10.4. LAPLACE TRANSFORMS 243

It remains to consider the other two integrals.

1k!

∫ t−δt k

0vke−v

(φ

( tvk

)−φ (t)

)dv+

1k!

∫∞

t+δt k

vke−v(

φ

( tvk

)−φ (t)

)dv (∗)

Now v→ vke−v is increasing for v < k. The first of these, on this interval, tvk ≤ t− δ and

so there is a constant C (t) such that∣∣φ ( tv

k

)−φ (t)

∣∣<C (t). Thus, using Stirling’s formula,the first integral is dominated by

C (t)1k!

∫ t−δt k

0vke−vdv ≤ 2C (t)

1√2πkk+1/2e−k

(t−δ

tk)(

t−δ

tk)k

e−(t−δ

t k)

< 2C (t)1√2π

√k(

eδt

(1− δ

t

))k

≡ 2C (t)1√2π

√krk

where r ≡ eδt

(1− δ

t

)which is less than 1 since δ < t. Then limk→∞

√krk = 0 by a use of

the root test. In the second integral of ∗,∣∣φ ( tv

k

)−φ (t)

∣∣≤C (t)em tvk . Then, simplifying this

second integral, it is dominated by C(t)k!∫

∞t+δ

t k vke−(1−mtk )vdv which converges to 0 as k→∞

by Lemma 10.4.7.I think the approach given above is especially interesting because it gives an explicit

description of φ (t) at most points1. I will next give a proof based on the Weierstrass approx-imation theorem to prove this major result which shows that the function is determined byits Laplace transform. I think it is easier to follow but lacks the explicit description givenabove. Later in the book is a way to actually compute the function with given Laplacetransform using a contour integral and the method of residues from complex analysis.

Lemma 10.4.9 Suppose q is a continuous function defined on [0,1] . Also suppose thatfor all n = 0,1,2, · · · ,

∫ 10 q(x)xndx = 0. Then it follows that q = 0.

Proof: By assumption, for p(x) any polynomial,∫ 1

0 q(x) p(x)dx = 0. Now let {pn (x)}be a sequence of polynomials which converge uniformly to q(x) by Theorem 6.10.2. Saymaxx∈[0,1] |q(x)− pn (x)|< 1

n Then

∫ 1

0q2 (x)dx =

∫ 1

0q(x)(q(x)− pn (x))dx+

=0︷ ︸︸ ︷∫ 1

0q(x) pn (x)dx

≤∫ 1

0|q(x)(q(x)− pn (x))|dx≤

∫ 1

0|q(x)|dx

1n

Since n is arbitrary, it follows that∫ 1

0 q2 (x)dx = 0 and by continuity, it must be the casethat q(x) = 0 for all x since otherwise, there would be a small interval on which q2 (x) ispositive and so the integral could not have been 0 after all.

Lemma 10.4.10 Suppose |φ (t)| ≤Ce−δ t for some δ > 0 and all t > 0 and also that φ

is continuous. Suppose that∫

∞

0 e−stφ (t)dt = 0 for all s > 0. Then φ = 0.

1If φ is piecewise continuous on every finite interval, this is obvious. However, the method will end up showingthis is true more generally.