Laplace approximation and deterministic quadrature

By [F2], the symmetric matrix $H$ represents a self-adjoint endomorphism of ${\mathbb{R}}^d$. By [F3], choose an orthonormal eigenbasis $q_1,\dots ,q_d$ with eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$. Positive definiteness [F4] gives ${\lambda }_j\mathrm{\parallel }{q}_j{\mathrm{\parallel }}_2^2=q_j^{\mathsf{T}}H{q}_j>0$, so each ${\lambda }_j>0$. Writing $Q$ for the matrix with these columns and $\mathrm{\Lambda }=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)$, we have $H=Q\mathrm{\Lambda }{Q}^{\mathsf{T}}$ and $Q^{\mathsf{T}}Q=I$ by orthonormality. By the Leibniz definition of determinant in [F5], $\det I=1$; multiplicativity and transpose invariance then give $(\det Q{)}^2=1$. Also $\det \mathrm{\Lambda }={\prod }_j{\lambda }_j>0$ by [F5], hence $\det H=(\det Q{)}^2\det \mathrm{\Lambda }={\prod }_j{\lambda }_j>0$.

The substitution $x=\sqrt{2}t$ in [F8], followed by [F7], gives the improper Riemann value ${\int }_{\mathbb{R}}{e}^{-x^2\mathrm{/}2}dx=\sqrt{2\pi }$. On each $[-m,m]$, $m\ge 1$, the continuous integrand has equal Riemann and Lebesgue integrals by [F14] and [F18]. The nonnegative measurable functions $e^{-x^2\mathrm{/}2}{\mathbf{1}}_{[-m,m]}(x)$ increase to $e^{-x^2\mathrm{/}2}$ on $\mathbb{R}$, so monotone convergence [F17] identifies $G_1:={\int }_{\mathbb{R}}{e}^{-x^2\mathrm{/}2}d{\lambda }_1(x)$ with the limit of these proper integrals, namely $\sqrt{2\pi }$.

By [F4], choose a lower-triangular $L$ with positive diagonal such that $H=L{L}^{\mathsf{T}}$. Then [F5] gives $\det H=(\det L{)}^2$ and $\det L>0$, so [F6] identifies $\sqrt{\det H}=\det L$. Since $\det L>0$, [F12] also shows that $L$ is invertible.

For $k\ge 1$ put $G_k:={\int }_{{\mathbb{R}}^k}{e}^{-\mathrm{\parallel }z{\mathrm{\parallel }}_2^2\mathrm{/}2}d{\lambda }_k(z)$. For each $k$, the spaces ${\mathbb{R}}^k$ and $\mathbb{R}$ are sigma-finite by [F9], and [F14] makes the product integrand product-measurable. By [F15]–[F17], its integral under ${\lambda }_{k+1}$ equals its integral under ${\lambda }_k\times {\lambda }_1$; Tonelli [F10] then gives $G_{k+1}=G_k{G}_1$. Starting at [step 1.2], induction yields $G_d=(2\pi {)}^{d\mathrm{/}2}$.

The linear map $T(u)=L^{\mathsf{T}}u$ is a $C^1$ diffeomorphism of ${\mathbb{R}}^d$ by step 2.1, with Jacobian determinant $\det (L^{\mathsf{T}})=\det L>0$ by [F5]. Since $u^{\mathsf{T}}Hu=\mathrm{\parallel }{L}^{\mathsf{T}}u{\mathrm{\parallel }}_2^2$, [F11] gives $G_d=\det L{\int }_{{\mathbb{R}}^d}{e}^{-u^{\mathsf{T}}Hu\mathrm{/}2}d{\lambda }_d(u)$.

Substituting [step 2.2] and $\det L=\sqrt{\det H}$ from [step 2.1] into [step 3.1] gives the asserted formula.

Define $L(t):=f(a)+(f(b)-f(a))(t-a)\mathrm{/}(b-a)$. Then $L$ is affine and has the required endpoint values; any affine function agreeing with $f$ at the distinct endpoints has the same slope and intercept, so it equals $L$. By [F1] and [F2], $f^{\mathrm{\prime }\mathrm{\prime }}$ is bounded on $[a,b]$, so $M$ is a finite nonnegative real.

The desired inequality is equality at $x=a$ and $x=b$. Fix an interior point $x\in (a,b)$ and put $c_x:=(f(x)-L(x))\mathrm{/}((x-a)(b-x))$. The denominator is positive. Define $g(t):=f(t)-L(t)-c_x(t-a)(b-t)$; then $g(a)=g(x)=g(b)=0$.

Rolle's theorem [F3] on $[a,x]$ and $[x,b]$ gives $s_1\in (a,x)$ and $s_2\in (x,b)$ with $g^{\mathrm{\prime }}(s_1)=g^{\mathrm{\prime }}(s_2)=0$. Applying [F3] to $g^{\mathrm{\prime }}$ on $[s_1,s_2]$ gives $\xi \in (s_1,s_2)$ with $g^{\mathrm{\prime }\mathrm{\prime }}(\xi )=0$. All applications are valid because $f$ is $C^2$ on a neighborhood of $[a,b]$.

Since $L^{\mathrm{\prime }\mathrm{\prime }}=0$ and $((t-a)(b-t){)}^{\mathrm{\prime }\mathrm{\prime }}=-2$, step 3.1 yields $0=g^{\mathrm{\prime }\mathrm{\prime }}(\xi )=f^{\mathrm{\prime }\mathrm{\prime }}(\xi )+2c_x$. Thus $\mathrm{\mid }{c}_x\mathrm{\mid }=\mathrm{\mid }{f}^{\mathrm{\prime }\mathrm{\prime }}(\xi )\mathrm{\mid }\mathrm{/}2\le M\mathrm{/}2$. Multiplying by the positive denominator in the definition of $c_x$ proves the claimed bound at this $x$; together with the endpoint cases in step 2.1 this proves the result.

Put $h_0:=h(\widehat{\theta })$. By [F12], $h$ is Lebesgue measurable on $K$; by [F11] and [F13], $a{e}^{nh}$ is nonnegative and measurable for each $n$. Since $h\le h_0$ and the exponential is increasing, $0\le a{e}^{nh}\le e^{n{h}_0}a$. Thus [F14] and ${\int }_{K}a<\mathrm{\infty }$ show that $I_n$ is finite.

Choose $0<\delta <\min \{r_0,r_1\}$ small enough that $a(\theta )\le a(\widehat{\theta })+1$ whenever $\mathrm{\parallel }\theta -\widehat{\theta }{\mathrm{\parallel }}_2<\delta$; this is possible by continuity of $a$ at $\widehat{\theta }$. Then $\bar{B}(\widehat{\theta },\delta )\subseteq U\cap K$, and the local quadratic bound holds there.

The set $A:=K\cap \{\theta :\delta \le \mathrm{\parallel }\theta -\widehat{\theta }{\mathrm{\parallel }}_2\le r_1\}$ is closed in $K$ by [F6], hence compact by [F5]. If $A$ is nonempty, [F5] gives a maximizer of $h$ on $A$; it is not $\widehat{\theta }$, so uniqueness of the maximizer on $K$ gives ${\sup }_{A}h<h_0$. Combining this gap with the assumed gap on $K\setminus B(\widehat{\theta },r_1)$ yields some ${\eta }^{\mathrm{\prime }}>0$ such that $h\le h_0-{\eta }^{\mathrm{\prime }}$ throughout $K\setminus B(\widehat{\theta },\delta )$; if $A$ is empty, take ${\eta }^{\mathrm{\prime }}=\eta$.

For each coordinate vector $e_j$, the one-variable function $t\mapsto h(\widehat{\theta }+t{e}_j)$ has a local maximum at $t=0$ and is differentiable there. Its right difference quotients are nonpositive and its left difference quotients are nonnegative, so the derivative is zero. Thus every component of $\mathrm{\nabla }h(\widehat{\theta })$ vanishes.

Define $F_n(\theta ):=a(\theta )e^{n(h(\theta )-h_0)}$ on $K\cap B(\widehat{\theta },\delta )$ and $F_n(\theta ):=0$ elsewhere in ${\mathbb{R}}^d$, and put $g_n(u):=F_n(\widehat{\theta }+u\mathrm{/}\sqrt{n})$. By [F13], $F_n$ and $g_n$ are nonnegative measurable. The affine change of variables $\theta =\widehat{\theta }+u\mathrm{/}\sqrt{n}$ is a $C^1$ diffeomorphism with Jacobian $n^{-d\mathrm{/}2}$, so [F8] gives $n^{d\mathrm{/}2}{e}^{-n{h}_0}{\int }_{K\cap B(\widehat{\theta },\delta )}a(\theta )e^{nh(\theta )}d{\lambda }_d(\theta )={\int }_{{\mathbb{R}}^d}{g}_n(u)d{\lambda }_d(u)$. Equivalently, $g_n(u)=a(\widehat{\theta }+u\mathrm{/}\sqrt{n})e^{n[h(\widehat{\theta }+u\mathrm{/}\sqrt{n})-h_0]}$ when $\mathrm{\parallel }u{\mathrm{\parallel }}_2<\delta \sqrt{n}$ and $g_n(u)=0$ otherwise. For all $u,n$, [given] and step 2.1 give $0\le g_n(u)\le (a(\widehat{\theta })+1)e^{-c\mathrm{\parallel }u{\mathrm{\parallel }}_2^2}$. This majorant is integrable by [F10] with $H=2cI$, which is positive definite by [F2] and $c>0$.

Fix $u\in {\mathbb{R}}^d$ and put $v_n:=u\mathrm{/}\sqrt{n}$. Eventually $\mathrm{\parallel }{v}_n{\mathrm{\parallel }}_2<\delta \mathrm{/}2$. By [F3] and step 3.2, $n(h(\widehat{\theta }+v_n)-h_0)\to -\frac{1}{2}{u}^{\mathsf{T}}Hu$, because $no(\mathrm{\parallel }{v}_n{\mathrm{\parallel }}_2^2)=o(1)$. For the stronger remainder bound, the finitely many third partial derivatives are bounded on the compact set $K\cap \bar{B}(\widehat{\theta },\delta \mathrm{/}2)$ by [F5]; apply [F4] on the open convex ball $B(\widehat{\theta },\delta \mathrm{/}2)$ with $k=2$ to get a remainder $O(\mathrm{\parallel }{v}_n{\mathrm{\parallel }}_2^3)$, which also vanishes after multiplication by $n$. Continuity of $a$ gives $a(\widehat{\theta }+v_n)\to a(\widehat{\theta })$.

On $K\setminus B(\widehat{\theta },\delta )$, step 3.1 gives $0\le n^{d\mathrm{/}2}{e}^{-n{h}_0}{\int }_{K\setminus B(\widehat{\theta },\delta )}a{e}^{nh}d{\lambda }_d\le n^{d\mathrm{/}2}{e}^{-n{\eta }^{\mathrm{\prime }}}{\int }_{K}ad{\lambda }_d⟶0$. Indeed, the ratio of consecutive terms $n^{d\mathrm{/}2}{e}^{-n{\eta }^{\mathrm{\prime }}}$ tends to $e^{-{\eta }^{\mathrm{\prime }}}<1$, so the sequence is eventually bounded by a geometric sequence tending to zero.

Steps 4.1 and 3.3 show pointwise convergence of $g_n$ to $a(\widehat{\theta })e^{-u^{\mathsf{T}}Hu\mathrm{/}2}$ and provide an integrable majorant. Applying dominated convergence [F9] and then [F10] yields $n^{d\mathrm{/}2}{e}^{-n{h}_0}{\int }_{K\cap B(\widehat{\theta },\delta )}a{e}^{nh}d{\lambda }_d⟶a(\widehat{\theta })(2\pi {)}^{d\mathrm{/}2}\mathrm{/}\sqrt{\det H}$.

By [F13] the two indicator-restricted integrands are measurable, and by [F14] the integral over $K$ is the sum of its integrals over $K\cap B(\widehat{\theta },\delta )$ and $K\setminus B(\widehat{\theta },\delta )$. Adding the local limit in step 5.1 and the vanishing off-mode term in step 4.2 gives $n^{d\mathrm{/}2}{e}^{-n{h}_0}{I}_n⟶a(\widehat{\theta })(2\pi {)}^{d\mathrm{/}2}\mathrm{/}\sqrt{\det H}>0$. Dividing by this positive limit proves the stated asymptotic equivalence. The argument establishes only the leading relative equivalent; it gives no $O(n^{-1})$ remainder.

Since $f^{\mathrm{\prime }\mathrm{\prime }}$ is continuous on $[a,b]$, [F3] shows it is bounded; set $M:={\sup }_{[a,b]}\mathrm{\mid }{f}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }<\mathrm{\infty }$. The function $f$ is also Riemann integrable.

By [F2], the endpoint derivatives $f^{\mathrm{\prime }}(a),f^{\mathrm{\prime }\mathrm{\prime }}(a),f^{\mathrm{\prime }}(b),f^{\mathrm{\prime }\mathrm{\prime }}(b)$ are the one-sided derivatives from within $[a,b]$. Define $\tilde{f}$ on $(a-1,b+1)$ by $\tilde{f}(x):=\{\begin{array}{ll}f(a)+f^{\mathrm{\prime }}(a)(x-a)+\frac{1}{2}{f}^{\mathrm{\prime }\mathrm{\prime }}(a)(x-a{)}^2,& x<a,\\ f(x),& a\le x\le b,\\ f(b)+f^{\mathrm{\prime }}(b)(x-b)+\frac{1}{2}{f}^{\mathrm{\prime }\mathrm{\prime }}(b)(x-b{)}^2,& x>b\mathrm{.}\end{array}$ At each join the polynomial piece matches the value and the first two one-sided derivatives of $f$. Since $f,f^{\mathrm{\prime }},f^{\mathrm{\prime }\mathrm{\prime }}$ are continuous on $[a,b]$, these matches make $\tilde{f}$ $C^2$ on the open interval $(a-1,b+1)$. It agrees with $f$ on $[a,b]$ and ${\sup }_{[a,b]}\mathrm{\mid }{\tilde{f}}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }=M$. For $j=0,\dots ,N-1$, let $x_j=a+jh$ and let $L_j$ be the affine interpolant between $f(x_j)$ and $f(x_{j+1})$. Put $q_j(x):=\frac{M}{2}(x-x_j)(x_{j+1}-x)$ on that panel. Applying [F4] to $\tilde{f}$ on each panel gives $\mathrm{\mid }f(x)-L_j(x)\mathrm{\mid }\le q_j(x)$, and $q_j\ge 0$.

Let $P_N$ be the continuous piecewise-affine function equal to $L_j$ on each panel, and let $q_N$ be the continuous piecewise function equal to $q_j$ on each panel; continuity of $q_N$ at a knot holds because both adjacent formulas vanish there. Step 2.1 gives $-q_N\le f-P_N\le q_N$ throughout $[a,b]$.

By [F8], ${\int }_{x_j}^{x_{j+1}}{L}_j=\frac{h}{2}(f(x_j)+f(x_{j+1}))$ and ${\int }_{x_j}^{x_{j+1}}{q}_j=\frac{M{h}^3}{12}$; the second identity follows by integrating $\frac{M}{2}t(h-t)$ on $[0,h]$. Repeated additivity [F7] therefore gives ${\int }_a^b{P}_N=T_N(f)$, because each interior knot occurs twice with half weight, and ${\int }_a^b{q}_N=NM{h}^3\mathrm{/}12$.

The functions $q_N$ and $f-P_N$ are continuous and hence integrable by [F3]. Applying [F5] to the two pointwise inequalities in step 3.1 and using [F6] yields $\mid {\int }_a^{b}f-{\int }_a^b{P}_N\mid \le {\int }_a^b{q}_N$.

Substituting the identities from step 3.2 into step 4.1 gives $\mid {\int }_a^{b}f-T_N(f)\mid \le NM{h}^3\mathrm{/}12=(b-a)h^{2}M\mathrm{/}12$, since $Nh=b-a$. This is the claimed bound.