340 CHAPTER 13. NORMS

Hence this scalar valued function equals a constant and so the constant must be (x, y)X .Hence for all x, y, (Φ (−t) Φ (t)x− x, y)X = 0 for all x, y and this is so in particular fory = Φ(−t) Φ (t)x− x which shows that Φ (−t) Φ (t) = I. ■

As a special case, suppose λ ∈ C and consider

∞∑k=0

tkλk

k!

where t ∈ R. In this case, Ak = tkλk

k! and you can think of it as being in L (C,C). Then thefollowing corollary is of great interest.

Corollary 13.4.4 Let

f (t) ≡∞∑k=0

tkλk

k!≡ 1 +

∞∑k=1

tkλk

k!

Then this function is a well defined complex valued function and furthermore, it satisfies theinitial value problem,

y′ = λy, y (0) = 1

Furthermore, if λ = a+ ib,|f | (t) = eat.

Proof: The first part is a special case of the above theorem. Note that for f (t) =u (t) + iv (t) , both u, v are differentiable. This is because

u =f + f

2, v =

f − f

2i.

Then from the differential equation,

(a+ ib) (u+ iv) = u′ + iv′

and equating real and imaginary parts,

u′ = au− bv, v′ = av + bu.

Then a short computation shows(u2 + v2

)′= 2uu′ + 2vv′ = 2u (au− bv) + 2v (av + bu) = 2a

(u2 + v2

)(u2 + v2

)(0) = |f |2 (0) = 1

Now in general, ify′ = cy, y (0) = 1,

with c real it follows y (t) = ect. To see this,

y′ − cy = 0

and so, multiplying both sides by e−ct you get

d

dt

(ye−ct

)= 0

and so ye−ct equals a constant which must be 1 because of the initial condition y (0) = 1.Thus (

u2 + v2)(t) = e2at

and taking square roots yields the desired conclusion. ■