Asymptotics of the Mean-Field Heisenberg Model

Abstract

We consider the mean-field classical Heisenberg model and obtain detailed information about the total spin of the system by studying the model on a complete graph and sending the number of vertices to infinity. In particular, we obtain Cramér- and Sanov-type large deviations principles for the total spin and the empirical spin distribution and demonstrate a second-order phase transition in the Gibbs measures. We also study the asymptotics of the total spin throughout the phase transition using Stein’s method, proving central limit theorems in the sub- and supercritical phases and a nonnormal limit theorem at the critical temperature.

Appendix

1.1 A.1 Calculus of Φ _β

Recall that the free energy is obtained by minimizing the function

$$ \varPhi_\beta(x) := \log\biggl(\frac{x}{\sinh(x)} \biggr)+x \coth(x)-1-\frac {\beta}{2} \biggl(\coth(x)-\frac{1}{x} \biggr)^2. $$

In the following lemmas, we explicitly identify the minima for all β>0, and obtain an estimate used in the proof of Theorem 10.

Lemma 21

If β≤3, then

$$\inf_{x\ge0} \biggl\{\log\biggl(\frac{x}{\sinh(x)} \biggr)+x \biggl(\coth(x)- \frac{1}{x} \biggr)-\frac{\beta}{2} \biggl (\coth(x)- \frac{1}{x} \biggr)^2 \biggr\}=0, $$

achieved only at x=0.

Proof

We show first that the expression to be minimized is increasing. Differentiating the expression in question yields

$$ \begin{aligned} & \biggl[\frac{\sinh(x)}{x} \biggr] \biggl[ \frac{\sinh(x)-x\cosh(x)}{\sinh^2(x)} \biggr]+ \biggl[\coth (x)-\frac{1}{x} \biggr]+x \biggl[ \frac{1}{x^2}-\frac{1}{ \sinh^2(x)} \biggr] \\ &\qquad{}-\beta\biggl[\coth(x)-\frac {1}{x} \biggr] \biggl[ \frac{1}{x^2}-\frac{1}{\sinh^2(x)} \biggr] \\ &\quad= \biggl[\frac{1}{x^2}-\frac{1}{\sinh^2(x)} \biggr] \biggl [x-\beta\biggl( \coth(x)-\frac{1}{x} \biggr) \biggr]. \end{aligned} $$

Expanding sinh(x) in a Taylor series,

$$\sinh^2(x)= \Biggl(x+\sum_{n=1}^\infty \frac{x^{2n+1}}{(2n+1)!} \Biggr)^2> x^2. $$

The problem is therefore reduced to showing that

$$x-\beta\biggl(\coth(x)-\frac{1}{x} \biggr)> 0, $$

or alternatively,

$$\beta<\frac{x^2}{x\coth(x)-1}. $$

Clearly showing this for β=3 suffices to prove the lemma. Rearranging yet again, this is equivalent to showing that

$$\coth(x)-\frac{1}{x}<\frac{x}{3}. $$

Expanding coth(x) in terms of e ^2x and rearranging terms, this is furthermore equivalent to showing that

$$\biggl(1-x+\frac{x^2}{3} \biggr)e^{2x}>1+x+\frac{x^2}{3}. $$

Expanding e ^2x in a Taylor series, the left-hand side of the inequality above is given by

$$ \begin{aligned} &1+x+\frac{x^2}{3}+\sum _{n=3}^\infty x^n \biggl[ \frac{2^n}{n!}-\frac{2^{n-1}}{ (n-1)!}+\frac{2^{n-2}}{3(n-2)!} \biggr] \\ &\quad= 1+x+\frac{x^2}{3}+\sum_{n=3}^\infty \frac{2^{n-2}x^n}{3n!} \biggl[ \biggl(n-\frac{7}{2} \biggr)^2- \frac{1}{4} \biggr]. \end{aligned} $$

It is easy to see that the n=3 and n=4 terms in the power series above are zero and that the rest are all positive, thus completing the proof that the expression to be minimized is increasing. Moreover, recall that $\lim_{x\to0} \coth(x)-\frac{1}{x}=0$ and $\lim_{x\to0}\frac{x}{\sinh(x)}=1$, so

$$\lim_{x\to0} \biggl\{\log\biggl(\frac{x}{\sinh(x)} \biggr)+x \biggl(\coth(x) -\frac{1}{x} \biggr)-\frac{\beta}{2} \biggl (\coth(x)- \frac{1}{x} \biggr)^2 \biggr\}=0. $$

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Lemma 22

For β>3, there is a unique value of x∈(0,∞) which minimizes

$$\log\biggl(\frac{x}{\sinh(x)} \biggr)+x \biggl(\coth(x) -\frac{1}{x} \biggr)-\frac{\beta}{2} \biggl(\coth(x)-\frac{1}{x} \biggr)^2 $$

over [0,∞).

Proof

From the previous proof, we have that the derivative of the expression to be minimized is

$$\biggl[\frac{1}{x^2}-\frac{1}{\sinh^2(x)} \biggr] \biggl[x-\beta \biggl( \coth(x)-\frac{1}{x} \biggr) \biggr]\sim\frac{(3-\beta)x}{9}, $$

for x near zero. For β>3, it follows that x=0 is a local maximum of the expression on [0,∞). As x tends to infinity, the expression to be minimized is asymptotic to log(x), and there is therefore at least one interior minimum. Since $\frac{1}{x^{2}}-\frac{1}{\sinh^{2}(x)}>0$, it must be the case that at this interior minimum,

$$x-\beta\biggl(\coth(x)-\frac{1}{x} \biggr)=0; $$

that is,

$$\beta=\frac{x}{\coth(x)-\frac{1}{x}}. $$

In fact, the function $g(x)=\frac{x}{\coth(x)-\frac{1}{x}}$ is strictly increasing on (0,∞), and this equation thus uniquely determines x in terms of β. First observe that

$$g'(x)=\frac{\coth(x)-\frac{2}{x}+\frac{x}{\sinh^2(x)}}{ (\coth(x) -\frac{1}{x} )^2}, $$

and it thus suffices to show that

$$\coth(x)-\frac{2}{x}+\frac{x}{\sinh^2(x)}>0 $$

for x>0; multiplying through by xsinh²(x), one could equivalently show that

$$x\sinh(x)\cosh(x)+x^2-2\sinh^2(x)>0. $$

Using the identities $\sinh(x)\cosh(x)=\frac{\sinh(2x)}{2}$ and $\sinh^{2}(x)=\frac{\cosh(2x)-1}{2}$, this is equivalent to showing that

$$\frac{x}{2}\sinh(2x)+x^2-\cosh(2x)+1>0. $$

Expanding the left-hand side in Taylor series yields

$$\sum_{n=2}^\infty x^{2n} \biggl[ \frac{2^{2n-1}}{2(2n-1)!}-\frac{2^{2n}}{(2n)!} \biggr]=\sum _{n=3}^\infty \frac{x^{2n}2^{2n-2}}{(2n)!}[2n-4], $$

all of whose terms are indeed positive. □

Lemma 23

Let k ₂ denote the unique value of x∈(0,∞) with

$$x-\beta\biggl(\coth(x)-\frac{1}{x} \biggr)=0. $$

Then

$$\beta\biggl(\frac{1}{k_2^2}-\frac{1}{\sinh^2(k_2)} \biggr)<1. $$

In particular, if

$$\varPhi_\beta(x):=\log\biggl(\frac{x}{\sinh(x)} \biggr)+x \biggl( \coth(x) -\frac{1}{x} \biggr)-\frac{\beta}{2} \biggl(\coth(x)- \frac{1}{x} \biggr)^2, $$

then $\varPhi_{\beta}'(k_{2})=0$ and $\varPhi_{\beta}''(k_{2})> 0$.

Proof

Let

$$f(x):=x-\beta\biggl(\coth(x)-\frac{1}{x} \biggr), $$

so that f(k ₂)=0; as was shown in the previous proof, this uniquely defines k ₂ in terms of β. Moreover, it was also shown that lim _x→0 f(x)=0, lim _x→∞ f(x)=∞, and lim _x→0 f′(x)<0. That is, f is initially decreasing from 0, and then becomes increasing eventually, crossing the x-axis exactly once in (0,∞). It must therefore be that there is an x<k ₂ such that f′(x)=0. Now,

$$f'(x)=1-\beta\biggl(\frac{1}{x^2}-\frac{1}{\sinh^2(x)} \biggr). $$

In fact, $g(x):=\frac{1}{x^{2}}-\frac{1}{\sinh^{2}(x)}$ is decreasing on (0,∞): observe first that

$$g'(x)=2\coth(x)\operatorname{csch}^2(x)-\frac{2}{x^3}, $$

and so the claim is true if $\coth(x)\operatorname {csch}^{2}(x)<x^{-3}$. By the definitions of the hyperbolic trigonometric functions, this is equivalent to

$$4x^3\bigl(e^{4x}+e^{2x}\bigr)< \bigl(e^{2x}-1\bigr)^3. $$

Expanding in power series, this is equivalent to

Letting α:=i−1, β:=j−1, and γ=k−1, the coefficient of x ⁿ on the right-hand side is

It is now easy to see that the coefficient of x ⁿ on the right-hand side is smaller than the one on the left for each n≥3, and so it is in fact true that $g(x):=\frac{1}{x^{2}}-\frac{1}{\sinh^{2}(x)}$ is decreasing on (0,∞). It follows that f′(x) is increasing on (0,∞), and so the previously identified x<k ₂ such that f′(x)=0 is in fact the only zero of f′, and f′(k ₂)>0.

Now, recall from the proof of the previous lemma that

$$\varPhi_\beta'(x)= \biggl[\frac{1}{x^2}- \frac{1}{\sinh^2(x)} \biggr] \biggl[x-\beta\biggl(\coth(x)-\frac{1}{x} \biggr) \biggr]. $$

The value k ₂ was in fact determined by the fact that $\varPhi_{\beta}'(k_{2})=0$. Moreover,

$$ \begin{aligned} \varPhi_\beta''(x)=& \biggl[\frac{1}{x^2}-\frac{1}{\sinh^2(x)} \biggr] \biggl[1-\beta \biggl( \frac{1}{x^2}-\frac{1}{\sinh^2(x)} \biggr) \biggr] \\ &{}+ \biggl[2\coth(x)\operatorname{csch}^2(x)-\frac{2}{x^3} \biggr] \biggl[x-\beta\biggl(\coth(x)-\frac{1}{x} \biggr) \biggr], \end{aligned} $$

and so

$$\varPhi_\beta''(k_2)= \biggl[ \frac{1}{k_2^2}-\frac{1}{\sinh^2(k_2)} \biggr] \biggl[1-\beta \biggl( \frac{1}{k_2^2}-\frac{1}{\sinh^2(k_2)} \biggr) \biggr] >0. $$

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1.2 A.2 Proof of Theorem 19

This section is devoted to the proof of the abstract approximation theorem used in Sect. 6. For convenience, we recall the statement.

Theorem 19

Let (W,W′) be an exchangeable pair of positive random variables. Suppose there exists a σ-field $\mathcal{F}\supseteq\sigma(W)$ and k>0 deterministic such that

$$\mathbb{E}\bigl[W'-W \bigl|\mathcal{F} \bigr]=3k \bigl(1-cW^2 \bigr)+E $$

and

$$\mathbb{E}\bigl[\bigl(W'-W\bigr)^2 \bigl|\mathcal{F} \bigr]=kW+E'. $$

Let X have density

Then there are constants C ₁, C ₂, C ₃ depending only on c such that for all $h\in C^{2}(\mathbb{R})$,

$$ \begin{aligned} \bigl|\mathbb{E}h(W)-\mathbb{E}h(X) \bigr|\le& \frac{C_1\|h\|_\infty}{k}\mathbb{E}|E|+ \biggl(\frac{C_2(\|h\|_\infty+ \|h'\|_\infty)}{k} \biggr)\mathbb{E}\bigl|E'\bigr| \\ &{}+ \biggl(\frac{C_3(\|h\|_\infty+\|h'\|_\infty+ \|h''\|_\infty)}{k} \biggr)\mathbb{E}\bigl|W'-W\bigr|^3. \end{aligned} $$

In any version of Stein’s method, a crucial component is the characterization of the distribution of interest by a linear operator. The following lemma identifies the characterizing operator for the random variable X defined above.

Lemma 24

Let Y be a positive random variable. Then Y has density

if and only if

$$ \mathbb{E}\bigl[Yf'(Y)+6\bigl(1-cY^2\bigr)f(Y) \bigr]=0 $$

(23)

for all f∈C ¹((0,∞)) such that $\int_{0}^{\infty}f(t)t^{5}e^{-3ct^{2}}\,dt<\infty$. That is, the characterizing operator T _p for the distribution with density p is defined by

$$[T_p f](x)=xf'(x)+6\bigl(1-cx^2 \bigr)f(x). $$

Not only is the random variable X characterized by the operator above, but this operator is invertible on $\{h:\mathbb{E}h(X)=0\}$, and the inverse has the following important boundedness properties.

Lemma 25

Let $h:\mathbb{R}\to\mathbb{R}$ be given. Suppose that

$$f=f_h:=\frac{1}{tp(t)}\int_0^t \bigl[h(s)-\mathbb{E}h(X) \bigr]p(s)\,ds, $$

with p as above. Then $[T_{p}f_{h}](x)=h(x)-\mathbb{E}h(X)$ and

• (a) ∥f _h ∥_∞≤5∥h∥_∞.

• (b) $\|f_{h}'\|_{\infty}\le42\sqrt{c}\|h\|_{\infty}+3\|h'\|_{\infty}$.

• (c) $\|f_{h}''\|_{\infty}\le C_{1}\|h\|_{\infty}+C_{2}\|h'\|_{\infty}+ C_{3}\|h''\|_{\infty}$, where C ₁, C ₂, C ₃ are constants depending only on c.

With these two lemmas, the proof of Theorem 19 is relatively straightforward.

Proof of Theorem 19

Given h, let f be the solution to the Stein equation described above. Then by exchangeability and the conditions on (W,W′),

$$ \begin{aligned} 0&=\mathbb{E}\bigl[\bigl(W'-W\bigr) \bigl(f \bigl(W'\bigr)+f(W)\bigr)\bigr] \\ &=\mathbb{E}\bigl[\bigl(W'-W\bigr) \bigl(f\bigl(W'\bigr)-f(W) \bigr)+2\bigl(W'-W\bigr)f(W)\bigr] \\ &=\mathbb{E}\bigl[\bigl(W'-W\bigr)^2f'(W)+E''+6k \bigl(1-cW^2\bigr)f(W)+2Ef(W)\bigr] \\ &=\mathbb{E}\bigl[kWf'(W)+E'f'(W)+E''+6k \bigl(1-cW^2\bigr)f(W)+2Ef(W)\bigr]. \end{aligned} $$

Then

$$\mathbb{E}\bigl[Wf'(W)+6\bigl(1-cW^2\bigr)f(W)\bigr]=- \frac{1}{k}\mathbb{E}\bigl[E'f'(W)+2Ef(W)+E'' \bigr], $$

and

$$\bigl|E''\bigr|\le\frac{\|f''\|_\infty}{2} \bigl|\bigl(W'-W \bigr)\bigr|^3. $$

The result is thus immediate from Lemma 25. □

We conclude by giving the proofs of the key lemmas.

Proof of Lemma 24

If Y has the density above, then the fact that Y satisfies (23) is just integration by parts.

For the reverse implication, one must solve the so-called Stein equation; i.e., given $h:(0,\infty)\to\mathbb{R}$, find f=f _h such that

$$tf'(t)+6\bigl(1-ct^2\bigr)f(t)=h(t)-\mathbb{E}h(X), $$

where X has the density above. The solution f is given by

$$ \begin{aligned}f(t)&=\frac{1}{tp(t)}\int_0^t \bigl[h(s)-\mathbb{E}h(X) \bigr]p(s)\,ds \\ &=-\frac{1}{tp(t)}\int_t^\infty\bigl[h(s)-\mathbb{E} h(X) \bigr]p(s)\,ds, \end{aligned} $$

where $p(s):=\frac{s^{5}}{z}e^{-3cs^{2}}$. Observe that

$$\frac{d}{dt} \bigl[tf(t)p(t)\bigr]=\bigl[f(t)+tf'(t) \bigr]p(t)+tf(t)p'(t)=\bigl[h(t)-\mathbb{E}h(X)\bigr]p(t), $$

so that

$$h(t)-\mathbb{E}h(X)=f(t)+tf'(t)+\frac{tf(t)p'(t)}{p(t)}=6 \bigl(1-ct^2\bigr)f(t)+tf'(t). $$

Here we have made use of the fact, frequently used below, that

$$\frac{tp'(t)}{p(t)}=5-6ct^2. $$

It is shown in the next lemma that f and f′ are both bounded, and so if Y satisfies (23), then if h is given and f solves the Stein equation,

$$\mathbb{E}h(Y)-\mathbb{E}h(X)=\mathbb{E}\bigl[Yf'(Y)+6\bigl(1-cY^2 \bigr)f(Y) \bigr]=0, $$

and so $Y\stackrel{d}{=}X$. □

Proof of Lemma 25

(a) By the first expression for f _h , if $t\le t_{o}:=\sqrt{\frac {5}{6c}}$, then

$$f(t)\le\frac{2\|h\|_\infty}{tp(t)} \biggl(\int_0^tp(s)\,ds \biggr)\le\frac{\| h\|_\infty t^5}{3zp(t)}\le\frac{e^{5/2}\|h\|_\infty}{3}\le5\|h\|_\infty. $$

By the second expression for f _h ,

$$\bigl|f(t)\bigr|\le\frac{2\|h\|_\infty\mathbb{P}[X\ge t]}{tp(t)}. $$

It is easy to show directly that $\mathbb{P}[X\ge t]\le\frac{p(t)}{6ct} (1+\frac{2}{3ct^{2}}+\frac {2}{9c^{2}t^{4}} )$, and so

$$\bigl|f(t)\bigr|\le\frac{2\|h\|_\infty}{6ct^2} \biggl(1+\frac {2}{3ct^2}+\frac {2}{9c^2t^4} \biggr)\le\frac{84\|h\|_\infty}{125} $$

for t≥t _o . This completes the proof.

(b) Recall that, because f solves the Stein equation,

$$tf'(t)=6f(t) \bigl(ct^2-1\bigr)+h(t)-\mathbb{E}h(X). $$

For t≤t _o , observe that

$$ \begin{aligned} h(t)-\mathbb{E}h(X)-6f(t)&=h(t)-\mathbb{E}h(X)-\frac {6}{tp(t)} \int_0^t\bigl[h(s)-\mathbb{E}h(X)\bigr]p(s)\,ds \\ &=\frac{6}{tp(t)}\int_0^t \biggl( \bigl[h(t)-\mathbb{E}h(X)\bigr]\frac {s^5p(t)}{t^5}-\bigl[h(s)-\mathbb{E}h(X)\bigr]p(s) \biggr)\,ds. \end{aligned} $$

Now,

$$ \begin{aligned} &\biggl\vert\frac{1}{tp(t)}\int _0^t\bigl[h(t)-\mathbb{E}h(X)\bigr] \biggl( \frac {s^5p(t)}{t^5}-p(s) \biggr)\,ds\biggr\vert \\ &\quad\le\frac{2\|h\|_\infty}{tp(t)}\int_0^t \biggl\vert1-\frac{s^5p(t)}{t^5p(s)}\biggr\vert p(s)\,ds=\frac{2\| h\|_\infty}{tp(t)}\int _0^t\bigl\vert1-e^{3c(s^2-t^2)}\bigr \vert p(s)\,ds \\ &\quad\le2\|h \|_\infty ct^2, \end{aligned} $$

making use of the fact that p(s)≤p(t) in this range. Also,

$$ \begin{aligned} &\biggl\vert\frac{1}{tp(t)}\int _0^t \bigl(\bigl[h(t)-\mathbb{E}h(X)\bigr]- \bigl[h(s)-\mathbb{E}h(X)\bigr] \bigr)p(s)\,ds\biggr\vert \\ &\quad\le\frac{\|h'\|_\infty}{tp(t)} \int_0^t(t-s)p(s)\,ds\le\frac{\|h'\|_\infty t}{2}, \end{aligned} $$

so that for t≤t _o ,

$$\frac{1}{t} \bigl|h(t)-\mathbb{E}h(X)-6f(t) \bigr|\le12\|h\| _\infty ct_o+3\bigl\|h'\bigr\| _\infty, $$

and thus

$$ \begin{aligned} \bigl|f'(t) \bigr|&\le6ct_o\|f \|_\infty+12\|h\|ct_o+3\bigl\|h'\bigr\|_\infty \\ &\le42ct_o \|h\|_\infty+3\bigl\|h' \bigr\|_\infty. \end{aligned} $$

If t≥t _o , then

$$\bigl|f'(t) \bigr|\le6ct\bigl|f(t)\bigr|+\frac{6\|f\|_\infty+2\| h\|_\infty}{t_o}. $$

By the second expression for f,

$$ct\bigl|f(t)\bigr|\le\frac{2c\|h\|_\infty\mathbb{P}[X\ge t]}{p(t)}\le\frac{\|h\|_\infty}{3t} \biggl(1+ \frac{2}{3ct^2}+\frac {2}{9c^2t^4} \biggr), $$

making use again of the estimate $\mathbb{P}[X\ge t]\le\frac{p(t)}{6ct} (1+\frac{2}{3ct^{2}}+\frac{2}{9c^{2}t^{4}} )$. Taking $t_{o}=\sqrt{\frac{5}{6c}}$ as before and making some trivial simplifying estimates completes the proof of this part.

(c) Differentiating both sides of the Stein equation gives that

$$ tf''(t)=12ctf(t)+ \bigl(6ct^2-7\bigr)f'(t)+h'(t). $$

(24)

If t≤t _o , then |12cf(t)|≤5c∥h∥_∞ and by the Stein equation, |6ctf′(t)|≤36c∥f∥_∞+12c∥h∥_∞≤192c∥h∥_∞. Also by the Stein equation,

$$f'(t)=6f(t) \biggl(ct-\frac{1}{t} \biggr)+ \frac{h(t)-\mathbb{E}h(X)}{t}. $$

The first term can be absorbed into the existing bound, so that

$$\bigl|f''(t)\bigr|\le C\|h\|_\infty+ \frac{1}{t}\biggl\vert h'(t)+\frac{42f(t)-7[h(t)-\mathbb{E} h(X)]}{t}\biggr \vert. $$

(From now on we will not bother to keep track of specific constants and their dependence on c.) Now,

$$ \begin{aligned} & h^\prime(t) -\frac{7[h(t)-\mathbb{E} h(X)-6f(t)]}{t} \\ &\quad=h^\prime(t)-\frac{42}{t^2p(t)}\int_0^t \biggl(\bigl[h(t)-\mathbb{E}h(X)\bigr]\frac{s^5p(t)}{t^5}-\bigl [h(s)-\mathbb{E}h(X)\bigr]p(s) \biggr)\,ds \\ &\quad=-\frac {42}{t^2p(t)}\int_0^t \biggl( \bigl[h(t)-\mathbb{E}h(X)\bigr]\frac{s^5p(t)}{t^5}-\bigl [h(s)-\mathbb{E}h(X)\bigr]p(s) \\ &\qquad{}- \frac {(t-s)s^5h^\prime(t)p(t)}{t^5} \biggr)\,ds \\ &\quad=-\frac{42}{t^2p(t)}\int_0^t \bigl( \bigl[h(t)-\mathbb{E}h(X)\bigr]e^{-3ct^2}-\bigl[h(s)-\mathbb {E}h(X)\bigr]e^{-3cs^2} \\ &{}\qquad{}-(t-s)h^\prime(t) e^{-3ct^2} \bigr)\frac {s^5}{z}\,ds. \end{aligned} $$

Let $H(t):=[h(t)-\mathbb{E}h(X)]e^{-3ct^{2}}$. Then

$$H'(t)= \bigl[h'(t) -6ct\bigl[h(t)-\mathbb{E}h(X)\bigr] \bigr]e^{-3ct^2}, $$

and so the equation above becomes

$$ \begin{aligned} & h^\prime(t)-\frac{7[h(t)-\mathbb{E}h(X)-6f(t)]}{t} \\ &\quad=\frac{42}{t^2p(t)}\int_0^t \bigl(H(s)-H(t)-(s-t)H^\prime(t)+6ct(t-s)\bigl[h(t)-\mathbb{E}h(X) \bigr]e^{-3ct^2} \bigr)\frac{s^5}{z}\,ds. \end{aligned} $$

It follows that

Now,

$$ H''(s)= \bigl[h''(s)-12csh'(s)-6c \bigl(6cs^2-1\bigr)\bigl[h(s)-\mathbb{E}h(X)\bigr] \bigr]e^{-3cs^2}, $$

so

$$\sup_{s\in(0,t)} \bigl|H''(s)\bigr|\le \bigl\|h''\bigr\|_\infty+12ct_o \bigl\|h'\bigr\|_\infty+24c\|h\| _\infty. $$

All together, this gives that for t≤t _o ,

$$\bigl|f''(t)\bigr|\le C_1 \|h\|_\infty+C_2 \bigl\|h'\bigr\|_\infty+ \bigl\|h'' \bigr\|_\infty, $$

where C ₁, C ₂, C ₃ are constants depending only on c.

For t>t _o , it follows from (24) and estimates already carried out that

$$ \begin{aligned}\bigl|f''(t)\bigr|&\le12c\|f \|_\infty+ \biggl(6ct+\frac{7}{t_o} \biggr)\bigl|f'(t)\bigr|+ \frac{\|h'\| _\infty}{t_o} \\ &\le C_4\|h\|_\infty+C_5\bigl\|h' \bigr\|_\infty+ct\bigl|f'(t)\bigr|. \end{aligned} $$

By the Stein equation,

$$ctf'(t)=6cf(t) \bigl(ct^2-1\bigr)+ch(t)-c\mathbb{E}h(X), $$

and

$$\bigl|c^2t^2f(t)\bigr|\le\frac{2c^2t\|h\|_\infty\mathbb{P}[X\ge t]}{p(t)}\le\frac{c\|h\|_\infty}{3} \biggl(1+\frac{2}{3ct^2}+\frac {2}{9c^2t^4} \biggr), $$

and so finally

$$\bigl|f''(t)\bigr|\le C_6\|h\|_\infty+C_4 \bigl\|h'\bigr\|_\infty. $$

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