# Garrett: `Bernstein's analytic continuation of complex powers' 2 Let f be a polynomial in x 1 ; : : : ; x n with real coecients. For complex s, let f

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1 1 Bernstein's analytic continuation of complex powers c1995, Paul Garrett, garrettmath.umn.edu version January 27, 1998 Analytic continuation of distributions Statement of the theorems on analytic continuation Bernstein's proofs Proof of the Lemma: the Bernstein polynomial Proof of the Proposition: estimates on zeros

2 Garrett: `Bernstein's analytic continuation of complex powers' 2 Let f be a polynomial in x 1 ; : : : ; x n with real coecients. For complex s, let f s + be the function dened by f s +(x) = f(x) s if f(x) 0 f s +(x) = 0 if f(x) 0 Certainly for <(s) 0 the function f s + is locally integrable. For s in this range, we can dened a distribution, denoted by the same symbol f s +, by f s +() := R n f s +(x) (x) dx where is in Cc 1 (R n ), the space of compactly-supported smooth real-valued functions on R n. The object is to analytically continue the distribution f+, s as a meromorphic (distribution-valued) function of s. This type of question was considered in several provocative examples in I.M. Gelfand and G.E. Shilov's Generalized Functions, volume I. (One should also ask about analytic continuation as a tempered distribution). In a lecture at the 1963 Amsterdam Congress, I.M. Gelfand rened this question to require further that one show that the `poles' lie in a nite number of arithmetic progressions. Bernstein proved the result in 1967, under a certain `regularity' hypothesis on the zero-set of the polynomial f. (Published in Journal of Functional Analysis and Its Applications, 1968, translated from Russian). The present discussion includes some background material from complex function theory and from the theory of distributions.

3 Garrett: `Bernstein's analytic continuation of complex powers' 3 1 Analytic continuation of distributions First we recall the nature of the topologies on test functions and on distributions. Let Cc 1 (U) be the collection of compactly-supported smooth functions with support inside a set U R n. As usual, for U compact, we have a countable family of seminorms on Cc 1 (U): (f) := sup jd fj x where for = ( 1 ; : : : ; n ) we write, as usual, D = 1 n : : : x 1 x n It is elementary to show that (for U compact) Cc 1 (U) is a complete, locally convex topological space (a Frechet space). To treat U not necessarily compact (e.g., R n itself) let U 1 U 2 : : : be compact subsets of U so that their union is all of U. Then Cc 1 (U) is the union of the spaces Cc 1 (U i ), and we give it with the locally convex direct limit topology. The spaces D (U) and D (R n ) of distributions on U and on R n, respectively, are the continuous duals of D(U) = Cc 1 (U) and D(R n ) = Cc 1 (R n ). For present purposes, the topology we put on the continuous dual space V of a topological vectorspace V is the weak- topology: a sub-basis near 0 in V is given by sets U v; := f 2 V : j(v)j < g In this context, a V -valued function f on an open subset of C is holomorphic on if, for every v 2 V, the C-valued function z! f(z)(v) on is holomorphic in the usual sense. This notion of holomorphy might be more pedantically termed `weak- holomorphy', since reference to the topology might be required. If z o 2 for an open subset of C and f is a holomorphic V -valued function on? z o, say that f is weakly meromorphic at z o if, for every v 2 V, the C-valued function z! f(z)(v) has a pole (as opposed to essential singularity) at z o. Say that f is strongly meromorphic at z o if the orders of these poles are bounded independently of v. That is, f is strongly meromorphic

4 Garrett: `Bernstein's analytic continuation of complex powers' 4 at z o if there is an integer n and an open set containing z o so that, for all v 2 V the C-valued function z! (z? z o ) n f(z)(v) is holomorphic on. If n is the least integer f so that (z?z o ) n f is holomorphic at z o, then f is of order?n at z o, etc. To say that f is strongly meromorphic on an open set is to require that there be a set S of points of with no accumulation point in so that f is holomorphic on? S, and so that f is strongly meromorphic at each point of S. For brevity, but risking some confusion, we will often say `meromorphic' instead of `strongly meromorphic'. 2 Statement of the theorems on analytic continuation Let O be the polynomial ring R[x 1 ; : : : ; x n ]. For z 2 R n, let O z be the local ring at z, i.e., the ring of ratios P=Q of polynomials where the denominator does not vanish at z. Let m z be the maximal ideal of O z consisting of elements of O z whose numerator vanishes at z. Let I z (depending upon f) be the ideal in O z generated by f ; : : : ; f x 1 x n A point z 2 R n is simple with respect to the polynomial f if f(z) = 0 for some N we have I z m N z There are i 2 m z so that f = P i i f xi Remarks: The second condition is equivalent to the assertion that O z =I z is nite-dimensional. The simplest situation in which the second condition holds is when I z = O z, i.e., some partial derivative of f is non-zero at z. The third condition does not follow from the rst two. For example, Bernstein points out that with f(x; y) = x 5 + y 5 + x 2 y 2 the rst two conditions hold but the third does not. Theorem (local version): If z is a simple point with respect to f, then there is a neighborhood U of x so that the distribution f s +;U() := f s +(x) (x) dx

5 Garrett: `Bernstein's analytic continuation of complex powers' 5 on test functions 2 Cc 1 (U) on U has an analytic continuation to a meromorphic element in the continuous dual of Cc 1 (U). Theorem (global version): If all real zeros of f(x) are simple (with respect to f), then f s + is a meromorphic (distribution-valued) function of s 2 C. 3 Bernstein's proof Let R z be the ring of linear dierential operators with coecients in O z. Note that R z is both a left and a right O-module: for D 2 R z, for f; g 2 O and a smooth function near z, the denition is (fdg)() := f D(g) Lemma: There is a dierential operator D 2 R z and a non-zero `Bernstein polynomial' H in a single variable so that D(f n+1 ) = H(n)f n for any natural number n. (Proof below.) Proof of Local Theorem from Lemma: Let U be a small-enough neighborhood of z so that on it all coecients of D are holomorphic on U. For suciently large natural numbers n the function f n+1 + is continuously dierentiable, so we have For each xed 2 C 1 c Df n+1 + = H(n)f n + (U) consider the function g(s) := (Df s+1 +? H(s)f s +)() The hypotheses of the proposition below are satised, so the equality for all large-enough natural numbers implies equality everywhere: This gives us Df s+1 + = H(s)f s + f s + = Df s+1 + H(s) Now we claim that for any 0 n 2 the distribution f s + on Cc 1 (U) is meromorphic for <(s) >?n. For n = 0 this is certain. The formula just derived then gives the induction step. Further, this argument makes clear that the `poles' of f s + restricted to C1 c (U) are concentrated on the nite collection of arithmetic progressions i ; i? 1; i? 2; : : :

6 Garrett: `Bernstein's analytic continuation of complex powers' 6 where the i are the roots of H(s). In particular, the order of the pole of f s + at a point s o is equal to the number of roots i so that s o lies among i ; i? 1; i? 2; : : : In particular, the distribution f s + really is (strongly) meromorphic. This proves the Theorem, granting the Lemma and granting the Proposition. Proposition (attributed by Bernstein to `Carlson'): If g is an analytic function for <s > 0 and jg(s)j < be c<s and if g(n) = 0 for all suciently large natural numbers n, then g 0. Proof of Global Theorem: Invoking the Local Theorem and its proof above, for each z 2 R n we choose a neighborhood U z of z in which f s + is meromorphic, so that U z is ariski-open, i.e., is the complement of a nite union of zero sets of polynomials. Indeed, writing f(x) = X i f with i 2 O z, as in the proof of the Local Theorem, let i = g i =h i with polynomials g i and h i, and take U z to be the complement of the union of the zero-sets of the denominators h i. Then Hilbert's Basis Theorem implies that the whole R n is covered by nitely-many U z1 ; : : : ; U zn of these ariski-opens. Then make a partition of unity P subordinate to this nite cover, i.e., take 1 ; : : : ; n so that i 0, i 1, and spt i U zi. Then f s + = X i if s + By choice of the neighborhoods U zi, the right-hand side is a nite sum of meromorphic (distribution-valued) functions. 4 Proof of the Lemma: the Bernstein polynomial Now we prove existence of the dierential operator D and the 'Bernstein polynomial' H. This is the most serious part of this proof. (The complex function theory proposition is not entirely trivial, but is approximately standard). Proof of Lemma: Let where the i 2 m z are so that P := X i 2 R z f = X i f 2 R z

7 Garrett: `Bernstein's analytic continuation of complex powers' 7 Also put Then we have S i := f P? f = f (P + 1)? Q P (f) = X i i f = f Thus, by Leibniz' formula, S i f = f f? f f = 0 P (f n ) = nf n S i (f n ) = 0 Sublemma: There is a non-zero polynomial M in one variable so that M(P ) can be written in the form M(P ) = X i for some J i 2 R z. Proof of Sublemma: Write, as usual, For a natural number m, write J i f jj = 1 + : : : + n P m = X jjm D m; where m; 2 O z. That is, we move all the coecients to the right of the dierential operators. That this is possible is easy to see: for example, x i? x i x j x j is 0 or?1 as i = j or not. Further, the coecients m; are polynomials in the i. Thus, Then taking M(P ) of the form m; 2 m jj z M(P ) = X mq b m P m = X m; D b m m;

8 Garrett: `Bernstein's analytic continuation of complex powers' 8 with b m 2 R, the condition of the sublemma will be met if X m b m m; 2 I z for all indices. If jj N, where I m N, then this condition is automatically fullled. Thus, there are nitely-many conditions X m b m m; = 0 where m; is the image of m; in O z =m N z. Since the latter quotient is, by hypothesis, a nite-dimensional vector space, the collection of such conditions gives a nite collection of homogeneous equations in the coecients b m. More specically, there are dim O z =m N z cardf : jj < Ng such conditions. By taking q large enough we assure the existence of a nontrivial solution fb m g. This proves the Sublemma. Returning to the proof of the Lemma: as an equation in R z Now put Then we have M(P )(P + 1) = X J i (P + 1) = X J i S i + X J i X D = J i H(P ) = M(P )(P + 1) X D(f n+1 ) = ( J i f)(f n ) = = H(P )(f n ) = H(n)f n f as desired. This proves the Lemma, constructing the dierential operator D. 5 Proof of the Proposition: estimates on zeros The result we need is a standard one from complex function theory, although it is not so elementary as to be an immediate corollary of Cauchy's Theorem: Proposition: If g is an analytic function for <s > 0 and jg(s)j < be c<s and if g(n) = 0 for all suciently large natural numbers n, then g 0. Proof of Proposition: Consider G(z) := e?c g( z + 1 z? 1 )

9 Garrett: `Bernstein's analytic continuation of complex powers' 9 Then g is turned into a bounded function G on the disc, with zeros at points (n? 1)=(n + 1) for suciently large natural numbers n. We claim that, for a bounded function G on the unit disc with zeros i, either G 0 or X (1? j i j) < +1 i If we prove this, then in the situation at hand the natural numbers are mapped to n := (n? 1)=(n + 1) = 1? 1 n + 1 so here X n (1? j n j) = X n 1 n + 1 = +1 Thus, we would conclude G 0 as desired. We recall Jensen's formula: for any holomorphic function G on the unit disc with G(0) 6= 0 and with zeros 1 ; : : :, for 0 < r < 1 we have jg(0)j jijr r 1 j i j = exp 2? log jg(re i )j d Granting this, our assumed boundedness of G on the disc gives us an absolute constant C so that for all N jg(0)j jijr r j i j C (We can harmlessly divide by a suitable power of z to guarantee that G(0) 6= 0.) Then, letting r! 1, j i j jg(0)j?1 C?1 For an innite product of positive real numbers j i j less than 1 to have a value > 0, it is elementary that we must have X i (1? j i j) < +1 as claimed. This proves the proposition. While we're here, let's recall the proof of Jensen's formula (e.g., as in Rudin's Real and Complex Analysis, page 308). Fix 0 < r < 1 and let H(z) := G(z) r2? z r(? z)? z where the rst product is over roots with jj < r and the second is over roots with jj = r. Then H is holomorphic and non-zero in an open disk of radius

10 Garrett: `Bernstein's analytic continuation of complex powers' 10 r + for some > 0. Thus, log jhj is harmonic in this disk, and we have the mean value property log jh(0)j = 1 log jh(re i )j d 2? On one hand, jh(0)j = jg(0)j r jj On the other hand, if jzj = r the factors have absolute value 1. Thus, where r 2? z r(? z) log jh(re i )j = log jg(re i )j? X As noted in Rudin (see below), e i arg = jj=r log j1? e i(?arg ) j 1 2? log j1? e i j d = 0 Therefore, the integral appearing in the assertion of the mean value property is unchanged upon replacing H by G. Putting this all together gives Jensen's formula. And let's do the integral computation, following Rudin. There is a function (z) on the open unit disc so that exp((z)) = 1? z since the disc is simply-connected. We uniquely specify this by requiring that (0) = 0. We have <(z) = log j1? zj and j=(z)j < 2 Let > 0 be small. Let? =? be the path which goes (counterclockwise) around the unit circle from e i to e (2?)i and let = be the path which goes (clockwise) around a small circle centered at 1 from e (2?)i to e i. Then log j1? e i 1 j d = lim!0 2 2? log j1? e i j d =

11 Garrett: `Bernstein's analytic continuation of complex powers' 11 1 = < 2i 1 = < 2i? (z) dz = z (z) dz z by Cauchy's theorem. Elementary estimates show that the latter integral has a bound of the form which goes to 0 as! 0. C log(1=)

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