⁿ⁾) is a Cauchy sequence in l_∞. Then, for each coordinate k, |x_k⁽ⁿ⁾−x_k^(m)| ≤ ||x⁽ⁿ⁾ − x^(m)||, and so (x_k⁽ⁿ⁾) is a Cauchy sequence of complex numbers, converging to x_k, say. Also |x_k⁽ⁿ⁾−x_k^(m)| < є for n and m greater than or equal to N_є, say. Letting m→∞ we get that |x_k⁽ⁿ⁾−x_k| ≤ є for n ≥ N_є, so (x⁽ⁿ⁾−x) ∈ l_∞ and hence x ∈ l_∞. Also ||x⁽ⁿ⁾−x|| ≤ є for n ≥ N_є, and so x⁽ⁿ⁾ → x.

 5  Clearly ⟨ y, x⟩=∑_k=1ⁿ w_k y_k x_k = ⟨ x,y⟩, and the properties ⟨ x+y, z⟩=⟨ x,z⟩ + ⟨ y, z⟩ and ⟨ λ x, y⟩ =λ ⟨ x, y⟩ are also straightforward to check. The norm produced is

||x||=⟨ x,x⟩1/2=

⎛ ⎜ ⎜ ⎝

n ∑ k=1

wk |xk|2

⎞ ⎟ ⎟ ⎠

1/2

,

which is strictly positive unless each x_k is zero. For the completeness, note that a Cauchy sequence (x^(m)) has the property that, for each є>0 there is a number M_є with ||x^(m)−x^(p)||<є for m, p ≥ M_є. That is,

n ∑ k=1

wk |xk(m)−xk(p)|2<є2. (107)

Thus w_k|x_k^(m)−x_k^(p)|² < є², which is enough to show that in the kth coordinate we have a Cauchy sequence of complex numbers.

Define a vector x ∈ ℂⁿ by x_k=lim_{m → ∞} x_k^(m) for each k. Now we have x^(m) → x, since

n ∑ k=1

wk |xk(m)−xk|2≤ є2,

for m ≥ M_є, as we see on letting p → ∞ in (107).

 6  Lin(e₁, e₂, …) = c₀₀ because a sequence is a finite linear combination of the e_i if and only if it has finitely many nonzero terms. Taking the closure we get all of l₂ since anything in l₂ is the limit of a sequence in c₀₀. This is so, because

||(x1, x2, …) − (x1, x2, …, xN, 0, 0, …)||2 =

∞ ∑ n=N+1

|xn|2,

which tends to zero as N → ∞. Finally CLin(e₂,e₃,…) is the same except that we only get sequences whose first term is zero, i.e. we get {x=(x_n) ∈ l₂: x₁ = 0}.

 7  Calculate ⟨ 1,t⟩, ⟨ 1,t²−1/3⟩ and ⟨ t,t²−1/3⟩: they are all zero. Clearly the set is a basis for P₂. Normalise the functions, to get e₁(t)=1/√2, e₂(t)=t√3/2 and e₃(t)=√45/8(t²−1/3), an orthonormal sequence. It now follows that, writing f(t)=t⁴, the best approximation is

g(t) =⟨ f,e1⟩ e1+⟨ f,e2⟩ e2+⟨ f,e3⟩ e3    = (2/5)(1/2)+(0)t+(45/8)(16/105)(t2−1/3) = (−3/35)+(6/7)t2.

As a cross-check, note that f−g is indeed orthogonal to g.

 8  ⟨ 1,1⟩ =2/3, so e₁(t)=√3/2. Now form f(t)=t−⟨ t,e₁⟩ e₁=t−⟨ t,1⟩ (3/2)=t−3/5. Since ⟨ f,f⟩ =8/175 we take e₂(t)=√175/8(t−3/5).

 9  The Gram–Schmidt process gives e₁=(1,1,0,0)/√2, then y₂=(1,0,0,−1)−(1,1,0,0)/2=(1/2,−1/2,0,−1), and then e₂=y₂/||y₂||=(1,−1,0,−2)/√6.

The plane P^⊥ consists of all vectors orthogonal to (1,1,0,0) and (1,0,0,−1), and hence is

{(x,y,z,w)∈ ℂ4: x+y=0, x−w=0},

with general solution (a,−a,b,a), and basis (1,−1,0,1) and (0,0,1,0), which are already orthogonal.

We can thus take e₃=(1,−1,0,1)/√3 and e₄=(0,0,1,0) as a basis for P^⊥.

Finally, (P^⊥)^⊥ consists of all vectors orthogonal to (1,−1,0,1) and (0,0,1,0), namely

{(x,y,z,w)∈ ℂ4: x−y+w=0, z=0},

to which the general solution is (−a+b,b,0,a), with basis (−1,0,0,1) and (1,1,0,0). We are clearly back at P.

 10  The given change of variables

x=

2π(t−a)   b−a

or     t=a+

(b−a)x  2π

takes t∈ [a,b] to x ∈ [0,2π]. We know that

1  2π

∫

2π 0

einxe−imx  d x=

⎧ ⎨ ⎩ 1 if   n=m,  0 if   n ≠ m,

so we obtain

1   2π

∫

b a

ein λ (t−a) e−im λ (t−a)

2 π  b−a

d t=

⎧ ⎨ ⎩ 1 if  n=m,  0 if   n ≠ m,

where λ = 2π/(b−a). Hence the functions e_n(t)=1 / √b−a e^{inλ t} form an orthonormal set in L₂[a,b]. They are even an orthonormal basis, since the same coordinate change shows that their closed linear span contains all f ∈ C[a,b] such that f(a)=f(b).

 11  By the Riesz–Fischer theorem, ∑c_ne_n converges to an L₂ function iff ∑|c_n|² < ∞. Here c_n=n^α√2π so we require ∑n^2α < ∞, i.e. α<−1/2.

 12  We calculate

⟨ f,en⟩ =

π ∫ −π

et e−int d t/

√

2π

= (−1)n

eπ−e−π   (1−in) √ 2π

.

Also, by Parseval’s identity

∞ ∑ n=−∞

|⟨ f,en⟩ |2 = ||f||22 =

π ∫ −π

e2t d t  = (e2π − e−2π)/2.

That is,

∞ ∑ n=−∞

(eπ−e−π)2  (1+n2)2π

=

(e2π−e−2π) 2

.

Thus

∞ ∑ n=−∞

1  1+n2

= π

e2π − e−2π   (eπ−e−π)2

,

which gives the result.

 13  To check that the sequence is orthonormal it is probably easiest to write

sn=

√

2

(en−e−n)/2i    and    cn=

√

2

(en+e−n)/2,

and use the orthonormality of (e_n) to calculate ⟨ s_n, s_m ⟩, ⟨ s_n, c_m ⟩ and ⟨ c_n, c_m ⟩.

Since exp(± int)=cos(nt)± isin(nt) and also cos(nt)=(exp(int)−exp(−int))/2 and sin(nt)=(exp(int)−exp(−int))/2i, the linear spans of (e_n) and {e₀,s₁,c₁,s₂,c₂,…} are the same. Hence their closed linear spans are the same, so {e₀,s₁,c₁,s₂,c₂,…} also forms an o.n.b.

As we have an orthonormal basis we have an expansion

f=⟨ f,e0⟩ e0 +

∞ ∑ 1

⟨ f,cn⟩ cn +

∞ ∑ 1

⟨ f,sn⟩ sn

converging in L₂. Hence

a0=(1/2π)

π ∫ −π

f(t)  d t,  an = (1/π)

π ∫ −π

f(t) cosnt  d t and  bn = (1/π)

π ∫ −π

f(t) sinnt  d t.

 14  Complex analysis method: Clearly there is a polynomial g such that f(z)=g′(z). By the fundamental theorem of the calculus, ∫g′ = 0 because the contour is closed. Or you could use Cauchy’s theorem.

Direct method:

∫

2π 0

f(eiθ) ieiθ d θ =

n ∑ k=0

ak

∫

2π 0

iei(k+1)θ d θ=0,

where f(z)=∑_k=0ⁿ a_kz^k.

Now

∫

|z|=1

z  d z =

∫

|z|=1

(1/z)  d z =

∫

2π 0

e−iθ ieiθ d θ= 2π i.

(Again, there are several ways of doing this, as above.) But if f_n → f uniformly then ∫f_n → ∫f, which is impossible if f_n are polynomials and f(z)=z.

 15  Since |α(f)|=|f(1/2)|≤ sup_[0,1]|f(x)|=||f||_∞, we have that ||α|| ≤ 1 (actually it equals 1) when we use the supremum norm.

Suppose now we define f_n (starting at n=2, say) to be zero except on

[1/2−1/n, 1/2+1/n],

piecewise linear on [1/2−1/n, 1/2] and [1/2, 1/2+1/n], with f_n(1/2)=A_n, where A_n is a positive number that we’ll choose in a minute. (The graph is going to be a thin steep triangle.) Now ||f_n||₂² ≤ (2/n) × A_n², since f_n is zero except on a set of length 2/n and always at most A_n (we could work it out exactly, but why bother?) So if we choose A_n=√n/2 we get ||f_n||₂ ≤ 1, and α(f_n)=A_n so α(f_n) → ∞, which means that α is unbounded in the L₂ norm.

 16  Clearly we get orthonormality—just compute ⟨ z^k, z^l ⟩. The fact that it is an orthonormal basis (i.e., complete) follows since the only function orthogonal to every z^k has all its coefficients zero, so is the 0 function.

Now α(∑a_nzⁿ)=∑a_n wⁿ, and this is ∑a_n b_n only if we take b_n=wⁿ for each n. Hence

g(z)=

∞ ∑ n=0

w

n zn = 1 /(1−

w

z).

Finally ||α||=||g||, so compute the H₂ norm of g to get

⎛ ⎜ ⎜ ⎝

∞ ∑ n=0

|

w

|2n

⎞ ⎟ ⎟ ⎠

1/2

=

⎛ ⎜ ⎜ ⎝

1  1−|w|2

⎞ ⎟ ⎟ ⎠

1/2

.

 17  The function J_x must be χ_[0,x], where

χ[0,x](t)=

⎧ ⎨ ⎩ 1 on   [0,x], 0 elsewhere.

It has L₂ norm equal to the square root of ∫₀^x 1²  d t, i.e., √x. Hence

|(Vf)(x)| ≤ ||χ[0,x]||2   ||f||2 =

√

x

||f||2.

Now integrate

∫

1 0

|(Vf)(x)|2  d x ≤

∫

1 0

x ||f||22   dx =

1   2

||f||22,  as required.

 18  The strategy here is to solve the equation ⟨ Ax, y⟩ = ⟨ x, A^* y⟩, etc.

(i) ⟨ Ax, y ⟩=x₁y₂/1+x₂y₃/2+…. This must be ⟨ x, A^* y⟩, and so

A* y=(y2/1, y3/2, y4/3, …).

(ii) ⟨ Rx, u ⟩ = ⟨ x,y ⟩  ⟨ z, u⟩ = ⟨ x, R^* u ⟩, where R^* u = ⟨ z, u⟩ y = ⟨ u, z ⟩ y.

(iii) Let’s take two vectors m+n and m′+n′, with m, m′∈ M and n, n′∈ M^⊥. Then

⟨ PM(m+n), m′+n′ ⟩ = ⟨ m, m′+n′⟩= ⟨ m,m′⟩.

This is the inner product ⟨ m+n,m′⟩ so P_M^*(m′+n′)=m′, which means that P_M^*=P_M again.

 19  Clearly

⟨ Uen, Uem⟩=

⎧ ⎨ ⎩ 1 if  m=n,  0 if  m ≠ n,

since U preserves the inner product (see notes). Also, if ⟨ x,Ue_n ⟩=0 for all n, then ⟨ U^*x, e_n ⟩=0 for all n, so U^*x=0, because (e_n) is an o.n.b., and so x=0. Hence (Ue_n) is an o.n.b.

To show that the bilateral right shift V is unitary, you can check any of the equivalent definitions. It is perhaps easiest just to observe that V is clearly a surjection and that ||Vx||=||x|| for all x. Alternatively, you can check that V^* is the bilateral left shift, i.e., V^*=V⁻¹, by using the identity

⟨ ⟨ ⟨ ⟨ ∞ ∑ n=−∞ xnen , V* ∞ ∑ m=−∞ ymem ⟩ ⟩ ⟩ ⟩ = ⟨ ⟨ ⟨ ⟨ V ∞ ∑ n=−∞ xnen , ∞ ∑ m=−∞ ymem ⟩ ⟩ ⟩ ⟩   = ⟨ ⟨ ⟨ ⟨ V ∞ ∑ n=−∞ xnen+1 ,  ∞ ∑ m=−∞ ymem ⟩ ⟩ ⟩ ⟩   = ∞ ∑ n=−∞ xn yn+1 ,

which tells you that

V*

∞ ∑ m=−∞

ymem =

∞ ∑ m=−∞

ym+1em.

We saw in the lectures that S is not unitary, since SS^* ≠ I.

 20  Use the definition of adjoint:

⟨ Mf g, h ⟩ = ⟨ g, Mf* h ⟩

for g, h ∈ L₂(−π,π), which means that

⟨ g, Mf* h ⟩ =

π ∫ −π

f(t) g(t)

h(t)

d t.

This is the inner product between g and the function taking values f(t)h(t), so that M_f^* g = M_f′g, where f′(t)=f(t).

Clearly M_f M_f^* g = M_f^* M_f g, and is the function whose value at t is f(t) f(t) g(t).

Now M_f is Hermitian if and only if M_f=M_f^*, or f=f′. So f must be real-valued.

Also M_f is unitary if and only if M_f^*=(M_f)⁻¹, which means that f(t)f(t)=1 for all t, i.e. |f(t)|=1 for all t.

 21  Calculate: (I+T+…+Tⁿ⁻¹)(I−T)=(I−T)(I+T+…+Tⁿ⁻¹)=I−Tⁿ=I. So we have an inverse for I−T.

Of course T/λ is also nilpotent, so I−T/λ is invertible, and so (multiplying by λ, which is nonzero), we have λ I−T invertible, and λ ∉σ(T).

The spectrum is nonempty, so can only be {0}; indeed it’s obvious that T is not invertible when Tⁿ=0. Hence r(T)=0 as well.

 22  (i) Since Te_k=λ_k e_k, where (e_n) is the usual orthonormal basis of l₂, we see that λ_k is an eigenvalue, with eigenvector e_k. Eigenvalues are always in the spectrum.

(ii) T−λ I takes (x_n) to ((λ_n−λ)x_n), and so its inverse must take (y_n) to (y_n/(λ_n−λ)). This is a bounded operator, since the sequence (λ_n−λ)⁻¹ is bounded when λ ∉Λ.

Now σ(T) is a closed set. It contains Λ, so contains Λ. Indeed σ(T)=Λ, as it contains no points outside Λ, by (ii). Thus any nonempty compact set is the spectrum of some operator!

 23  Note that STS⁻¹−λ I = S(T−λ I)S⁻¹, and so STS⁻¹−λ I is invertible if and only if T−λ I is invertible—indeed in that case its inverse would be S(T−λ I)⁻¹S⁻¹. Hence σ(T)=σ(STS⁻¹).

Also, if Tu=λ u, then STS⁻¹ (Su)=STu=λ Su, and Su ≠ 0 if u ≠ 0. Thus any eigenvector u of T corresponds to an eigenvector Su of STS⁻¹, and vice-versa.

 24  The fact that U is a bijection and an isometry follows from the fact that the functions e_n(t)=e^int/√2π, n ∈ ℤ form an orthonormal basis of L₂(−π,π) (see notes), so that a function f is in L₂(−π,π) if and only if f(t)=∑_−∞^∞a_n e_n, where (a_n) ∈ l₂, and also ||f||₂=||(a_n)||₂ (Parseval).

Now UVU⁻¹f= UVU⁻¹ ∑_n=−∞^∞a_n e_n = ∑_n=−∞^∞a_n e_n+1, because V is the shift.

But if f(t)=∑_n=−∞^∞a_n e_n(t), then the function ∑_n=−∞^∞a_n e_n+1(t) is just f(t)e^it, since e_n(t)e^it=e_n+1(t) for all t.

Now we work with the operator T=UVU⁻¹ = M_e on L₂(−π,π), where e(t)=e^it. Using Question 21, we see that this is a unitary operator and so σ(T) ⊆ T, but we can argue more directly.

The operator (T−λ I) is multiplication by e^it−λ and its inverse, if it exists, is multiplication by h_λ(t)=1/(e^it−λ). For λ ∉T, h_λ∈ C[−π,π] and so T−λ I has a bounded inverse. However, if λ ∈ T, then multiplication by h_λ does not give a bounded operator (indeed, M_hλ e₀=h_λ/√2π, which is not even in L₂). Hence σ(V)=σ(T)=T.

Also T has no eigenvalues, as, no matter which λ ∈ ℂ we choose, there will be no nonzero function f such that f(t)e^it=λ f(t) for all t. Hence V has no eigenvalues either, by Question 24.

 25  If T₁ and T₂ are compact, and (x_n) is bounded, then we can find a subsequence (x_n(k)) of (x_n) such that (T₁x_n(k)) converges, and a further subsequence (x_n(k(l))) such that both (T₁x_n(k(l))) and (T₂x_n(k(l))) converge. Then ((a₁T₁+a₂T₂)x_n(k(l))) converges for any a₁, a₂ ∈ ℂ, so a₁T₁+a₂T₂ is compact. Since the norm limit of compact operators is compact, they form a closed subspace.

Given (x_n) bounded, we can find a subsequence (x_n(k)) such that (Tx_n(k)) converges, and hence so does (ATx_n(k)), since A is continuous; hence AT is compact. Also (Ax_n) is bounded so there is a subsequence of (TAx_n) that converges, and TA is compact.

 26  ∑_n=1^∞||Ae_n||² = ∑_n=1^∞∑_m=1^∞|⟨ Ae_n, f_m ⟩|², since (f_m) is an o.n.b. This equals

∞ ∑ n=1

∞ ∑ m=1

|⟨ en, A*fm ⟩|2,

or, summing over n first, ∑_m=1^∞||A^*f_m||², since (e_n) is also an o.n.b. Since the right hand side of the displayed formula doesn’t mention (e_n) it clearly makes no difference if we replace (e_n) by a different o.n.b. It is also clear that ||A||_HS=||A^*||_HS as the LHS is just ||A||_HS² and the RHS is ||A^*||_HS².

 27  (a) ⟨ T^*Tx,y⟩=⟨ Tx, Ty⟩=⟨ x,T^*T y⟩, so T^*T is Hermitian.

Also ⟨ TT^*x,y⟩=⟨ T^*x,T^*y⟩ = ⟨ x,T^**T^*y⟩=⟨ x,TT^*y⟩, since T=T^**, and hence TT^* is Hermitian. Both are compact, since the product of a compact operator and a bounded operator is always compact (by Question 1).

(b) The point is that Te_n=α_nf_n, and so ∑||Te_n||² = ∑|α_n|²< ∞ if and only if (α_n) ∈ l₂.

If α_n → 0, then T is the limit of finite rank operators T_mx=∑_n=1^m α_n ⟨ x,e_n⟩ f_n (cf. what we proved in the course about diagonal operators), and if α_n ¬→0, then, for some δ>0, ||Te_n(k)||=|α_n(k)| ≥ δ, and (Te_n(k)) has no convergent subsequence—again, see how we did this for diagonal operators.

⟨ Ten, fm ⟩ = ⟨ en, T* fm ⟩=

⎧ ⎨ ⎩ αn if  n=m,  0 otherwise.

Hence T^* maps f_m to α_me_m, so T^*x=∑_m=1^∞α_m⟨ x,f_m⟩ e_m.

This gives

T*Tx=

∞ ∑ n=1

αn ⟨ x,en⟩ T*fn=

∞ ∑ n=1

|αn|2 ⟨ x,en⟩ en,

and

TT*x=

∞ ∑ m=1

αm

⟨ x,fm⟩ Tem=

∞ ∑ m=1

|αm|2⟨ x,fm⟩ fm.

 28  The Neumann series is

(I−λ T)−1=1+λ T + λ T2 + …,

valid for sufficiently small λ (e.g. |λ|  ||T|| < 1).

Taking f(x)=x, we find that (Tf)(x)=∫₀¹ xy³   dy=x/4, and in general (Tⁿf)(x)=x/4ⁿ.

The solution we obtain is φ=(1−λ T)⁻¹f, which gives

φ(x)=x+λ x/4 + λ2 x2/16 + …,

which converges to

φ(x)=x/(1−λ/4)=4x/(4−λ),

at least for |λ| < 4. It is easily seen that this solution is valid for all λ ≠ 4.

 29  Calculate

(Tφk)(x)=

π ∫ −π

h(x−y) eiky  d y.

Make the change of variables t=x−y to get

(Tφk)(x)=

∫

x+π x−π

h(t)eik(x−t)  d t = 2π ak eikx,

using orthogonality and periodicity properties, so that φ_k is an eigenvector with eigenvalue 2π a_k.

Now T is a Hilbert–Schmidt operator with an orthonormal basis of eigenvectors, namely (e_k)=(φ_k)/√2π. We can now work with either the (e_k) or the (φ_k). If φ has Fourier series ∑_n=−∞^∞d_n e^int, then φ−λ Tφ has Fourier series

∞ ∑ n=−∞

cneint=

∞ ∑ n=−∞

dn(1−λ λn) eint,

where

λn=	⎧ ⎨ ⎩ 0 if  n=0, 2π /|n| if n ≠ 0,

so the solution is

φ(t)=

∞ ∑ n=−∞

cn   1−λ λn

eint.

 30  Take e₁=t/||t||, and ||t||²=∫₋₁¹ t²  d t = 2/3, so e₁(t)=√3/2 t.

Let w₂(t)=t²−⟨ t²,e₁⟩ e₁=t², and normalize to get e₂(t)=√5/2 t².

Let w₃(t)=t⁴−⟨ t⁴,e₁⟩ e₁ − ⟨ t⁴,e₂⟩ e₂ = t⁴ − 0 − (2/7)(5/2)t² = t⁴− 5t²/7. Now

||w3||2=

∫

1 −1

(t8−10t6/7+25t4/49)  d t= (2/9)−(20/49)+(10/49)= 8/441,

so we take e₃(t)=(21/√8)(t⁴−5t²/7).

The best approximation in X to f is g=∑_k=1³ ⟨ f,e_k⟩ e_k, giving

g(t)=0 + (2/3)(5/2)t2 + (−8/105)(441/8)(t4−5t2/7),

which reduces to g(t)=14t²/3−21t⁴/5.

As a check, note that f−g is orthogonal to each of the functions t, t² and t⁴.

 31  We see that φ_n(x)=⟨ x, u_n⟩, where u_n=(1,1,…,1,0,0,…), with n nonzero terms. Now ||φ_n||=||u_n||=√n.

 32  To get the adjoint calculate

⟨ (A+iB)x,y⟩ = ⟨ Ax,y⟩ + i⟨ Bx,y⟩ = ⟨ x,Ay⟩ + i⟨ x,By⟩ = ⟨ x, (A−iB)y ⟩,

so T^*=A−iB.

Now A=(T+T^*)/2 and B=(T−T^*)/(2i) (very like the formulae for real and imaginary parts of a complex number).

Since these formulae do define self-adjoint operators A and B, it is clear that every operator T has a unique decomposition as T=A+iB.

Note

T*T=(A−iB)(A+iB)=A2−iBA+iAB+B2

and

TT*=(A+iB)(A−iB)=A2+iBA−iAB+B2,

so that T^*T−TT^*=2i(AB−BA), and T is normal if and only if AB=BA.

 33  All we need to do is look for eigenvalues, as the spaces are finite-dimensional.

(i) Df=λ f is impossible unless λ=0, since the degree of Df is lower than the degree of f. So σ(D)={0}. Indeed Dⁿ⁺¹=0, so D is nilpotent, which also implies that its spectrum is {0}, see earlier examples sheets. The only eigenvectors in P_n are constant functions, so we do not get a basis of eigenvectors.

(ii) D(e^ikt)=ike^ikt, so σ(D)={0,± i, ± 2i, …, ± ni}. Now D has (2n+1) distinct eigenvalues, and T_n has an orthonormal basis of eigenvectors, namely (e^ikt/√2π)_k=−nⁿ.

 34  We get φ=(I−λ T)⁻¹f=f+λ Tf + λ² T² f + ….

Now Tf(x)=∫₀^x t²  d t=x³/3, (T²f)(x)=∫₀^x (t⁵/3)  d t=x⁶/18, ….

In general (Tⁿ f)(x)=x³ⁿ/(3ⁿ n!). Summing the series we find that

φ(x)=exp(λ x3/3),

and the series converges for all λ ∈ ℂ.