Watson’s lemma and Laplace’s method

1 Watson’s lemma

Our proof of Watson’s lemma follows Miller.

Theorem 1 (Watson’s lemma).

Suppose that $T\mathrm{>}\mathrm{0}$, that $\varphi \mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{C}$ belongs to $L^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{[}\mathrm{0}\mathrm{,}T\mathrm{]}\mathrm{)}$, that $\sigma \mathrm{>}\mathrm{-}\mathrm{1}$, and that $g\mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}{t}^{\mathrm{-}\sigma }\mathit{}\varphi \mathit{}\mathrm{(}t\mathrm{)}$ is $C^{\mathrm{\infty }}$ on some neighborhood of $\mathrm{0}$. Then $F\mathrm{:}\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{)}\mathrm{\to }\mathrm{C}$ defined by

$$F(\lambda )={\int }_0^T{e}^{-\lambda t}\varphi (t)𝑑t$$

satisfies

$$F(\lambda )\sim \sum _{n=0}^{\mathrm{\infty }}\frac{g^{(n)}(0)\mathrm{\Gamma }(\sigma +n+1)}{n!{\lambda }^{\sigma +n+1}},\lambda \to \mathrm{\infty }.$$

Proof.

Take $g$ to be $C^{\mathrm{\infty }}$ on some interval with left endpoint and right endpoint $s$, . For $p$ a nonnegative integer and $\lambda >1$, define

$$F_p(\lambda )={\int }_0^s{e}^{-\lambda t}{t}^{\sigma +p}𝑑t,$$

which satisfies, doing the change of variable $\tau =\lambda t$,

$F_p(\lambda )$ $=$ ${\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-\lambda t}{t}^{\sigma +p}𝑑t-{\displaystyle {\int }_s^{\mathrm{\infty }}}{e}^{-\lambda t}{t}^{\sigma +p}𝑑t$

$=$ ${\lambda }^{-(\sigma +p+1)}{\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-\tau }{\tau }^{\sigma +p}𝑑\tau -{\displaystyle {\int }_s^{\mathrm{\infty }}}{e}^{-\lambda t}{t}^{\sigma +p}𝑑t$

$=$ ${\lambda }^{-(\sigma +p+1)}\mathrm{\Gamma }(\sigma +p+1)-{\displaystyle {\int }_s^{\mathrm{\infty }}}{e}^{-\lambda t}{t}^{\sigma +p}𝑑t.$

Using the Cauchy-Schwarz inequality,

${\displaystyle {\int }_s^{\mathrm{\infty }}}{e}^{-\lambda t}{t}^{\sigma +p}𝑑t$ $=$ ${\displaystyle {\int }_s^{\mathrm{\infty }}}{e}^{-\lambda t/2}{e}^{-\lambda t/2}{t}^{\sigma +p}𝑑t$

$\le$ ${\left({\displaystyle {\int }_s^{\mathrm{\infty }}}{e}^{-\lambda t}𝑑t\right)}^{1/2}{\left({\displaystyle {\int }_s^{\mathrm{\infty }}}{e}^{-\lambda t}{t}^{2\sigma +2p}𝑑t\right)}^{1/2}$

$=$ $e^{-\lambda s/2}{\lambda }^{-1/2}{\left({\displaystyle {\int }_s^{\mathrm{\infty }}}{e}^{-\lambda t}{t}^{2\sigma +2p}𝑑t\right)}^{1/2}$

$e^{-\lambda s/2}{\left({\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-t}{t}^{2\sigma +2p}𝑑t\right)}^{1/2}$

$=$ $e^{-\lambda s/2}\mathrm{\Gamma }{(2\sigma +2p+1)}^{1/2}.$

For any nonnegative integer $m$ we have $e^{-\lambda s/2}=o_m({\lambda }^{-(\sigma +m+1)})$ as $\lambda \to \mathrm{\infty }$, hence, dealing with $\mathrm{\Gamma }(2\sigma +2p+1)$ merely as a constant depending on $p$,

$$F_p(\lambda )={\lambda }^{-(\sigma +p+1)}\mathrm{\Gamma }(\sigma +p+1)+o_{m,p}({\lambda }^{-(\sigma +m+1)})$$ (1)

as $\lambda \to \mathrm{\infty }$.

Write

$$F(\lambda )={\int }_0^s{e}^{-\lambda t}\varphi (t)𝑑t+{\int }_s^T{e}^{-\lambda t}\varphi (t)𝑑t.$$

One the one hand,

$$\left|{\int }_s^T{e}^{-\lambda t}\varphi (t)𝑑t\right|\le {\int }_s^T{e}^{-\lambda t}|\varphi (t)|𝑑t\le e^{-\lambda s}{\int }_s^T|\varphi (t)|𝑑t\le e^{-\lambda s}{\parallel \varphi \parallel }_{L^1},$$

which shows that for any nonnegative integer $n$,

$${\int }_s^T{e}^{-\lambda t}\varphi (t)𝑑t=o_n({\lambda }^{-(\sigma +n+1)})$$

as $\lambda \to \mathrm{\infty }$.

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

$$r_N(t)=g(t)-\sum _{n=0}^N\frac{g^{(n)}(0)}{n!}{t}^n,t\in (r,s),$$

satisfies

$$|r_N(t)|\le sup|g^{(N+1)}(\tau )|\cdot \frac{{|t|}^{N+1}}{(N+1)!},$$

where the supremum is over those $\tau$ strictly between $0$ and $t$. Then for $t\in (0,s)$,

Using the definition of $r_N$,

${\displaystyle {\int }_0^s}{e}^{-\lambda t}\varphi (t)𝑑t$ $=$ ${\displaystyle {\int }_0^s}{e}^{-\lambda t}{t}^{\sigma }{\displaystyle \sum _{n=0}^N}{\displaystyle \frac{g^{(n)}(0)}{n!}}{t}^{n}dt+{\displaystyle {\int }_0^s}{e}^{-\lambda t}{t}^{\sigma }{r}_N(t)𝑑t$

$=$ ${\displaystyle \sum _{n=0}^N}{\displaystyle \frac{g^{(n)}(0)}{n!}}{F}_n(\lambda )+{\displaystyle {\int }_0^s}{e}^{-\lambda t}{t}^{\sigma }{r}_N(t)𝑑t$

and using the inequality for $r_N(t)$,

$\left|{\displaystyle {\int }_0^s}{e}^{-\lambda t}{t}^{\sigma }{r}_N(t)𝑑t\right|$ $\le$ ${\displaystyle {\int }_0^s}{e}^{-\lambda t}{t}^{\sigma }|r_N(t)|𝑑t$

$\le$

$=$

Using this and (1),

$${\int }_0^s{e}^{-\lambda t}{t}^{\sigma }{r}_N(t)𝑑t=O_N({\lambda }^{-(\sigma +N+2)}).$$

Putting together what we have shown, for any nonnegative integer $N$, as $\lambda \to \mathrm{\infty }$,

$F(\lambda )$ $=$ ${\displaystyle \sum _{n=0}^N}{\displaystyle \frac{g^{(n)}(0)}{n!}}{F}_n(\lambda )+O_N({\lambda }^{-(\sigma +N+2)})+O_N({\lambda }^{-(\sigma +N+2)})$

$=$ ${\displaystyle \sum _{n=0}^N}{\displaystyle \frac{g^{(n)}(0)}{n!}}{\lambda }^{-(\sigma +n+1)}\mathrm{\Gamma }(\sigma +n+1)+{\displaystyle \sum _{n=0}^N}{\displaystyle \frac{g^{(n)}(0)}{n!}}\cdot o_{N,n}({\lambda }^{-(\sigma +N+1)})$

$+O_N({\lambda }^{-(\sigma +N+2)})$

$=$ ${\displaystyle \sum _{n=0}^N}{\displaystyle \frac{g^{(n)}(0)}{n!}}{\lambda }^{-(\sigma +n+1)}\mathrm{\Gamma }(\sigma +n+1)+o_N({\lambda }^{-(\sigma +N+1)}),$

which proves the claim. ∎

2 Laplace’s method for an interval

Theorem 2.

Suppose that , that $f\mathrm{\in }{C}^{\mathrm{2}}\mathit{}\mathrm{(}\mathrm{[}a\mathrm{,}b\mathrm{]}\mathrm{,}\mathrm{R}\mathrm{)}$, and that there is a unique $x_{\mathrm{0}}\mathrm{\in }\mathrm{[}a\mathrm{,}b\mathrm{]}$ at which $f$ is equal to its supremum over $\mathrm{[}a\mathrm{,}b\mathrm{]}$. Suppose also that and that . Then

$${\int }_a^b{e}^{Mf(x)}𝑑x\sim e^{Mf(x_0)}\sqrt{\frac{2\pi }{-M{f}^{\prime \prime }(x_0)}}.$$

as $M\mathrm{\to }\mathrm{\infty }$.

Proof.

We remark first that $f^{\prime }(x_0)=0$ 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 $M\to \mathrm{\infty }$. We shall prove that the liminf and the limsup of this ratio are both 1, which will prove the claim. Let $ϵ>0$. Because $f^{\prime \prime }:[a,b]\to ℝ$ is continuous, there is some $\delta >0$ such that implies $f^{\prime \prime }(x)\ge f^{\prime \prime }(x_0)-ϵ$; we take $\delta$ small enough that $(x_0-\delta ,x_0+\delta )\subset [a,b]$. Writing

$$f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+R_1(x)=f(x_0)+R_1(x),x\in [a,b],$$

Taylor’s theorem tells us that for each $x\in [a,b]$ there is some ${\xi }_x$ strictly between $x_0$ and $x$ such that

$$R_1(x)=\frac{f^{\prime \prime }({\xi }_x)}{2}{(x-x_0)}^2.$$

Thus for we have , so

$$f(x)\ge f(x_0)+\frac{f^{\prime \prime }(x_0)-ϵ}{2}{(x-x_0)}^2.$$

Using this inequality, which applies for any $x\in (x_0-\delta ,x_0+\delta )$, and because the integrand in the following integral is positive,

${\displaystyle {\int }_a^b}{e}^{Mf(x)}𝑑x$ $\ge$ ${\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{Mf(x)}𝑑x$

$\ge$ ${\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{M\left(f(x_0)+\frac{f^{\prime \prime }(x_0)-ϵ}{2}{(x-x_0)}^2\right)}𝑑x$

$=$ $e^{Mf(x_0)}{\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{-M\frac{-f^{\prime \prime }(x_0)+ϵ}{2}{(x-x_0)}^2}𝑑x.$

Changing variables, keeping in mind that ,

$${\int }_{x_0-\delta }^{x_0+\delta }{e}^{-M\frac{-f^{\prime \prime }(x_0)+ϵ}{2}{(x-x_0)}^2}𝑑x={\int }_{-\delta \sqrt{M\frac{-f^{\prime \prime }(x_0)+ϵ}{2}}}^{\delta \sqrt{M\frac{-f^{\prime \prime }(x_0)+ϵ}{2}}}{e}^{-y^2}{\left(M\frac{-f^{\prime \prime }(x_0)+ϵ}{2}\right)}^{-1/2}𝑑y.$$

Thus

$$\frac{{\int }_a^b{e}^{Mf(x)}𝑑x}{e^{Mf(x_0)}{\left(\frac{-M{f}^{\prime \prime }(x_0)}{2}\right)}^{-1/2}}$$ (2)

is lower bounded by

$${\left(\frac{-f^{\prime \prime }(x_0)+ϵ}{-f^{\prime \prime }(x_0)}\right)}^{-1/2}{\int }_{-\delta \sqrt{M\frac{-f^{\prime \prime }(x_0)+ϵ}{2}}}^{\delta \sqrt{M\frac{-f^{\prime \prime }(x_0)+ϵ}{2}}}{e}^{-y^2}𝑑y,$$

so we get that the liminf of (2) as $M\to \mathrm{\infty }$ is lower bounded by

$${\left(\frac{-f^{\prime \prime }(x_0)+ϵ}{-f^{\prime \prime }(x_0)}\right)}^{-1/2}\sqrt{\pi }.$$

But this is true for all $ϵ>0$ and (2) and its liminf do not depend on $ϵ$, so the liminf of (2) as $M\to \mathrm{\infty }$ is lower bounded by $\sqrt{\pi }$. In other words,

$$\underset{M\to \mathrm{\infty }}{lim\; inf}\frac{{\int }_a^b{e}^{Mf(x)}𝑑x}{e^{Mf(x_0)}{\left(\frac{-M{f}^{\prime \prime }(x_0)}{2\pi }\right)}^{-1/2}}\ge 1.$$

Let $ϵ>0$ with ; this is possible because . Because $f^{\prime \prime }:[a,b]\to ℝ$ is continuous there is some $\delta >0$ such that implies that $f^{\prime \prime }(x)\le f^{\prime \prime }(x_0)+ϵ$; we take $(x_0-\delta ,x_0+\delta )\subset [a,b]$. Taylor’s theorem tells us that for any $x\in [a,b]$ there is some ${\xi }_x$ strictly between $x_0$ and $x$ such that

$$f(x)=f(x_0)+\frac{f^{\prime \prime }({\xi }_x)}{2}{(x-x_0)}^2.$$

Therefore, as implies that ,

$$f(x)\le f(x_0)+\frac{f^{\prime \prime }(x_0)+ϵ}{2}{(x-x_0)}^2.$$ (3)

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

$$C=\underset{x\in [a,x_0-\delta ]\cup [x_0+\delta ,b]}{sup}f(x).$$

Because $x_0$ is not in this union of intervals, by hypothesis we know that , and we define $\eta =f(x_0)-C>0$. This means that for all $x\in [a,x_0-\delta ]\cup [x_0+\delta ,b]$, $f(x)\le f(x_0)-\eta$. Then

${\displaystyle {\int }_a^b}{e}^{Mf(x)}𝑑x$ $=$ ${\displaystyle {\int }_a^{x_0-\delta }}{e}^{Mf(x)}𝑑x+{\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{Mf(x)}𝑑x+{\displaystyle {\int }_{x_0+\delta }^b}{e}^{Mf(x)}𝑑x$

$\le$ ${\displaystyle {\int }_a^{x_0-\delta }}{e}^{MC}𝑑x+{\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{Mf(x)}𝑑x+{\displaystyle {\int }_{x_0+\delta }^b}{e}^{MC}𝑑x$

$=$ $(b-a-2\delta )e^{MC}+{\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{Mf(x)}𝑑x$

$(b-a)e^{MC}+{\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{Mf(x)}𝑑x.$

For the integral over $(x_0-\delta ,x_0+\delta )$,

${\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{Mf(x)}𝑑x$ $\le$ ${\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{M\left(f(x_0)+\frac{f^{\prime \prime }(x_0)+ϵ}{2}{(x-x_0)}^2\right)}𝑑x$

$=$ $e^{Mf(x_0)}{\displaystyle {\int }_{x_0-\delta }^{x_0+\delta }}{e}^{M\frac{f^{\prime \prime }(x_0)+ϵ}{2}{(x-x_0)}^2}𝑑x$

$e^{Mf(x_0)}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{M\frac{f^{\prime \prime }(x_0)+ϵ}{2}{(x-x_0)}^2}𝑑x.$

Changing variables, and keeping in mind that ,

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{M\frac{f^{\prime \prime }(x_0)+ϵ}{2}{(x-x_0)}^2}𝑑x$ $=$ ${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-y^2}{\left(-{\displaystyle \frac{M}{2}}(f^{\prime \prime }(x_0)+ϵ)\right)}^{-1/2}𝑑y$

$=$ ${\left(-{\displaystyle \frac{M}{2\pi }}(f^{\prime \prime }(x_0)+ϵ)\right)}^{-1/2}.$

Therefore

which we rearrange as

As $M\to \mathrm{\infty }$ the first term on the right-hand side tends to $0$, because $\eta >0$. Therefore,

$$\underset{M\to \mathrm{\infty }}{lim\; sup}\frac{{\int }_a^b{e}^{Mf(x)}𝑑x}{e^{Mf(x_0)}{\left(-\frac{M}{2\pi }(f^{\prime \prime }(x_0)+ϵ)\right)}^{-1/2}}\le 1.$$

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

$$\underset{M\to \mathrm{\infty }}{lim\; sup}\frac{{\int }_a^b{e}^{Mf(x)}𝑑x}{e^{Mf(x_0)}{\left(\frac{-M{f}^{\prime \prime }(x_0)}{2\pi }\right)}^{-1/2}}\le 1,$$

completing the proof. ∎