Ensembles And Limits

On a compact interval $K\subset J$, let $Z_N$, $D_N$, and $Q_N$ be the first $N$ partial sums of ${\sum }_i{g}_i{e}^{-\beta E_i}$, $-{\sum }_i{g}_i{E}_i{e}^{-\beta E_i}$, and ${\sum }_i{g}_i{E}_i^2{e}^{-\beta E_i}$. Since $\mathrm{\mid }{E}_i{\mathrm{\mid }}^r\le 1+\mathrm{\mid }{E}_i{\mathrm{\mid }}^2$ for $r=0,1,2$, each tail is bounded uniformly on $K$ by the corresponding tail of ${\sum }_i{a}_i$, which tends to zero [A1]. Thus all three sequences converge uniformly; call their limits $Z,D,Q$. Each finite partial sum has the displayed derivatives by the chain rule [F5] and the exponential derivative [F2]. Each partial sum is continuous, so the uniform limits are continuous by the $\epsilon \mathrm{/}3$ definition of continuity.

For an interior $\beta$ and sufficiently small $h\mathrm{\ne }0$, apply [F4] to $Z_N$ on the interval with endpoints $\beta$ and $\beta +h$: its difference quotient is $D_N({\xi }_{N,h})$ for some intermediate ${\xi }_{N,h}$. Choose $N=N(h)$ so that ${\sup }_K\mathrm{\mid }{Z}_N-Z\mathrm{\mid }<h^2$ and ${\sup }_K\mathrm{\mid }{D}_N-D\mathrm{\mid }<\mathrm{\mid }h\mathrm{\mid }$. The difference quotient for $Z$ then differs from $D(\beta )$ by at most $2\mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }h\mathrm{\mid }+{\sup }_{\mathrm{\mid }t-\beta \mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }}\mathrm{\mid }D(t)-D(\beta )\mathrm{\mid }$, which tends to zero by continuity of $D$. Hence $Z^{\mathrm{\prime }}=D$. Applying the same argument to $D_N$ and $Q_N$ proves $D^{\mathrm{\prime }}=Q$, so the displayed derivatives hold.

The summands of $Z$ are positive and at least one is present, hence $0<Z<\mathrm{\infty }$ on $K$; summing the definition of $p_i$ gives ${\sum }_i{p}_i=1$. The envelope gives finite first and second moments, and ${\sum }_i{p}_i(E_i-\bar{E}{)}^2\le 2{\sum }_i{p}_i{E}_i^2+2{\bar{E}}^2<\mathrm{\infty }$. Differentiating $\log Z$ using [F3] yields $-Z^{\mathrm{\prime }}\mathrm{/}Z=\bar{E}$ and $Z^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}Z-(Z^{\mathrm{\prime }}\mathrm{/}Z{)}^2=⟨E^2⟩-{\bar{E}}^2={\mathrm{Var}}_{\beta }(E)$ after expanding the defining variance.

Since $p_i\mathrm{/}{g}_i=e^{-\beta E_i}\mathrm{/}Z$, its logarithm is $-\beta E_i-\log Z$. The finite first moment makes ${\sum }_i{p}_i\mathrm{\mid }\log (p_i\mathrm{/}{g}_i)\mathrm{\mid }\le \mathrm{\mid }\beta \mathrm{\mid }{\sum }_i{p}_i\mathrm{\mid }{E}_i\mathrm{\mid }+\mathrm{\mid }\log Z\mathrm{\mid }{\sum }_i{p}_i<\mathrm{\infty }$; summing the identity gives $H_g=\beta \bar{E}+\log Z$.

The sequence is norm bounded, say $\mathrm{\parallel }{x}^{(n)}\mathrm{\parallel }\le M$. For each coordinate, $\mathrm{\mid }{x}_i^{(n)}-x_i^{(m)}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(n)}-x^{(m)}\mathrm{\parallel }$, so [F2] gives a unique limit $x_i$. For every finite $F\subset I$, passing to the limit in its finite sum gives ${\sum }_{i\in F}\mathrm{\mid }{x}_i{\mathrm{\mid }}^2\le M^2$. Taking the finite-subset supremum proves $x\in {\mathrm{\ell }}^2(I)$. No selection of nonunique coordinate limits is involved.

If $x,y\in D(H_E)$ and $a,b\in \mathbb{C}$, then for every finite $F$, ${\sum }_{i\in F}\mathrm{\mid }{E}_i(a{x}_i+b{y}_i){\mathrm{\mid }}^2\le 2\mathrm{\mid }a{\mathrm{\mid }}^2{\sum }_{i\in F}\mathrm{\mid }{E}_i{x}_i{\mathrm{\mid }}^2+2\mathrm{\mid }b{\mathrm{\mid }}^2{\sum }_{i\in F}\mathrm{\mid }{E}_i{y}_i{\mathrm{\mid }}^2\mathrm{.}$ The right side is bounded independently of $F$, so $ax+by\in D(H_E)$; the zero vector is included. The coordinate action is linear and has square-summable output by the domain definition. Every finite-support vector belongs to this domain because each $E_i$ is finite.

Given $ϵ>0$, choose $N$ with $\mathrm{\parallel }{x}^{(n)}-x^{(m)}\mathrm{\parallel }\le ϵ$ for all $n,m\ge N$. For fixed $n\ge N$ and finite $F$, let $m\to \mathrm{\infty }$ to obtain ${\sum }_{i\in F}\mathrm{\mid }{x}_i^{(n)}-x_i{\mathrm{\mid }}^2\le {ϵ}^2$. Taking the supremum over $F$ gives $\mathrm{\parallel }{x}^{(n)}-x\mathrm{\parallel }\le ϵ$. Thus the space is complete and is Hilbert by [F3].

For any $x\in {\mathrm{\ell }}^2(I)$ and $ϵ>0$, the finite-subset supremum supplies a finite $F$ whose square sum differs from the total by less than ${ϵ}^2$. For every finite $G\subset I\setminus F$, its square sum is at most that difference, by applying the total bound to $F\cup G$. Hence the norm of the omitted tail is at most $ϵ$. Finite-coordinate truncations therefore approximate $x$ and lie in $D(H_E)$ by step 1.2, proving density. If $I$ is empty all spaces and maps are zero, and the same statements hold directly.

By [F5] the ambient space is Hilbert and the operator domain and action are linear. Every finitely supported sequence belongs to $D(H_E)$. For $x\in {\mathrm{\ell }}^2(I)$, the tails of its square-sum can be made arbitrarily small, so finite-coordinate truncations converge in norm to $x$; finite-support vectors are therefore dense in ${\mathrm{\ell }}^2(I)$.

For $x,y\in D(H_E)$, Cauchy–Schwarz [F4] makes ${\sum }_i\mathrm{\mid }{E}_i{x}_i\overline{y_i}\mathrm{\mid }$ finite, and reality of $E_i$ gives $⟨H_{E}x,y⟩={\sum }_i{E}_i{x}_i\overline{y_i}=⟨x,H_{E}y⟩$. Thus $H_E$ is symmetric.

If $y\in D(H_E^{\ast })$, there is $z\in {\mathrm{\ell }}^2(I)$ such that $⟨H_{E}x,y⟩=⟨x,z⟩$ for every $x\in D(H_E)$. Taking $x=u_i$ gives $E_i\overline{y_i}=\overline{z_i}$, hence $z_i=E_i{y}_i$ for every $i$; since $z\in {\mathrm{\ell }}^2(I)$, $y\in D(H_E)$ and $H_E^{\ast }y=H_{E}y$. Conversely symmetry in step 1.2 shows every $y\in D(H_E)$ is in $D(H_E^{\ast })$ with that value. Therefore $D(H_E^{\ast })=D(H_E)$ and $H_E^{\ast }=H_E$.

For every finite real $a$, $\{f>a\}={\bigcup }_n\{f_n>a\}$, since the sequence increases to $f$. These threshold sets generate the order-topology Borel sigma-algebra in [F1], so $f$ is measurable. The simple minorants of $f_n$ are also minorants of $f_{n+1}$, so their integral suprema increase. Each is at most $\int f$; hence $L:={\lim }_n\int f_n$ exists in $[0,\mathrm{\infty }]$ and $L\le \int f$.

Fix a nonnegative simple $s\le f$ and $0<t<1$. The measurable sets $A_n:=\{f_n\ge ts\}$ increase and cover $\{s>0\}$, since $f_n(x)\to f(x)\ge s(x)>ts(x)$ there. Write $s={\sum }_{j=1}^m{c}_j{\mathbf{1}}_{B_j}$ in disjoint level form. For each $c_j>0$, $A_n\cap B_j\uparrow B_j$; countable additivity, applied to successive disjoint differences, gives $\mu (A_n\cap B_j)\uparrow \mu (B_j)$. The finite-sum definition [F2] then gives $\int s{\mathbf{1}}_{A_n}\uparrow \int s$. As $f_n\ge ts{\mathbf{1}}_{A_n}$, monotonicity and homogeneity of the simple integral imply $L\ge t\int s$. Let $t\uparrow 1$ and take the supremum over $s\le f$; [F1, F3] give $L\ge \int f$. Together with step 1.1 this proves monotone convergence.

For nonnegative measurable $f$, define $s_n:=2^{-n}{\sum }_{k=1}^{n{2}^n}{\mathbf{1}}_{\{f\ge k{2}^{-n}\}}$. Each $s_n$ is a finite-valued measurable simple function. Every threshold counted in $s_n$ is also counted at the even-index threshold in $s_{n+1}$, so $s_n\le s_{n+1}$. If $f(x)<\mathrm{\infty }$, once $n>f(x)$ the mesh gives $0\le f(x)-s_n(x)<2^{-n}$; if $f(x)=\mathrm{\infty }$, then $s_n(x)=n$. Thus $s_n\uparrow f$. Define $t_n:=2^{-n}{\sum }_{k=1}^{n{2}^n}{\mathbf{1}}_{\{g\ge k{2}^{-n}\}}$; the same threshold and mesh argument shows $t_n\uparrow g$. By [F4], $f+g$ and $s_n+t_n$ are measurable with the defined nonnegative extended addition. Apply the monotone-convergence conclusion of step 2.1 to $s_n$, $t_n$, and $s_n+t_n$. Refine their finitely many level sets to the finite partition of intersections $B_j\cap C_k$; regrouping the resulting sums by [F2] gives $\int (s_n+t_n)=\int s_n+\int t_n$. Passing to the three monotone limits proves $\int (f+g)=\int f+\int g$.

For $c>0$, simple minorants of $cf$ are exactly $c$ times simple minorants of $f$, so the supremum definition gives $\int cf=c\int f$; for $c=0$ both sides are zero. If an integrable real function is expressed as $u-v$ with $u,v\ge 0$ integrable, the value $\int u-\int v$ is independent of the expression: $u-v=u^{\mathrm{\prime }}-v^{\mathrm{\prime }}$ implies $u+v^{\mathrm{\prime }}=u^{\mathrm{\prime }}+v$, and step 3.1 equates the finite sums of integrals. The definition using positive and negative parts is one such expression, so signed integrals are linear.

Set $h_k:={\inf }_{n\ge k}{f}_n$. For each real $a$, $\{h_k<a\}={\bigcup }_{n\ge k}\{f_n<a\}$, so $h_k$ is measurable by the sigma-algebra closure in [F1]. The sequence $h_k$ is increasing and its pointwise limit is ${\mathrm{lim\hspace{0.17em}inf}}_n{f}_n$, which is measurable since $\{{\lim }_k{h}_k>a\}={\bigcup }_k\{h_k>a\}$.

For every $n\ge k$, monotonicity gives $\int h_k\le \int f_n$, so $\int h_k\le {\inf }_{n\ge k}\int f_n$. Taking $k\to \mathrm{\infty }$ and applying monotone convergence to $h_k$ yields $\int \mathrm{lim\hspace{0.17em}inf}{f}_n={\lim }_k\int h_k\le {\lim }_k{\inf }_{n\ge k}\int f_n={\mathrm{lim\hspace{0.17em}inf}}_n\int f_n$.

The set on which $(f_n)$ is Cauchy is measurable: write it as the countable intersection/union/intersection of sets $\{\mathrm{\mid }{f}_n-f_k\mathrm{\mid }<1\mathrm{/}m\}$ using [F4]. On this set the finite real and imaginary limits exist; their measurable limit functions, set to zero on its complement, define the measurable $f_0$. The assumed almost-everywhere convergence gives $f_0=f$ outside a measurable null set. Enlarge that set by the countably many measurable domination exceptions to a measurable null set $N$. Set $f_n$ and $f_0$ to zero on $N$, which preserves their measurability. A nonnegative measurable function supported on $N$ has integral zero by the simple-minorant definition, so these changes preserve every defined integral. Convergence and domination now hold everywhere, and $\mathrm{\mid }{f}_0\mathrm{\mid }\le g$ proves integrability. This does not redefine the potentially nonmeasurable $f$ and then claim it was measurable.

First take all functions real. The nonnegative functions $f_n+g$ converge pointwise to $f_0+g$. Fatou gives $\int (f_0+g)\le {\mathrm{lim\hspace{0.17em}inf}}_n\int (f_n+g)=\int g+{\mathrm{lim\hspace{0.17em}inf}}_n\int f_n$, so $\int f_0\le {\mathrm{lim\hspace{0.17em}inf}}_n\int f_n$. Applying Fatou to $g-f_n\ge 0$ gives $\int (g-f_0)\le {\mathrm{lim\hspace{0.17em}inf}}_n\int (g-f_n)=\int g-{\mathrm{lim\hspace{0.17em}sup}}_n\int f_n$, hence ${\mathrm{lim\hspace{0.17em}sup}}_n\int f_n\le \int f_0$. Additivity of integrable signed integrals follows by writing each as a difference of its positive and negative parts and using [F3]. The inequalities imply convergence of integrals.

For complex $f_n$, apply step 2.1 separately to their real and imaginary parts, each dominated by $g$; their integral limits combine to $\int f_0$. To prove $L^1$ convergence, apply Fatou to the nonnegative functions $2g-\mathrm{\mid }{f}_n-f_0\mathrm{\mid }$, which converge to $2g$; it gives $2\int g\le 2\int g-{\mathrm{lim\hspace{0.17em}sup}}_n\int \mathrm{\mid }{f}_n-f_0\mathrm{\mid }$, so the latter limsup is zero. Therefore $\int \mathrm{\mid }{f}_n-f_0\mathrm{\mid }\to 0$ as claimed.

For each fixed parameter, measurable energy/observable coordinates and [F7] make every finite linear combination, exponential weight, weighted first/second moment and parameter difference quotient measurable; their displayed pointwise derivative limits are the same measurable compositions. For $\beta$ in the interior of a compact interval $K\subset J$, the difference quotient of $e^{-\beta E(x)}$ tends pointwise to $-E(x)e^{-\beta E(x)}$. The mean-value theorem [F5] bounds its absolute value, for small $h$, by $\mathrm{\mid }E\mathrm{\mid }{e}^{-\theta E}$ for an intermediate $\theta \in K$; this is at most $G_K$ because $\mathrm{\mid }E\mathrm{\mid }\le 1+\mathrm{\mid }E{\mathrm{\mid }}^2$. Apply dominated convergence [F2] along any sequence of nonzero increments tending to zero; by [F1] the integral of the difference quotient is the difference quotient of $Z$. Every such sequence has the same limit, so $Z^{\mathrm{\prime }}=-\int E{e}^{-\beta E}d\mu$. Applying the same argument to $-E{e}^{-\beta E}$, dominated by $G_K$, gives $Z^{\mathrm{\prime }\mathrm{\prime }}=\int E^2{e}^{-\beta E}d\mu$; for each of the derivative kernels $-E{e}^{-\gamma E}$ and $E^2{e}^{-\gamma E}$, pointwise continuity in $\gamma$ and the same envelope let dominated convergence prove continuity of both derivatives.

The density $p_{\beta }$ is nonnegative and integrates to one. The envelope makes its first and second energy moments finite. Differentiating $\log Z$ using [F4] gives $-Z^{\mathrm{\prime }}\mathrm{/}Z=\int E{p}_{\beta }d\mu$ and $Z^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}Z-(Z^{\mathrm{\prime }}\mathrm{/}Z{)}^2=\int E^2{p}_{\beta }d\mu -(\int E{p}_{\beta }d\mu {)}^2$. The latter is the variance by expanding its defining square.

Since $\log p_{\beta }=-\beta E-\log Z$ almost everywhere and $\int \mathrm{\mid }E\mathrm{\mid }{p}_{\beta }d\mu <\mathrm{\infty }$, the bound $p_{\beta }\mathrm{\mid }\log p_{\beta }\mathrm{\mid }\le \mathrm{\mid }\beta \mathrm{\mid }\mathrm{\mid }E\mathrm{\mid }{p}_{\beta }+\mathrm{\mid }\log Z\mathrm{\mid }{p}_{\beta }$ makes the entropy integrand absolutely integrable. Substitution and normalization yield $-\int p_{\beta }\log p_{\beta }d\mu =\beta ⟨E{⟩}_{\beta }+\log Z$. This entropy depends on the chosen reference measure $\mu$.

Apply [F2] to $s_B$ at $u_0$ with increment $-E_i$. It gives $s_B(u_0-E_i)=s_B(u_0)-\beta E_i+r_i$ where $\mathrm{\mid }{r}_i\mathrm{\mid }\le M{E}_i^2\mathrm{/}2\le \delta$.

The marginal is therefore $p_i=q_i{e}^{r_i}\mathrm{/}C$, where $C:={\sum }_j{q}_j{e}^{r_j}$. The bounds $e^{-\delta }\le e^{r_j}\le e^{\delta }$ and ${\sum }_j{q}_j=1$ imply $e^{-\delta }\le C\le e^{\delta }$, so $e^{-2\delta }\le p_i\mathrm{/}{q}_i\le e^{2\delta }$.