Limiting derivative and concentration

At every $\gamma \in J$, finite differentiation gives $p_L^{\mathrm{\prime }}(\gamma )=-L^{-d}{\mathbb{E}}_{\gamma ,L}{H}_L$ and $p_L^{\mathrm{\prime }\mathrm{\prime }}(\gamma )=L^{-d}{\mathrm{Var}}_{\gamma ,L}(H_L)\ge 0$; these derivatives are continuous since the finite partition sum is smooth and positive. Apply [F5] to $p_L^{\mathrm{\prime }}$ on $[a,b]\subset J$: $p_L^{\mathrm{\prime }}(b)-p_L^{\mathrm{\prime }}(a)=p_L^{\mathrm{\prime }\mathrm{\prime }}(c)(b-a)\ge 0$, so $p_L^{\mathrm{\prime }}$ is nondecreasing. For $a<b<c$ in $J$, apply [F5] to $p_L$ on $[a,b]$ and $[b,c]$. Their secant slopes are derivative values respectively below and above $b$, hence the left slope is no greater than the right. Rearrangement gives $p_L(b)\le [(c-b)p_L(a)+(b-a)p_L(c)]\mathrm{/}(c-a)$, the convexity inequality. Passing this inequality to the pointwise limit proves convexity of $p$.

Fix $t>0$ small enough that $\beta \pm t\in J$. Since ${\exp }^{\mathrm{\prime }}=\exp >0$ and $\exp (0)=1$, [F5] gives $\exp (u)\ge 1$ for $u\ge 0$. For the upper tail, the pointwise inequality ${\mathbf{1}}_{\{H_L\ge L^d(e+\epsilon )\}}\le e^{t(H_L-L^d(e+\epsilon ))}$ and finite summation give $\mathbb{P}(H_L\mathrm{/}{L}^d\ge e+\epsilon )\le \exp \{L^d[p_L(\beta -t)-p_L(\beta )-t(e+\epsilon )]\}$. Differentiability of $p$ and $e=-p^{\mathrm{\prime }}(\beta )$ imply $p(\beta -t)=p(\beta )+te+o(t)$; choose $t$ so $\mathrm{\mid }p(\beta -t)-p(\beta )-te\mathrm{\mid }\le t\epsilon \mathrm{/}4$ and $\mathrm{\mid }p(\beta +t)-p(\beta )+te\mathrm{\mid }\le t\epsilon \mathrm{/}4$. Thus the upper bracket's limit is at most $-3t\epsilon \mathrm{/}4$. Pointwise convergence at $\beta -t$ and $\beta$ then makes the bracket at most $-t\epsilon \mathrm{/}2$ for all sufficiently large $L$.

For any small $h>0$ with $\beta \pm h\in J$, apply [F5] to $p_L$ on $[\beta -h,\beta ]$ and $[\beta ,\beta +h]$. Each secant slope is a value of $p_L^{\mathrm{\prime }}$ on its respective open interval, so the monotonicity in step 1.1 gives $[p_L(\beta )-p_L(\beta -h)]\mathrm{/}h\le p_L^{\mathrm{\prime }}(\beta )\le [p_L(\beta +h)-p_L(\beta )]\mathrm{/}h$. Take lower and upper limits in $L$, use [F1], then let $h\downarrow 0$; differentiability of $p$ forces $p_L^{\mathrm{\prime }}(\beta )\to p^{\mathrm{\prime }}(\beta )$. The identity in step 1.1 yields the claimed mean limit.

For the lower tail use ${\mathbf{1}}_{\{H_L\le L^d(e-\epsilon )\}}\le e^{-t(H_L-L^d(e-\epsilon ))}$; its expectation is $\exp \{L^d[p_L(\beta +t)-p_L(\beta )+t(e-\epsilon )]\}$. For the same $t$ chosen in step 1.2, the bracket has limit at most $-3t\epsilon \mathrm{/}4$; increasing $L_0$ therefore makes this at most $e^{-L^{d}t\epsilon \mathrm{/}2}$. Adding both tail bounds yields the result with $c=t\epsilon \mathrm{/}2$.

Steps 1.2–2.2 bound the two tails whose union is $\{\mathrm{\mid }{H}_L\mathrm{/}{L}^d-e\mathrm{\mid }\ge \epsilon \}$; the theorem is conditional on differentiability of $p$ and does not assert convergence of second derivatives or equivalence for arbitrary observables.