Spectral theory of generalized second order differential operators and its applications to Markov processes

Theorem 1. Following three conditions are mutually equivalent; - (i) mn(x)•¨m0(x)(n•¨•‡) at each continuity point x e [0,•‡) of m0(x). - (ii) ƒÕn(x,-s)•¨ƒÕ0(x,-s)(n•¨•‡), x e [0, •‡), s>0. - (iii) hn(s)•¨h0(s)(n•¨•‡), s>0. PROOF. (i)\_??\_(ii). Let s be a positive number and A (<l0) be a con tinuity point of m0(x). Then by (3), $$(14) 0 \leq \varphi_n(x, -s) \leq \exp\{sAm_n(A)\}, x \in [0, A].$$ It is easy to show that (15) $$\frac{d}{dx}\varphi_n(x,-s) \leq sm_n(A) \exp\{sAm_n(A)\}, \qquad x \in [0,A].$$ Since mn(A) converges to m0(A) (<oo) by assumption, it follows that, for some sufficiently large N, {(ƒÕn(x,-s)}•‡n=N is uniformly bounded and equi continuous in xe [0, A]. Appealing to the Ascoli-Arzela's theorem, we can choose a subsequence {nj} and a function ƒÕ\(x,-s) defined on [0, A] such that ƒÕnj(x,-s)•¨ƒÕ\(x,-s) uniformly in xe [0, A]. Since $$\varphi_{n_j}(x,-s)=1+s\int_{[0,x)}(x-\xi)\varphi_{n_j}(\xi,-s)\ dm_{n_j}(\xi), \qquad x\in[0,A],$$ letting j to infinity, we have (16) $$\varphi^(x,-s)=1+s\int_{[0,x)}(x-\xi)\varphi^(\xi,-s)\ dm_0(\xi), \qquad x\in[0,A].$$ By the uniqueness of the solutions of the integral equation, we have ƒÕ\*(x ,-s)=ƒÕ0(x,-s). Hence, a standard argument provides us with ƒÕn(x,-s)•¨ƒÕ0(x,-s) uniformly in x e [0, A]. A being arbitrary, we obtain (17) $$\lim_{n \to \infty} \varphi_n(x, -s) = \varphi_0(x, -s), \quad s > 0, \quad x \in [0, \ell_0]$$ When x e (l0, •‡), we choose y so that l0<y<x. Then $$\lim_{n\to\infty} m_n(y) (= m_0(y)) = \infty.$$ Since (18) $$\varphi_n(x,-s) = 1 + s \int_{[0,x)} m_n(\xi) d\xi$$ $$\geq 1 + s(x-y)m_n(y),$$ we have, (19) $$\lim_{n\to\infty} \varphi_n(x,-s) = \infty = \varphi_0(x,-s), \qquad x \in (\ell_0,\infty),$$ (ii)•Ë(iii). Note that there exists c(>0) such that (20) $$\varphi_0(c,-1) = 1 + \int_{[0,c)} m_0(\xi) d\xi > 1.$$ Setting s=-1, x=cin (3), we have, (21) $$\varphi_n(c,-1) \leq \exp\left\{\int_0^c m_n(\xi) d\xi\right\} \qquad (n=0,1,2,\cdots).$$ Since we assume $$\varphi_n(c,-1) \rightarrow \varphi_0(c,-1) \quad (>1)$$ there exist positive constant ƒÂ and N such that $$(22) 1 + \delta \leq \exp m_n(c), n \geq N$$ Hence, mn(c)•†ƒÁ0, n>N holds for some suitably chosen constant ƒÁ(>0) . Using (3) again, (23) $$\varphi_n(x,-s) \ge 1 + s \int_0^x m_n(\xi) d\xi \ge 1 + s(x-c), \qquad n \ge N.$$ Hence, appealing to the dominated convergence theorem, (24) $$\lim_{n\to\infty} \int_0^\infty \frac{dx}{\varphi_n(x,-s)^2} = \int_0^\infty \frac{dx}{\varphi_0(x,-s)^2}, \quad s>0.$$ This proves hn(s)•¨h0(s). (iii)•Ë(i). The family of left-continuous, non-decreasing functions {arctg mn(x)}•‡n=1 is uniformly bounded and therefore we can choose an inex tensible measure dm\*(x) and a subsequence {mnj} of an arbitrary sub sequence {mnj}•¼{mn} nnsuch that (25) $$m_n(x) \rightarrow m^*(x)$$ at each continuity point $x \in [0, \infty)$ of $m^*(x)$ . Since we have already proved that (i) implies (iii), it holds that hnj(s) con verges to h\(s), h\(s) being the characteristic function of m\(x). Of course h\(s)=h0(s) by the assumption (iii). Hence m\*(x)=m0(x) by the uniqueness part of the above Krein's theorem. This completes our theorem.

Lemma 1. Let an order preserving homeomorphism f(x) of [0 , •‡) to itself vary regularly with exponent ƒÏ(0<ƒÏ<•‡) at •‡ [0]. Then the inverse function f-1(x) of f(x) varies regularly at •‡ [0] with exponent 1/ƒÏ. PROOF. If lim f-1(Ax)> A'MA for some 2>0, we can choose a f \_1(x) positive constant ƒÂ and a sequence $$x_1 < x_2 < x_3 < \cdots < x_n < \cdots \rightarrow \infty \quad [x_1 > x_2 > \cdots > x_n > \cdots \rightarrow 0]$$ ] so that f-1(ƒÉxn)>(ƒÉ1/ƒÏ+ƒÂ)f-1(xn)(n=1,2, •c). Since f is order preserving, $$\lambda = \frac{\lambda x_n}{x_n} = \frac{f(f^{-1}(\lambda x_n))}{f(f^{-1}(x_n))}$$ $$\geq \frac{f((\lambda^{1/\rho} + \delta)f^{-1}(x_n))}{f(f^{-1}(x_n))} \rightarrow (\lambda^{1/\rho} + \delta)^{\rho} \qquad (>\lambda).$$ This is a contradiction. Hence, (28) $$\overline{\lim} \frac{f^{-1}(\lambda x)}{f^{-1}(x)} \leq \lambda^{1/\rho} \qquad (\lambda > 0).$$ Similarly we have (29) $$\underline{\lim} \frac{f^{-1}(\lambda x)}{f^{-1}(x)} \ge \lambda^{1/\rho} \qquad (\lambda > 0).$$

Lemma 2. Let h(s) be the characteristic function of m(x) . Then a-lh(cs) is that o f acm(ax) for arbitrary constants a and c. PROOF. This can be proved by changing the integral parameters .

Theorem 2. Let L(x) be a continuous function varying slowly at •‡ [0] such that tƒ¿L(t) is increasing for large [small] t and let fƒ¿(x) (0•…ƒ¿•…1) denote the inverse function o f tƒ¿L(t). (When ƒ¿=0, we assume that lira L(t)=•‡ [0] in order to assure the existence of the inverse function .) Then, (i) fƒ¿(x) varies regularly at •‡ [0] with exponent 1/ƒ¿. (ii) $$h(s) \sim D_a s^{-a} L(1/s) \qquad as \ s \to 0 \ [\infty]$$ if and only if (31) $$\lim_{\lambda \to \infty[0]} \frac{m(x\lambda)}{f_{\epsilon}(\lambda)} = x^{1/\alpha - 1} \qquad (x > 0, x \neq 1),$$ where Dƒ¿ denotes the constant defined in Example 1. - (iii) If, in particular, ƒ¿•'0, the following three conditions are mutually equivalent; - (A) h(s)•`Dƒ¿s-ƒ¿L(1/s) as s•¨0+[•‡]. (B) $$\sigma(\lambda) \sim \{\alpha(1-\alpha)\}^{1-\alpha} \{\Gamma(2-\alpha)\}^{-2} \lambda^{1-\alpha} L(1/\lambda) \text{ as } \lambda \to 0 + [\infty].$$ (C) $$m(x) \sim x^{1/\alpha-1}K_{\alpha}(x)$$ as $x \to \infty[0+]$ , where Kƒ¿(x) is the slowly varing function defined by Kƒ¿(x)=x-1/ƒ¿fa(x). PROOF. (i) This was already proved in Lemma 1. (ii) We first remark that (30) is equivalent to (30)' $$\lim_{A\to\infty\atop [0]} \frac{h(s/A)}{A^{\alpha}L(A)} = D_{\alpha}s^{-\alpha}, \quad s>0.$$ In fact, (30) follows from (30)•L by setting s=1, while (30)•L is obtained from (30) as follows; $$\begin{split} &\lim_{\stackrel{A\to\infty}{=[0]}} \frac{h(s/A)}{A^{\alpha}L(A)} = &\lim_{\stackrel{A\to\infty}{=[0]}} \frac{h(s/A)}{(A/s)^{\alpha}L(A/s)} \frac{s^{-\alpha}L(A/s)}{L(A)} \\ &= &\lim_{\stackrel{t\to0}{=[\infty]}} \frac{h(t)}{t^{-\alpha}L(1/t)} \lim_{\stackrel{A\to\infty}{=[0]}} \frac{L(A/s)}{L(A)} s^{-\alpha} \\ &= &D.s^{-\alpha}. \end{split}$$ On the other hand, (31) is equivalent to (31)' $$\lim_{\substack{A \to \infty \\ 101}} \frac{A^{\alpha}L(A)}{A} m(A^{\alpha}L(A)x) = x^{1/\alpha - 1}.$$ By Lemma 2, 1h(s/A) is the characteristic function of A aL(A) Aƒ¿L(A) m(Aƒ¿L(A)x . Since Dƒ¿s-ƒ¿ is the characteristic function of x1/ƒ¿-1 A (30)•L and (31)•L are equivalent to each other by Theorem 1, and therefore the assertion is proved. (iii) This is easily shown by (ii) and the Hardy-Littlewood Tauberian theorem for the Stieltjes transform. REMARK. When ƒ¿=0, it does not necessarily follow from (31) that m(x)•`x-1fƒ¿(x). For instance, take ex for fƒ¿ and xƒÏex (ƒÏ•†0) for m(x). However, it still holds that m(x) varies regularly with exponent •‡ if (31) is valid when ƒ¿=0. In fact, since fƒ¿(x) varies regularly with exponent •‡ by (ii), we have, $$\lim_{\substack{x \to \infty \\ [0]}} \frac{m(\lambda x)}{m(x)} = \lim_{\substack{x \to \infty \\ [0]}} \frac{\sqrt{\lambda} x m(\lambda x)}{f_{\alpha}(\sqrt{\lambda} x)} \frac{f_{\alpha}(\sqrt{\lambda} x)}{\sqrt{\lambda} x m(x)}$$ $$= \lim_{\substack{y \to \infty \\ [0]}} \frac{y m(\sqrt{\lambda} y)}{f_{\alpha}(y)} \lim_{\substack{x \to \infty \\ [0]}} \frac{f_{\alpha}(\sqrt{\lambda} x)}{\sqrt{\lambda} x m(x)}$$ $$= \lambda^{\infty}.$$ It is easy to see that, in the theorem above, L(t)=C-ƒ¿ifK(x)=C (posi tive constant). Hence, as the corollary of Theorem 2 (iii), we have the following;

Theorem 3. Let a=inf {x;m(x)>0}. Then, $$m(x) = (x-a)^{\beta} \{C + o(1)\} \quad (C, \beta > 0) \quad as \quad x \to \infty[a+]$$ if and only if $$\sigma(t) = C^{-1/(\beta+1)}B(\beta)t^{\beta/(\beta+1)} + o(t^{\beta/(\beta+1)}) \quad as \quad t \to 0[\infty]$$ where $$B(\beta) = \{\beta(1+\beta)^{-2}\}^{\beta/(\beta+1)} \{\Gamma((2\beta+1)/(\beta+1))\}^{-2}$$ The 'only if' part of Theorem 3 has been obtained by I. S. Kac [8]. Finally, we remark that the method used in the proof of Theorem A of I. S. Kac [8] can be generalized easily as follows;

Theorem 4. Let f(x) be a positive increasing function on [0, •‡) and f-1(x) be its inverse function. Then; - (i) If xm(x)•†Cf(x) [xm(x)•…Cf(x)] (C>0) for all x>0, then h(s)•… 2f-1(1/ (Cs)) [h(s)•†f-1(1/Cs))/2] for all s>0. - (ii) If h(s)•…Cf-1(1/s) [h(s)•†Cf-1(1/s)] (C>0) for all s>0, then xm(x) •† f(x/(2C))[xm(x)•…f(2x/C)] for all x>0. - (iii) (i) and (ii) still hold replacing "for all x" and "for all s" with "for all sufficiently large x" and "for all sufficiently small s" or with "for all sufficiently small x" and "for all sufficiently large s". PROOF. If xm(x)•†Cf(x), using the I. S. Kac's inequality (cf. [8]) (32) $$\frac{1}{sm(x)+1/x} \le h(s) \le x + \frac{1}{sm(x)}, \quad s > 0, \quad x > 0,$$ we have (33) $$h(s) \leq x + \frac{1}{sm(x)} \leq x \left(1 + \frac{1}{Csf(x)}\right).$$ Set x=f-1(1/(Cs)). The remains are easy to prove by similar arguments. COROLLARY. Let L(x) and Ka(x) be the same as in Theorem 2. Then we have, (i) $$0 < \lim_{x \to \infty \atop x = 0} \frac{m(x)}{x^{1/\alpha - 1} K_{\alpha}(x)} \quad \text{if and only if} \quad \overline{\lim}_{s \to 0 \atop s = 0} \frac{h(s)}{s^{-\alpha} L(1/s)} < \infty$$ (ii) $$\overline{\lim_{x \to \infty \atop [0]}} \frac{m(x)}{x^{1/\alpha - 1} K_{\alpha}(x)} < \infty \quad \text{if and only if} \quad 0 < \underline{\lim_{s \to 0 \atop [\infty]}} \frac{h(s)}{s^{-\alpha} L(1/s)}.$$

Lemma 3. For the generalized diffusion process associated with the inextensible measure dm(x), the condition (D. K) holds if and only if l=•‡. REMARK. The condition l=•‡ is equivalent, probabilistically, that the process is recurrent in the sense that the particle can hit each fixed point in Em with probability one.

Theorem 5. Let {Xt}t•†0 be the generalized diffusion process associated with the inextensible measure dm(x), V(s) be a non-negative Borel function with compact support in [0, l) such that $$(45) 0 < C = \int V(x) dm(x) < \infty.$$ Then; (i) If m(x) varies regularly at •‡ with exponent p (0<ƒÀ•…•‡), then (46) $$\lim_{t\to\infty} P_x \left\{ \frac{1}{Ch(1/t)} \int_0^t V(X_{\tau}) d\tau < u \right\} = g_{1/(\beta+1)}(u), \qquad x \in E_m.$$ (ii) Conversely, if we can choose some normalizing function u(t) (lira u(t)=•‡) such that $$\lim_{t \to \infty} P_x \left\{ \frac{1}{u(t)} \int_0^t V(X_{\tau}) d\tau < u \right\} = G(u)$$ holds for some x e Em, where G(u) is an appropriate non-degenerate dis tribution, then m(x) varies regularly with exponent p for some (0<ƒÀ•…•‡), and hence $$G(u) = g_{1/(\beta+1)}(au)$$ for suitable $a > 0$ . PROOF. (i) By lemma 3, the condition (D. K) is satisfied by taking h(s) for v(s). Since m(x) varies regularly with exponent ƒÀ, it follows from Theorem 2 that h(s) has the representation (48) h(s)=s-1/(ƒÀ+1)L(1/s) where L(t) is a slowly varying function. Hence the assertion follows from above Darling and Kac's theorem. (ii) If l<•‡, it is not difficult to see that, with probability one, $$\lim_{t \to t} X_t = \sup \{x \; ; \; x \in E_m\}$$ where ƒÄ denotes the life time. Consequently, •çt0V(XƒÑ) dƒÑ is finite with prob ability one, and therefore (47) does not hold for any u(t) such that lira u(t) =•‡. Hence l=•‡ and the condition (D. K) is satisfied. By the Darling and Kac's theorem, we have that h(1/s) varies regularly with exponent ƒÀ =1/ƒ¿-1. This completes our theorem. The first half of Theorem 5 has been proved, except the case of ƒÀ=•‡, by C. Stone [11] in another method. EXAMPLE 2. Let V and Xt be the same as in Theorem 5. The necessary and sufficient condition that (50) $$\lim_{t \to \infty} P \left\{ \frac{1}{u(t)} \int_0^t V(X_t) \, d\tau > x \right\} = e^{-x} \qquad (x > 0)$$ holds for an appropriate u(t) (lim u(t)=•‡ is that (51) $$\lim_{x \to \infty} \frac{m(\lambda x)}{m(x)} = \begin{cases} \infty & 1 < \lambda \\ 0 & 0 < \lambda < 1. \end{cases}$$ In particular, if m(x)•`xƒÏex,x•¨•‡, then (52) $$\lim_{t\to\infty} P\left\{\frac{1}{C\log t} \int_0^t V(X_{\tau}) d\tau > x\right\} = e^{-x}$$ where $$C = \int V(x) dm(x)$$ .

Theorem 6. Let m0(x) be strictly increasing in x and suppose that one of the following conditions is satisfied; - (i) Xn(t) is J-covergent to X0(t) with probability one. - (ii) Mn(t)=sup X(v) is J-covergent to M0(t)=supX0(Ą) with prob ability one. - (iii) ƒÐnx=inf [t;Xn(t)>x] converges in distribution to ƒÐ0x=inf [t;X0(t) >x]. Then, - (55) mx(x)•¨>m0(x) at each continuity point x e [0, •‡) of mo(x). PROOF. It is easy to see that (ii) follows from (i), and using the strong Markov property of X0(t), we can show that (ii) implies (iii). Therefore we have to prove that (55) follows from (iii). For each x e [0, •‡), we define xn= inf [y;mn(y)>mn(x)]. Since X0(t) is assumed to be a diffusion process, ƒÐ0x, is strictly decreasing in x. Hence xn converges to x if (iii) holds. Now note that (56) $$E(e^{-s\sigma_x^n}) = E(e^{-s\sigma_{x_n}^n}) = \frac{1}{\varphi_n(x_n, -s)}, \quad s > 0.$$ Since E(e-sƒÐnx) converges to E(e-sƒÐ0x) by (iii), it follows that (57) $$\varphi_n(x_n, -s)$$ converges to $\varphi_0(x, -s)$ . By a similar argument used in the proof of theorem 1, we can show that {ƒÕn(x,-s)}n is equi-continuous in x on each compact set. Hence it easily follows from (57) that ƒÕn(x,-s) converges to ƒÕ0(x,-s), which is equivalent to (55) by Theorem 1. We next study the limit processes of generalized diffusion processes 1/u(c)X(ct) as c•¨•‡ , where u(c) is some normalizing function. Let Xa(t) denotes the diffusion process with the speed measure x1/ƒ¿-1. Xƒ¿(t) is obtain ed by changing the scale of the Bessel diffusion process of index 2(1-ƒ¿). THEOREM B (C. Stone). Let X(t) be a generalized diffusion process associated with an inexten sible measure dm(x) with the characteristic function h(s). If m(x) varies regularly with exponent 1/ƒ¿-1(0<ƒ¿<1) at •‡, then we have as c•¨•‡, the following: - (B. 1) 1/u(c) X(ct) is J-convergent to Xƒ¿(t) with probability one where u(c) =D-1ƒ¿h(1/c). - (B. 2) 1/u(c) sup X(cƒÑ) is J-convergent to sup Xƒ¿(ƒÑ) with probability one. - (B. 3) inf [t;1/u(c)X(ct)>x] converges in distribution to inf [t;Xƒ¿(t)>x]. We can prove the converse of Theorem B as follows.

Theorem 7. Suppose one of the following three conditions holds for an appropriate normalizing function u(c) and for some non-degenerate generalized diffusion process Y(t) (i.e., the speed measure is not supported on a single point); - (i) 1/u(c)X(ct) is J-convergent to Y(t) with probability one as c•¨•‡ - (ii) 1/u(c) sup X(cĄ) is J-convergent to sup Y(Ą) with probability one as - (iii) inf [t;1/u(c)X(ct)>x] converges, in probability, to inf [t;Y(t)>x]. - (58) Y(t)=CXƒ¿(t) for some C>0, Then, for some 0<ƒ¿<1, - (59) m(x) varies regularly at •‡ with exponent 1/ƒ¿-1, and therefore, - (60) u(c)•`coast. h(1/c) as c•¨•‡, where h(s) is the characteristic function of X(t). PROOF. If (i) holds, we have 1/u(c)X(cat)•¨Y(at), and 1/u(ca)X(cat)•¨ Y(t). Hence, there exists the limit; (61) $$u^*(a) = \lim_{c \to \infty} \frac{u(ac)}{u(c)}.$$ Consequently, we have u\*(a)=aƒ¿ for some a, and therefore Y(at) and aƒ¿Y(t) have the same finite-dimensional distributions . We remark that the speed measure of Y(at) and aƒ¿Y(t) are given by a-1m(x) and a-ƒ¿m(a-ƒ¿x) where m(x) is the speed measure of Y(t) (cf. [11]) . Hence a-lm(x) =a-ƒ¿m(a-ƒ¿x), which provides us with (62) $$\tilde{m}(x) = \tilde{m}(1)x^{1/\alpha - 1} \qquad (0 < \alpha < 0)$$ Since we have already remarked that sup Y(Ą) and inf [t;Y(t)>x] are the characteristics of Y(t), we can similarly show that both (ii) and (iii) imply (62). Appealing to Theorem 6, we now obtain that c-lu(c)m(u(c)x) , which is the speed measure of 1 X(ct), converges to m(x)=m(1)x1/ƒ¿-1. Set u(c) x=1 and let w(t) be the inverse function of u(c) , we have (63) $$\lim_{t \to \infty} \frac{t}{w(t)} m(t) = \tilde{m}(1).$$ Since by (61), u(c) varies regularly with exponent a, w(x) varies regularly with exponent 1/ƒ¿. Hence, m(x) varies regularly at •‡ with exponent 1/ ƒ¿ -1 , and appealing to Theorem 6, we have the assertion of our theorem .