11.2. LAPLACE TRANSFORM 267

Now it follows from that section that the integrand converges to −te−stφ (t). Therefore, itonly is required to obtain an estimate and use the dominated convergence theorem. Thene−st e−ht−1

h φ (t) =∣∣∣∣e−(s+h)t 1− eht

hφ (t)

∣∣∣∣≤Ceλ t∣∣∣∣1− eht

h

∣∣∣∣e−(s+h) ≤Cte−(s+h−λ )t∣∣∣∣1− eht

th

∣∣∣∣Now

∣∣∣e−(s+h−λ )t∣∣∣ = e−(Re(s+h)−λ )t ≤ e−εt whenever |h| is small enough because of the

assumption that Res > λ . eht −1 =∫ t

0 hehudu and so the last expression is∣∣∣∣∫ t0 hehudu

th

∣∣∣∣≤ 1t

∣∣∣∣∫ t

0ehudu

∣∣∣∣≤ 1t

∫ t

0e|h|udu≤ et|h|,

and so for |h| small enough, say smaller than ε/2, the integrand in 11.1 is no larger thanCte−εte

ε2 t =Cte−

ε2 t which is in L1 and so the dominated convergence theorem applies and it

follows that f ′ (s) =∫

∞

0 (−t)e−stφ (t)dt whenever Re(s)> λ . Continuing similarly, yieldsall the derivatives. However, the existence of all the derivatives will follow from generalresults on analytic functions presented later. ■

The whole approach for Laplace transforms in differential equations is based on theassertion that if L ( f ) = L (g) , then f = g. However, this is not even true because if youchange the function on a set of measure zero, you don’t change the transform. However, iff ,g are continuous, then it will be true. Actually, it is shown here that if L ( f ) = 0, andf is continuous, then f = 0. The approach here is based on the Weierstrass approximationtheorem or rather a case of it.

Lemma 11.2.3 Suppose q is a continuous function defined on [0,1] . Also suppose thatfor all n = 0,1,2, · · · , that

∫ 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 Corollary 3.1.3. Saymaxx∈[0,1] |q(x)− pn (x)|< 1

n . Then from this uniform convergence, of pn to q,

∫ 1

0q2 (x)dx = lim

n→∞

∫ 1

0q(x) pn (x)dx = 0.

By continuity, it must be the case that q(x) = 0 for all x since otherwise, there would be asmall interval on which q2 (x) is positive and so the integral could not have been 0 after all.■

Lemma 11.2.4 Suppose | f (t)| ≤Ce−δ t for some δ > 0 and all t > 0 and also that f iscontinuous. Suppose that

∫∞

0 e−st f (t)dt = 0 for all s > 0. Then f = 0.

Proof: First note that limt→∞ | f (t)| = 0. Next change the variable letting x = e−t andso x ∈ [0,1]. Then this reduces to

∫ 10 xs−1 f (− ln(x))dx. Now if you let q(x) = f (− ln(x)) ,

it is not defined when x = 0, but x = 0 corresponds to t → ∞. Thus limx→0+ q(x) = 0.Defining q(0)≡ 0, it follows that it is continuous and letting s−1 be various integers, forall n = 0,1,2, · · · ,

∫ 10 xnq(x)dx = 0 and so q(x) = 0 for all x from Lemma 11.2.3. Thus

f (− ln(x)) = 0 for all x ∈ (0,1] and so f (t) = 0 for all t ≥ 0. ■