Which Moments of a Logarithmic Derivative Imply Quasiinvariance ?

In many special contexts quasiinvariance of a measure under a one-parameter group of transformations has been established. A remarkable classical general result of A.V. Skorokhod [6] states that a measure on a Hilbert space is quasiinvariant in a given direction if it has a logarithmic derivative in this direction such that e aj j is -integrable for some a > 0. In this note we use the techniques of [7] to extend this result to general one-parameter families of measures and moreover we give a complete characterization of all functions : [0;1) ! [0;1) for which the integrability of (j j) implies quasiinvariance of . If is convex then a necessary and su cient condition is that log (x)=x 2 is not integrable at 1. 1991 Mathematics Subject Classi cation: 26 A 12, 28 C 20, 60 G 30

The paper is divided into two parts.The rst part does not mention quasiinvariance at all.It treats only one-dimensional functions and, implicitly, one-dimensional measures.The reason is as follows: A measure on R has a logarithmic derivative if and only if has an absolutely continuous Lebesgue density f, and is given by (x) = f 0 f (x) -a.e.. Then the -integrability of (j j) is equivalent to the Lebesgueintegrability of (j f 0 f j)f.The quasiinvariance of is equivalent to the statement that f(x) 6 = 0 Lebesgue-a.e.. Therefore in the case of one-dimensional measures, a function allows a quasiinvariance criterion, as indicated in the abstract, i for all absolutely continuous functions f 0, the integrability of (j f 0 f j)f implies that f is strictly positive.The main result of the rst part, Theorem 1, gives necessary and su cient reformulations of this property which are easier to check.The most simple of these reformulations is the divergence of the integrals of the length of this interval and of the integral R I (j f 0 f j)fdx.Finally we give an example showing that the introduction of the lower convex hull in these results really is necessary.
The second part of the paper then proves that the one-dimensional situation is typical.The quasiinvariance criterion works on the real axis if and only if it works for the transport of a measure under an arbitrary measurable ow, or even more generally for general one parameter families of measures which are di erentiable in the sense of 7].If this criterion applies then one gets even the typical Cameron-Martin type formula for the Radon-Nikodym-densities between the members of such a family (cf.e.g.3], 1], 5], 7]).In the situation of Skorokhod's result mentioned in the summary, we see that the exponential functions (x) = e ax ; a > 0 can be replaced by exp( x log x ) but not by exp( x (log x) 2 ).This shows that Skorokhod's exponential criterion is not strictly optimal but it gives the optimal power of log .
2 A class of one-dimensional functions Theorem 1: For a measurable function : 0; 1) !0; 1) the following six conditions are equivalent: (A) Let f : R ! 0; 1) be absolutely continuous such that Z and f 6 = 0.Then f(x) > 0 for Lebesgue-all x 2 R. (A 0 ) Let f : R ! 0; 1) be absolutely continuous such that x 7 !(j f 0 f (x)j) f(x) is locally Lebesgue integrable and f 6 = 0.Then f(x) > 0 for all x 2 R. (B) For some a > 0 the following implication holds .
(C) Let be the largest nondecreasing convex function und suppose (c) > 0.
x 2 dx = 1: (3) (C 0 ) Similarly, lim x!1 (x) = 1, and for d in the range of log , In particular the conditions (A) (B) hold for if and only if they hold for .If is convex and nondecreasing and some power p with p > 0 satis es one of the conditions then the same is true for .
Therefore by assumption (A) the integral in (1) diverges.Hence (B 0 ) =) (C 0 ) : Let h(t) = (log ) 1 (t).De ne the number z i by z i = 1 h(i) .From (B) it follows easily that (x)   x ! 1 as x ! 1.Thus the same holds for .Since is convex and increasing the function 1=h is continuous and decreasing.Therefore for the proof of (4) it is su cient to prove that the sum P 1 i=1 z i diverges.
Suppose, on the contrary, that P 1 i=1 z i < 1. Choose y i 2z i such that y i ( 1 yi ) c i + 1 where c i = inf x : This is possible by de nition of this in mum c i .The a ne function l i : x 7 !c i x ci 2z i is since it is negative on 0; 1 2z i ), and on 1 2z i ; 1) even the larger function x 7 !c i x is bounded by .Therefore, from the de nition of , we get We apply (B 0 ) with a = 1 and use the summability of the z i and hence of the y i to Now ( 1 z i ) = e i by construction of the z i , thus (5) gives which is a contradiction.
(C) () (C 0 ) : Both, ( 3) and ( 4), imply that is continuous, nondecreasing and unbounded at in nity.Therefore there is some c such that is even strictly increasing on c; 1), and the assertion follows from lemma 1 below, applied to ' = log .
. Presumably, this is the most useful implication.We formulate the main part of the proof as the separate Theorem 2 since it involves only integrals over nite intervals and can be applied also to functions which do not satisfy the conditions of the theorem.In order to deduce our implication from Theorem 2 assume (4) and let (x) = (x) (0).Then lim x!1 log (x) log (x) = 0 and hence using the equivalence of ( 3) and ( 4) we get Now if f is absolutely continuous and x 7 !(j f 0 f (x)j)f(x) is locally integrable then also the function x 7 !(j f 0 f (x)j)f(x) is locally integrable and hence (9) below gives a lower bound for the values of f on any interval s; t] such that f(s) > 0. The case f(t) > 0 follows by re ection.In particular f is strictly positive which is the assertion of (A 0 ).Finally we prove the last statement.Let be convex and nondecreasing and suppose that p satis es one of the conditions.If p < 1 then max(1; p ) and using criterion (B) it follows that satis es the same condition.If p > 1 then p , by Jensen's inequality, is also convex nondecreasing and hence = and p = ( p ) .
Since log p = p log , the criterion (C) carries over from p to .
In the proof we have used the following elementary fact.
i.e. both integrals converge at the same time and if they do (6) holds.
Proof: The change of variables y = '(x) gives Since '(T ) T > 0 for large T the left-hand side of (6) dominates the right-hand side.
For the converse inequality assume that the integral on the right-hand side of ( 6) is nite.The inde nite integral on the left-hand side of (7) is monotone in T, so it has a nite or in nite limit.Therefore by (7) the limit lim T !1 '(T ) T exists and it must be 0 because otherwise the integral on the right-hand side of (6) would be in nite.This implies (6).
The following result gives a quantitative version of the implication (C 0 ) =) (A 0 ) in Theorem 1.
Theorem 2: Let : R ! 0; 1) be a convex even function with (0) = 0. Let f : s; t] !0; 1) be absolutely continuous such that f(s) > 0. Then (log ) 1 (x) dx for y 0. If the range of F contains the number I f(s) + t s (which certainly is true if F(y) ! 1 for y ! 1) then (8) can be rewritten as f(t) f(s) exp F 1 ( I f(s) + t s) : (9) This gives a lower estimate of the uctuation of the function f in terms of the integral I and the length of the interval s; t].
Remark: In the special case (x) = e ax there is an elegant more abstract proof of property (A) of Theorem 1, see 4], prop.2.18.That proof does not give a lower bound for the values of f in terms of I but on the other hand it works also in higher dimensions whereas our method is strictly one-dimensional.
Proof: Both sides of (8) remain unchanged if f is multiplied by some positive constant.Therefore we may and shall, for notational convenience, assume f(s) = 1.For a > 0; i 2 N 0 let x (a)  i := inffy s : f(y) = e ai g; We also introduce the numbers z (a)  i := x (a) i x (a) i 1 , z (a)   i := a (log ) 1 (ai) for i 2 N and nally N a := supfn 2 N : x (a) n tg: We apply Jensen's inequality in the second step of the following estimates Since is convex and (0) = 0 the function y 7 ! (y)  y is increasing on 0; 1).Example 1 For every 0 < p < 1 there is a convex increasing function : 0; 1) !0; 1) which satis es the conditions of Theorem 1, but p does not.x 2 dx = 1 but (C) does not hold for p .This shows that in (C) (and in (C 0 )) the convex lower envelope cannot be replaced by the function itself.Switching roles of and p , the example also shows that in the last statement of the theorem the convexity of cannot be replaced by the convexity of p for p > 1.With some additional e ort one could modify the example in such a way that for no p < 1 the function p satis es the conditions of Theorem 1.

This function then satis es
Construction: We write q instead of 1 p .We start by setting b 0 = 0, 0 = 0, 0 = 1: We shall choose recursively points a 1 < b 1 < a 2 < b 2 < : : : and real numbers k ; k ; k ; k 2 N and set So the function alternates between a ne and exponential type.The constants are chosen in such a way that at the points a k the graph of is bent upwards, while at the points b k the two one-sided derivatives agree.Assume that all numbers a i ; b i ; i ; i ; i with i < k are already chosen such that (10) gives a continuous convex increasing function on some interval 0; b k 1 + "] which is di erentiable with the possible exception of the points a i for i < k.In the case k = 1 let a 1 = 1.For k > 1 we then know that a k 1 1 and, comparing logarithmic derivatives of and of x q , respectively, we see that (x) > x q on the interval (a k 1 ; b k 1 ], in particular k 1 b k 1 + k 1 > b q k 1 .Since q > 1 this implies that there is a solution > b k 1 of the equation k 1 x + k 1 = x q (11) which we choose as a k .Then is de ned on b k 1 ; a k ] by the third part of (10).Choose k such that k e qa k = a q k ; i.e. k = a q k e qa k : Let b k = 3 2 a k and de ne on a k ; b k ] according to the second part of (10).The numbers k and k are determined by the equation of the (left) tangent of at b k .
Verification: By construction, 0 (a k ) = k 1 = 0 (b k 1 ) = q (b k 1 ) < q (a k ) = 0 (a k +): i.e. this extension of continues to be convex and continuous.Moreover, (b k ) = k e qb k = a q k e qa k e q 3 2 a k = a q k e q 2 a k : (12) The sequence (a k ) is certainly unbounded by the choice of the b k .By construction of a k , at this point the slope of y = x q is bigger than k 1 .Thus, qa q 1 k > k 1 = 0 (b k 1 ) > q (a k 1 ) = qa q k 1 and hence a k a k 1 > (a k 1 ) 1 q 1 : This implies a k a k 1 !1: Because of ( 12) and ( 13) x 2 dx = (q log a k + q 2 a k ) 1 for eventually all k.Together with the convexity this shows that satis es condition (C).
On the other hand, for z k = 1 a k , (13) implies and, therefore, P 1 k=1 z k p ( 1 z k )e i < 1: So p does not have property (B).This concludes the discussion of the example.
3 Logarithmic derivatives and quasiinvariance Let M(E) be the linear space of nite signed measures on a measurable space (E; B), equipped with the total variation norm k k .Let C be a linear space of bounded test functions on E which is normde ning for M(E), k k = supf R ' d : ' 2 C; k'k 1 1g for all 2 M(E).Typical examples of spaces C with this property are the space of bounded continuous functions for a topology for which B is the class of Baire sets, i.e. the -eld generated by C, or the space of smooth cylindrical functions on a Hilbert space.Let I be a real interval and let ( t ) t2I be a family of elements of M(E).We call this map C -di erentiable at t 2 I with logarithmic derivative t 2 L 1 ( t ) if for every '  di erentiability condition at one (and then at all) t we call di erentiable along T with logarithmic derivative = 0 .In this case the logarithmic derivative for general t is given by t (x) = (T t x): (15) This extends the concept of the di erentiability of a measure on a linear space in a certain direction which was the main subject of 2] and the relevant parts of 6].
The more general aspects have been studied, starting with 3], in 5] and 7], for a comparison with concepts of the Gross-Malliavin calculus see e.g.8].
We need two results from 7]: (a) Suppose that t exists for all t 2 I, and that Then there are a probability measure on B and B B(I)-measurable functions g; g 0 on E I such that t (dx) = g(t; x) (dx); 0 t (dx) = g 0 (t; x) (dx) and thus t (x) = g 0 (t;x) The condition (18) clearly is necessary for (19) to make sense.But how can one verify it ?The interaction of the Radon-Nikodym derivatives t for varying t may be complicated.Therefore, it seems desirable to have su cient conditions for the equivalence of the t in terms of the onedimensional laws of the t with respect to the measures t .The following continuation of the main result of the rst section provides an answer of this type.Proof: 1. Suppose that has the indicated property.We want to show that condition (A 0 ) of Theorem 1 is ful lled.Let f be an absolutely continuous nonnegative function on the real axis for which x 7 !(j f 0 f (x)j) f(x) is locally integrable and such that f does not vanish everywhere.We have to show that f is strictly positive.Otherwise there are two points a; b with f(a) > 0 and f(b) = 0. Without loss of generality a < b.Let t (dx) = f(x + t)dx.In order to apply our condition, we have to make sure that these measures are nite.For this, rede ne f on b; 1) by f(x) = 0 and on 1; a) by f(x) = f(a) exp(x a) Then Z a 1 (j f 0 f (x)j)f(x) dx = Z a 1 (1)f(a) exp(x a) dx < 1 and, similarly, R 1 b (j f 0 f (x)j)f(x) dx = 0 < 1.The modi ed f still satis es (1) and it is certainly Lebesgue integrable.Thus, we have the ow situation mentioned above with T t x = x t.The family t is di erentiable (even for the topology induced by the total variation norm) with t (x) = f 0 f (x + t).Then the local integrability assumption and the two tail estimates imply R (j t (x)j) t (dx) = R (j 0 (x)j) (dx) < 1 for all t.Therefore, the condition (20) is satis ed.By our assumption on this implies that the measures t are all equivalent, i.e. the function f cannot vanish on a half-line as our f does.This contradiction shows that f must be strictly positive.Hence has property (A 0 ). 2. Suppose, conversely, that is a function of the type considered in Theorem 1.Let t be a C -di erentiable and k k-bounded family ( t ) t2I ; I R of signed measures on a measurable space and let a; b 2 I with (20) be given.First we claim that (16) holds.In fact from condition (C) in Theorem 1 we nd positive constants u; v such that v (y) y for all y > u.Then k t k 1; t = Z E j t (x)j d t v Z j tj>u (j t j) dj t j + Z E u dj t j vk (j t j)k 1; t + uk t k: Since the measures are k k-bounded (20) implies ( 16).Therefore we can choose g; g 0 and with the properties listed after (16).Then (20) can be rewritten as Z b a Z E (j g 0 (t; x) g(t; x) j)g(t; x) (dx) dt < 1: By Fubini, there is a -nullset N such that R b a (j g 0 t gt (x)j)g t (x) dt < 1 for every x 2 E n N. Then extending t 7 !g(t; x) outside of the interval a; b] by exponential tails (or zero) as in the rst part of this proof we can apply condition (A) in Theorem 1 and conclude that for each x 2 E nN either g(t; x) > 0 for all t 2 a; b] or g(t; x) = 0 for all t 2 a; b].This implies that the measures t ; t 2 a; b] are all equivalent.3. Moreover the function g( ; x) is continuous by (17) and therefore it is bounded away from 0 by some constant (a; b; x) on the interval a; b] for a -(and b -) almost all x 2 E. Then (17) and the representation t (x) = g 0 (t;x) g(t;x) show that (18) and hence also (19) hold.
In particular, we get the following version of Skorokhod's theorem for the function given by (y) = exp( y j log(y)j ) for y > 0. Corollary 4 Let be a probability measure on a measurable space E and let T = (T t ) t2R be a measurable ow on E. Suppose is C -di erentiable along T with logarithmic derivative .If satis es the following integrability condition Z E exp j (x)j j log(j (x)j)j (dx) < 1 then is quasiinvariant under the ow T and the corresponding Radon-Nikodym derivatives are given by (19).But, even for translation families on the real axis, the quasiinvariance is not implied by the weaker condition Z E exp j (x)j log(j (x)j) 2 (dx) < 1: Proof: We consider the function (y) = exp( y j log(y)j ) for y > 0. Then it is easily veri ed that is convex and increasing for su ciently large y and, thus, it satis es the criterion (C).Because of (15) we have k (j t j)k 1; t = Z (j j) T 1 t d t = Z (j j) d for all t, and hence our integrability assumption implies (20).
On the other hand (y) = exp( y log(y) 2 ) for y > 0 de nes a function which does not satisfy the condition (C).The function f used in the proof of (A) =) (B) in Theorem 1 then satis es (1) for this function but f has compact support.Therefore the logarithmic derivative = f 0 f of the measure 2 M(R) whose density is f, satis es the weakened integrability condition of our Corollary, but this measure is not quasiinvariant under the ow of translations.

R 1 c
log (x)=x 2 dx where is the lower nondecreasing convex envelope of .Moreover we give, for every with this property, explicit lower estimates for the values of f on an interval I, in terms Documenta Mathematica 3 (1998) 261{272 )j) f(x)dx = 1 which proves (B) with a = 1.(B) () (B 0 ) : Denote by (B a ) the statement that (B) holds with the constant a.Clearly, (B b ) =) (B c ) if c b:We prove B a =) B 2a : Suppose P 1 i=1 z i < 1; z i > 0 and let y 2j = y 2j 1 = z j for j 2 N: Then P ai and hence the last sum can be further estimated from a#0 N a a = lim a#0 log f x (a) Na = min s x t log f(x); the proof is complete.
x)j ds < 1 j a j + j b j a:e: (18) holds then all measures t ; a t b are equivalent and we have the 'abstract Cameron-Martin' formula d b The measure t t is the derivative of the M(E)-valued curve ( t ) with respect to the topology C = (M(E); C) and is denoted by 0 t .An important special class of examples are families ( t ) t2R which are induced by a measurable ow: If T = (T t ) t2R is a one-parameter group of bimeasurable bijections of E, and 2 M(E) is a xed measure, one considers the family of measures t = T1 t .If ( t ) satis es the above 7] O. G. Smolyanov and H. v. Weizs acker.Di erentiable families of measures.Journal of Functional Analysis, 118:454{476, 1993.8] O. G. Smolyanov and H. v. Weizs acker.Di erentiable measures and a Malliavin-Stroock theorem.submitted, 1997.