Watson’s lemma and Laplace’s method

Jordan Bell
June 17, 2014

1 Watson’s lemma

Our proof of Watson’s lemma follows Miller.11 1 Peter D. Miller, Applied Asymptotic Analysis, p. 53, Proposition 2.1.

Theorem 1 (Watson’s lemma).

Suppose that T>0, that ϕ:RC belongs to L1([0,T]), that σ>-1, and that g(t)=t-σϕ(t) is C on some neighborhood of 0. Then F:(0,)C defined by

F(λ)=0Te-λtϕ(t)𝑑t

satisfies

F(λ)n=0g(n)(0)Γ(σ+n+1)n!λσ+n+1,λ.
Proof.

Take g to be C on some interval with left endpoint <0 and right endpoint s, 0<s<T. For p a nonnegative integer and λ>1, define

Fp(λ)=0se-λttσ+p𝑑t,

which satisfies, doing the change of variable τ=λt,

Fp(λ) = 0e-λttσ+p𝑑t-se-λttσ+p𝑑t
= λ-(σ+p+1)0e-ττσ+p𝑑τ-se-λttσ+p𝑑t
= λ-(σ+p+1)Γ(σ+p+1)-se-λttσ+p𝑑t.

Using the Cauchy-Schwarz inequality,

se-λttσ+p𝑑t = se-λt/2e-λt/2tσ+p𝑑t
(se-λt𝑑t)1/2(se-λtt2σ+2p𝑑t)1/2
= e-λs/2λ-1/2(se-λtt2σ+2p𝑑t)1/2
< e-λs/2(0e-tt2σ+2p𝑑t)1/2
= e-λs/2Γ(2σ+2p+1)1/2.

For any nonnegative integer m we have e-λs/2=om(λ-(σ+m+1)) as λ, hence, dealing with Γ(2σ+2p+1) merely as a constant depending on p,

Fp(λ)=λ-(σ+p+1)Γ(σ+p+1)+om,p(λ-(σ+m+1)) (1)

as λ.

Write

F(λ)=0se-λtϕ(t)𝑑t+sTe-λtϕ(t)𝑑t.

One the one hand,

|sTe-λtϕ(t)𝑑t|sTe-λt|ϕ(t)|𝑑te-λssT|ϕ(t)|𝑑te-λsϕL1,

which shows that for any nonnegative integer n,

sTe-λtϕ(t)𝑑t=on(λ-(σ+n+1))

as λ.

One the other hand, for each nonnegative integer N, Taylor’s theorem tells us that the function rN:(r,s) defined by

rN(t)=g(t)-n=0Ng(n)(0)n!tn,t(r,s),

satisfies

|rN(t)|sup|g(N+1)(τ)||t|N+1(N+1)!,

where the supremum is over those τ strictly between 0 and t. Then for t(0,s),

|rN(t)|sup0<τ<s|g(N+1)(τ)|tN+1(N+1)!.

Using the definition of rN,

0se-λtϕ(t)𝑑t = 0se-λttσn=0Ng(n)(0)n!tndt+0se-λttσrN(t)𝑑t
= n=0Ng(n)(0)n!Fn(λ)+0se-λttσrN(t)𝑑t

and using the inequality for rN(t),

|0se-λttσrN(t)𝑑t| 0se-λttσ|rN(t)|𝑑t
sup0<τ<s|g(N+1)(τ)|1(N+1)!0se-λttσ+N+1𝑑t
= sup0<τ<s|g(N+1)(τ)|1(N+1)!FN+1(λ)dt.

Using this and (1),

0se-λttσrN(t)𝑑t=ON(λ-(σ+N+2)).

Putting together what we have shown, for any nonnegative integer N, as λ,

F(λ) = n=0Ng(n)(0)n!Fn(λ)+ON(λ-(σ+N+2))+ON(λ-(σ+N+2))
= n=0Ng(n)(0)n!λ-(σ+n+1)Γ(σ+n+1)+n=0Ng(n)(0)n!oN,n(λ-(σ+N+1))
+ON(λ-(σ+N+2))
= n=0Ng(n)(0)n!λ-(σ+n+1)Γ(σ+n+1)+oN(λ-(σ+N+1)),

which proves the claim. ∎

2 Laplace’s method for an interval

Theorem 2.

Suppose that a<b, that fC2([a,b],R), and that there is a unique x0[a,b] at which f is equal to its supremum over [a,b]. Suppose also that a<x0<b and that f′′(x0)<0. Then

abeMf(x)𝑑xeMf(x0)2π-Mf′′(x0).

as M.

Proof.

We remark first that f(x0)=0 because f is equal to its supremum over [a,b] at this point, which is not a boundary point. The claim says that a ratio has limit 1 as M. We shall prove that the liminf and the limsup of this ratio are both 1, which will prove the claim. Let ϵ>0. Because f′′:[a,b] is continuous, there is some δ>0 such that |x-x0|<δ implies f′′(x)f′′(x0)-ϵ; we take δ small enough that (x0-δ,x0+δ)[a,b]. Writing

f(x)=f(x0)+f(x0)(x-x0)+R1(x)=f(x0)+R1(x),x[a,b],

Taylor’s theorem tells us that for each x[a,b] there is some ξx strictly between x0 and x such that

R1(x)=f′′(ξx)2(x-x0)2.

Thus for |x-x0|<δ we have |ξx-x0|<δ, so

f(x)f(x0)+f′′(x0)-ϵ2(x-x0)2.

Using this inequality, which applies for any x(x0-δ,x0+δ), and because the integrand in the following integral is positive,

abeMf(x)𝑑x x0-δx0+δeMf(x)𝑑x
x0-δx0+δeM(f(x0)+f′′(x0)-ϵ2(x-x0)2)𝑑x
= eMf(x0)x0-δx0+δe-M-f′′(x0)+ϵ2(x-x0)2𝑑x.

Changing variables, keeping in mind that f′′(x0)<0,

x0-δx0+δe-M-f′′(x0)+ϵ2(x-x0)2𝑑x=-δM-f′′(x0)+ϵ2δM-f′′(x0)+ϵ2e-y2(M-f′′(x0)+ϵ2)-1/2𝑑y.

Thus

abeMf(x)𝑑xeMf(x0)(-Mf′′(x0)2)-1/2 (2)

is lower bounded by

(-f′′(x0)+ϵ-f′′(x0))-1/2-δM-f′′(x0)+ϵ2δM-f′′(x0)+ϵ2e-y2𝑑y,

so we get that the liminf of (2) as M is lower bounded by

(-f′′(x0)+ϵ-f′′(x0))-1/2π.

But this is true for all ϵ>0 and (2) and its liminf do not depend on ϵ, so the liminf of (2) as M is lower bounded by π. In other words,

lim infMabeMf(x)𝑑xeMf(x0)(-Mf′′(x0)2π)-1/21.

Let ϵ>0 with f′′(x0)+ϵ<0; this is possible because f′′(x0)<0. Because f′′:[a,b] is continuous there is some δ>0 such that |x-x0|<δ implies that f′′(x)f′′(x0)+ϵ; we take (x0-δ,x0+δ)[a,b]. Taylor’s theorem tells us that for any x[a,b] there is some ξx strictly between x0 and x such that

f(x)=f(x0)+f′′(ξx)2(x-x0)2.

Therefore, as |x-x0|<δ implies that |ξx-x0|<δ,

f(x)f(x0)+f′′(x0)+ϵ2(x-x0)2. (3)

Furthermore, f:[a,b] is continuous, so it makes sense to define

C=supx[a,x0-δ][x0+δ,b]f(x).

Because x0 is not in this union of intervals, by hypothesis we know that C<f(x0), and we define η=f(x0)-C>0. This means that for all x[a,x0-δ][x0+δ,b], f(x)f(x0)-η. Then

abeMf(x)𝑑x = ax0-δeMf(x)𝑑x+x0-δx0+δeMf(x)𝑑x+x0+δbeMf(x)𝑑x
ax0-δeMC𝑑x+x0-δx0+δeMf(x)𝑑x+x0+δbeMC𝑑x
= (b-a-2δ)eMC+x0-δx0+δeMf(x)𝑑x
< (b-a)eMC+x0-δx0+δeMf(x)𝑑x.

For the integral over (x0-δ,x0+δ),

x0-δx0+δeMf(x)𝑑x x0-δx0+δeM(f(x0)+f′′(x0)+ϵ2(x-x0)2)𝑑x
= eMf(x0)x0-δx0+δeMf′′(x0)+ϵ2(x-x0)2𝑑x
< eMf(x0)-eMf′′(x0)+ϵ2(x-x0)2𝑑x.

Changing variables, and keeping in mind that f′′(x0)+ϵ<0,

-eMf′′(x0)+ϵ2(x-x0)2𝑑x = -e-y2(-M2(f′′(x0)+ϵ))-1/2𝑑y
= (-M2π(f′′(x0)+ϵ))-1/2.

Therefore

abeMf(x)𝑑x<(b-a)eMC+eMf(x0)(-M2π(f′′(x0)+ϵ))-1/2,

which we rearrange as

abeMf(x)𝑑xeMf(x0)(-M2π(f′′(x0)+ϵ))-1/2<(b-a)e-Mη(-M2π(f′′(x0)+ϵ))1/2+1.

As M the first term on the right-hand side tends to 0, because η>0. Therefore,

lim supMabeMf(x)𝑑xeMf(x0)(-M2π(f′′(x0)+ϵ))-1/21.

This is true for all ϵ>0, so it holds that

lim supMabeMf(x)𝑑xeMf(x0)(-Mf′′(x0)2π)-1/21,

completing the proof. ∎