Local maximal operators on fractional Sobolev spaces

Transcription

1 Local maximal operators on fractional Sobolev spaces Antti Vähäkangas joint with H. Luiro University of Helsinki April 3, / 19

2 Let G R n be an open set. For f L 1 loc (G), the local Hardy Littlewood maximal function is defined at x G by M G f(x) = sup r M G f : G [0, ] 1 f(y) dy, B(x, r) B(x,r) where the supremum is taken over all 0 < r < dist(x, G), Theorem of Hardy, Littlewood and Wiener implies that, if 1 < p, then f L p (G) c(n, p) f L p (G) for every f L p (G). This is the typical size -estimate for maximal operators. 2 / 19

3 We are interested in smoothing properties of local maximal operators M G. A motivating observation: if f L 1 loc (G) is non-negative a.e. then M G f = f f is superharmonic. This suggests applications in potential theory and PDE s, where the following toolkit is standard by now: (1) Let B be a ball in G such that 3B G. Then 1 f(y) dy 2 n inf B M Gf. B B (2) Boundary values: If f C 0 (G), then M G f C 0 (G). If f W 1,p 0 (G) with 1 < p <, then M G f W 1,p 0 (G). (3) M G f 2M G f a.e. in G if f W 1,p (G) with 1 < p. (4) M G is continuous in W 1,p (G) if 1 < p < as shown by Luiro (PEMS 2010). Points (2) and (3) above are due to Kinnunen and Lindqvist (Crelle 1998). 3 / 19

4 Let us recall the fractional order Sobolev spaces in an open set G in R n. For 1 p < and s (0, 1) we let W s,p (G) be the collection of all functions f L p (G) with f W s,p (G) := f L p (G) + f W s,p (G) <, where Remarks: ( f W s,p (G) := G G f(x) f(y) p ) 1/p x y n+sp dx dy. (1) The Besov space F s pp(r n ) coincides with the Sobolev space W s,p (R n ). (2) Luiro proved the boundedness of M G on the trace space F s pp(g) when 0 < s < 1 and 1 < p <. (3) The trace space F s pp(g) need not coincide with W s,p (G)! If G is a domain, then this is true if and only if G is regular: B(x, r) G r n for every x G and 0 < r < 1. (Zhou, to appear in TAMS) 4 / 19

5 Concerning the boundedness of local maximal operators on W s,p (G), we prove: Theorem (Luiro-V. 2014) Let 1 < p < and 0 < s < 1. If G R n is an open set, then for every f W s,p (G). M G f W s,p (G) C(n, s, p) f W s,p (G) As a corollary, we see that the local maximal operator is bounded on W s,p (G). I will return to the proof in the 2nd hour; 5 / 19

6 As an application, we study Whitney localization of (s, p)-hardy inequality ; We adapt the classical case by Lehrbäck Shanmugalingam (Pot. Anal., to appear). For a compact set K G, we write cap s,p (K, G) = inf u u p W s,p (G), where the infimum is taken over all real-valued functions u C 0 (G) such that u(x) 1 for every x K. An open set G R n admits an (s, p)-hardy inequality if there exists a constant c > 0 such that dist(x, G) sp dx c cap s,p (K, G) (1) K for every compact set K G. 6 / 19

7 Theorem (Luiro-V. 2014, Dyda-V. 2013) Let 0 < s < 1 and 1 < p < be such that sp < n. A bounded open G in R n admits an (s, p)-hardy inequality if and only if (A) There exists a constant c > 0 such that for every Q W(G). l(q) n sp c cap s,p (Q, G) (2) (B) There exists a constant N > 0 such that ( ) cap s,p (Q, G) Ncap s,p Q, G. (3) Q E Q E for every finite subfamily E W(G) of Whitney cubes. 7 / 19

8 Proof of : Let us fix a compact set K G and a test function u for cap s,p (K, G). By replacing u with max{0, min{u, 1}} we may assume that 0 u 1. We partition W(G) as W 1 W 2, where W 1 = {Q W(G) : u Q := W 2 = W(G) \ W 1. Q u < 1/2}, Write K dist(x, G) sp dx { = Q W 1 + Q W 2 } dist(x, G) sp dx. K Q 8 / 19

9 For every Q W 1 and every x K Q, 1 2 = < u(x) u Q = u(x) u Q. Thus, by Jensen s inequality, dist(x, G) sp dx l(q) sp u(x) u Q p dx Q W K Q 1 Q W Q 1 u(x) u(y) p dy dx Q W 1 l(q) n sp Q W 1 Q u p W s,p (G). Q Q Q u(x) u(y) p x y n+sp dy dx 9 / 19

10 In order to estimate the series associated with W 2, we need observations: (1) Let us consider a cube Q W 2 and a point x Q; Observe that int(q) B(x, diam(q)) B(x, dist(x, G)). Hence, M G u(x) u(y) dy 1 2. (4) (2) The support of M G u is compact in G since G is bdd and u C 0 (G). (3) Since u C 0 (G), we find that M G u is continuous. Concluding from (1) (3) we find that there is ρ > 0 such that ρm G u is an admissible test function for cap s,p ( Q W2 Q, G). Q 10 / 19

11 By condition (B) and the inequality (4), dist(x, G) sp dx Q W 2 K Q The last term is dominated by l(q) n sp Q W 2 c cap s,p (Q, G) Q W 2 ( ) cncap s,p Q, G Q W 2 cnρ p G G C(n, s, p, N, c, ρ) u p W s,p (G). M G u(x) M G u(y) p x y n+sp dy dx. 11 / 19

12 We need the following lower dimensional maximal functions. Fix 1 < p < and F L p (R 2n ). For almost every (x, y) R 2n, we write M 00 (F )(x, y) = F (x, y) M 01 (F )(x, y) = sup r>0 M 10 (F )(x, y) = sup r>0 M 11 (F )(x, y) = sup r>0 B(y,r) B(x,r) B(0,r) F (x, z) dz F (z, y) dz F (x + z, y + z) dz By Fubini s theorem, operators M ij are bounded on L p (R 2n ). 12 / 19

13 If g L p (G), then we write S(g)(x, y) = S G,n,s,p (g)(x, y) = χ G(x)χ G (y) g(x) g(y) x y n p +s for almost every (x, y) R 2n. Observe that g W s,p (G) = Sg Lp (R 2n ). Theorem (Luiro-V. 2014) Let G R n be an open set, 1 < p < and 0 < s < 1. Then there is a constant C = C(n, p, s) > 0 such that, for almost every (x, y) R 2n, S(M G f)(x, y) C ( Mij (M kl (Sf))(x, y) + M ij (M kl (Sf))(y, x) ) (5) i,j,k,l if f L p (G) is such that Sf L p (R 2n ). 13 / 19

14 By replacing the function f with f we may assume that f 0. We may restrict ourselves to points (x, y) G G for which both x and y are Lebesgue points of f and both M G f(x) and M G f(y) are finite. By symmetry, we may further assume that M G f(x) > M G f(y). There are 0 r(x) dist(x, G) and 0 r(y) dist(y, G) such that S(M G f)(x, y) = M Gf(x) M G f(y) x y n p +s = B(x,r(x)) f B(y,r(y)) f x y n p +s B(x,r(x)) f B(y,r f 2) x y n p +s for any given number 0 r 2 dist(y, G). 14 / 19

15 Case r(x) x y. Let us denote r 1 = r(x) and choose r 2 = 0. If r 1 = 0, then we get that Suppose then that r 1 > 0. Now 1 S(M G f)(x, y) x y n p +s 1 = x y n p +s We have shown that B(x,r 1) S(M G f)(x, y) S(f)(x, y). B(x,r 1) B(x,r 1) f(z) dz B(y,r 2) f(z) f(y) dz f(z) dz χ G (z)χ G (y) f(z) f(y) z y n p +s dz M 10 (Sf)(x, y). S(M G f)(x, y) S(f)(x, y) + M 10 (Sf)(x, y) 15 / 19

16 Case r(x) > x y. Let us denote r 1 = r(x) > 0 and choose r 2 = r(x) x y > 0. We then have f(z) dz f(z) dz = f(x + z) f(y + r 2 z) dz B(x,r 1) B(y,r 2) B(0,r 1) r 1 ( ) = f(x + z) f(a) da B(0,r 1) B(y+ r 2 r1 z,2 x y ) G ( + f(a) da f(y + r ) 2 z) dz B(y+ r 2 r1 z,2 x y ) G r 1 A 1 + A 2, where we have written ( A 1 = A 2 = B(0,r 1) B(0,r 1) ( B(y+ r 2 r1 z,2 x y ) G ) f(x + z) f(a) da dz, f(y + r 2 z) f(a) da B(y+ r 2 r1 z,2 x y ) G r 1 ) dz. 16 / 19

17 Recall that r 2 = r 1 x y and fix z B(0, r 1 ). It holds that This in turn implies y + r 2 r 1 z (x + z) = y x + (r 2 r 1 ) r 1 z y x + x y r 1 z 2 y x. B(y + r 2 r 1 z, 2 x y ) B(x + z, 4 x y ). (6) Moreover, since r 1 > x y and {y + r2 r 1 z, x + z} B(x, r 1 ) G, we obtain the norm equivalences B(y + r 2 r 1 z, 2 x y ) G x y n B(x + z, 4 x y ) G. (7) Here the implied constants depend only on n. 17 / 19

18 Hence A 1 = B(0,r 1) B(x,r 1) ( ( f(x + z) f(a) da B(x+z,4 x y ) G f(z) f(a) da B(z,4 x y ) G ) dz. ) dz By observing that both z and a in the last double integral belong to G and using (7) again, we can continue as follows: ( ) A 1 χ G (z)χ G (a) f(z) f(a) x y n p +s B(x,r 1) B(z,4 x y ) z a n p +s da dz ( ) = S(f)(z, a) da dz B(x,r 1) B(z,4 x y ) B(x,r 1) ( S(f)(z, a) da B(z+y x,5 x y ) ) dz. 18 / 19

19 Applying the lower dimensional maximal operators we find that A 1 M x y n p +s 01 (Sf)(z, z + y x) dz B(x,r 1) = M 01 (Sf)(x + z, y + z) dz M 11 (M 01 (Sf))(x, y). B(0,r 1) We have shown that x y n p s A 1 M 11 (M 01 (Sf))(x, y). The estimate for A 2 is very similar, resulting in A 2 M x y n p +s 01 (Sf)(z, z + x y) dz B(y,r 2) = M 01 (Sf)(y + z, x + z) dz M 11 (M 01 (Sf))(y, x). B(0,r 1) 19 / 19