NOTES ON SCHAUDER ESTIMATES. r 2 x y 2

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1 NOTES ON SCHAUDER ESTIMATES CRISTIAN E GUTIÉRREZ JULY 26, 2005 Lemma 1 If u f in B r y), then ux) u + r2 x y 2 B r y) B r y) f, x B r y) Proof Let gx) = ux) Br y) u r2 x y 2 Br y) f We have g = u + Br y) f f + B r y) f 0, that is, g is subharmonic in B ry) Then Br y) g = Br y) g = 0, so g 0 in B ry) the lemma follows Lemma 2 If u is a solution to u = f in B r y) v solves v = 0 v = u on B r y), then r 2 x y 2 in particular, for all x B r y) inf f vx) ux) r2 x y 2 B r y) ux) vx) r2 x y 2 f, B r y) B r y) Proof Since v u) = f, the inequality immediately follows from Lemma 1 Lemma 3 Let 0 < α < 1 There exist positive constants C 0, ɛ 0 0 < λ < 1 such that for any f any solution to u = f in u L ) 1 f L ) ɛ 0 there exists a second degree harmonic polynomial qx) = 1 Ax, x + B x + C such that 2 1) ux) qx) λ 2+α, for x λ, 2) A + B + C C 0 Proof Let v be the harmonic function in the statement of Lemma 2 By the maximum principle B1 0) v u 1 Since v is harmonic 3) D β vx) Cn, β ) v Cn, β ), for x 1/2 1 f,

2 2 C E GUTIÉRREZ JULY 26, 2005 We shall prove that the second order Taylor polynomial of v about 0, qx) = 1 2 D2 v0)x, x + Dv0) x + v0), satisfies 1) 2) In fact, 2) follows from 3) Also, qx) = traced 2 qx) = traced 2 v0) = v0) = 0, so q is harmonic We have vx) = qx) + 1 [ x D) 3 vx) ] 3!, x=ξ ξ an intermediate point between 0 x Then from Lemma 2 we get ux) qx) ux) vx) + vx) qx) 1 x 2 f + 1 [ x D) 3 vx) ] 3! x=ξ 1 x 2 f + C n x 3 D β vz), for x 1/2 z 1/2, β =3 We write 1 x 2 f + C n x 3 = I + II, from 3) ) x 2+α II = C n x 1 α λ 2+α λ C n x 1 α λ 2+α, if x λ if we pick λ = 1 2 λ2+α 1 2C n ) 1/1 α) With this value of λ, we next want I 1 2 λ2+α If ɛ 0 n λ 2+α, we then have we are done I = 1 x 2 f 1 f 1 ɛ λ2+α Theorem 4 Suppose u C 2 ) C ), u = f, f is Hölder continuous at 0, ie, f x) f 0) [ f ] α,0 = < x 1 x α Then there exists a second degree polynomial px, 0) = 1 Ax, x + B x + C such that 2 4) ux) px, 0) C 1 x 2+α, for x 1/2,

3 NOTES ON SCHAUDER ESTIMATES July 26, C 1 C 0 [ f ]α,0 + f L B 1 ) + u L B 1 )), A + B + C C 0 [ f ]α,0 + f L B 1 ) + u L B 1 )) Proof We may assume that i) f 0) = 0, ii) [ f ] α,0 + f L B 1 ) ɛ 0, iii) u L 1 Indeed, if we let vx) = ux) x 2 f 0), hx) = ɛ 0 f x) f 0) [ f ] α,0 + f L B 1 ) + v L B 1 ) vx) ūx) = ɛ 0, [ f ] α,0 + f L B 1 ) + v L B 1 ) then ū = h in B 1, h satisfies i) ii), ū satisfies iii) Claim: there exists a sequence of harmonic polynomials p k x) = 1 2 A kx, x + B k x + C k such that 5) ux) p k x) λ 2+α)k, for x λ k 6) A k A k+1 C λ αk, B k B k+1 C λ α+1)k, C k C k+1 C λ α+2)k, for k = 1, where C is a universal constant In view of ii) iii) above we can apply Lemma 3, we let p 1 x) be the polynomial in that lemma Suppose p k x) is constructed We will construct p k+1 Let We have wx) = wx) = u p k)λ k x) λ α+2)k 1 [ λ 2k u)λ k x) λ 2k p λ α+2)k k )λ k x) ] = 1 λ f αk λk x) = g k x) From 5), w L B 1 ) 1 from i) ii) above, g k L B 1 ) ɛ 0 Hence by application of Lemma 3 to w, we get a harmonic polynomial q k x) -depending on g k - such that 7) wx) q k x) λ 2+α, for x λ

4 4 C E GUTIÉRREZ JULY 26, 2005 From the definition of w 7) we then get Therefore, if we take uλ k x) p k λ k x) λ 2+α)k q k x) λ 2+α)k+1), for x λ 8) p k+1 x) = p k x) + λ 2+α)k q k x/λ k ), then p k+1 satisfies 5) k replaced by k + 1 Writing q k x) = 1 2 A k x, x + B k x + C k, from 8) we get A k+1 = A k + λ αk A k, B k+1 = B k + λ α+1)k B k, C k+1 = C k + λ α+2)k C k, for k = 1, 2, then 6) follows from 2) This completes the proof of the claim Next, we notice that from 6), 2), since 0 < λ < 1, it follows that A k, B k, C k are Cauchy sequences therefore we let px, 0) = 1 2 A x, x + B x + C, where A, B, C are the corresponding limits We show that px, 0) satisfies 4) Given x 1/2, let k be a positive integer such that λ k+1 < x λ k Hence from 5) we obtain p k x) px, 0) C λ αk x 2 + λ α+1)k x + λ α+2)k C x 2+α ux) px, 0) ux) p k x) + p k x) px, 0) C x 2+α + C x 2+α we are done Suppose f C α ) u is a solution to u = f in Let y r < disty, ) Define gx) = r 2 f y + rx) vx) = uy + rx) for x We have that v is a solution to v = g in [g] α,0 = x 1 gx) g0) x α = r 2 x 1 f y + rx) f y) x α = r 2+α z r f y + z) f y) z α From Theorem 4 applied to v, there exists a quadratic polynomial px, 0) such that 9) vx) px, 0) C 1 x 2+α, for x 1/2, 10) C 1 C 0 [g]α,0 + g L ) + v L )) From 9) the definition of v we get that uz) pz y)/r, 0) C 1 r 2+α z y 2+α, for z y r/2

5 NOTES ON SCHAUDER ESTIMATES July 26, We have g L ) = r 2 f L B r y)), v L ) = u L B r y)) If we let qx, y) = px y)/r, 0), r = disty, ), then ux) qx, y) C 1 x y 2+α, for x y disty, )/2, C 1 = C 1 disty, ) 2+α C 0 z disty, ) f y + z) f y) z α +disty, ) α f L B r y)) + disty, ) 2 α u L B r y))) In particular, we obtain that if u = f in, then for each y B 1/2 0) there exists a quadratic polynomial px, y) such that ux) px, y) C x y 2+α, for x y 1/4, C C 0 [ f ]α,b1 0) + f L ) + u L )) Lemma 5 Suppose u C 2 Ω) is such that there exist constants C 1 > 0 0 < α < 1 so that for each y Ω there exists a quadratic polynomial px, y) such that 11) ux) px, y) C 1 x y 2+α, for all x By, disty, Ω)/2) Then 12) px, y) = 1 2 x y)t D 2 uy)x y) + Duy) x y) + uy), 13) D ij ux 1 ) D ij ux 2 ) C 1 x 1 x 2 α for all x 1, x 2 Ω distx i, Ω) > diamω)/2 Proof We use the following result: if f C 2 I) where I is an open interval, then 14) lim h 0 f a + h) + f a h) 2 f a) h 2 = f a), for a I Notice that the converse to this result is not true, take f x) = x x at a = 0) From 11) it follows immediately that py, y) = uy) On the other h, uh e j + y) uy) D j uy) = lim, h 0 h

6 6 C E GUTIÉRREZ JULY 26, 2005 uh e j + y) uy) h ph e j + y, y) py, y) h C 1 h e j 2+α 0, h as h 0 So D x py, y) = Duy) If η is a nonzero vector in R n g η t) = ut η + y), then g η t) = n i,j=1 η i η j u ij t η + y) In particular, g e k 0) = u kk y), g e k +e l 0) = u kk y) + 2u kl y) + u ll y) Hence u kl y) = 1 { g e 2 k +e l 0) g e k 0) g e l 0) }, from 14) we get where Also Since u kl y) = 1 2 lim kl uh, y) + kl u h, y) 2 kl u0, y) h 0 h 2 kl uh, y) = uhe k + e l ) + y) uhe k + y) uhe l + y) p kl y) = 1 2 lim kl ph, y) + kl p h, y) 2 kl p0, y) h 0 h 2 kl uh, y) phe k + e l ) + y, y) phe k + y, y) phe l + y, y)) 4 C 1 h 2+α, kl u h, y) p he k + e l ) + y, y) p he k + y, y) p he l + y, y)) 4 C 1 h 2+α, kl u0, y) = uy) = py, y) we obtain that D 2 uy) = D 2 xpy, y) then 12) is proved To prove 13) we use the following lemma of Calderón-Zygmund, [CZ61, Lemma 26] Lemma 6 Given an integer m 0, there exists a function ϕ C 0 Rn ) port in the unit ball such that ϕ ɛ P = P for each ɛ > 0 every polynomial P of degree m As usual, ϕ ɛ x) = ɛ n ϕx/ɛ) Let x 1, x 2 Ω be such that distx i, Ω) > diamω)/2 write ux) = ux) px, x 1 ) + px, x 1 ) ux) = ux) px, x 2 ) + px, x 2 ), convolving these expressions ϕ ɛ using Lemma 6 m = 2 we get u ɛ x) = [u p, x 1 )] ϕ ɛ x) + px, x 1 ) u ɛ x) = [u p, x 2 )] ϕ ɛ x) + px, x 2 ),

7 NOTES ON SCHAUDER ESTIMATES July 26, for distx, Ω) > ɛ, taking derivatives Hence D ij u ɛ x) = [u p, x 1 )] D ij ϕ ɛ x) + D ij ux 1 ) D ij u ɛ x) = [u p, x 2 )] D ij ϕ ɛ x) + D ij ux 2 ) D ij ux 1 ) D ij ux 2 ) = [u p, x 2 )] D ij ϕ ɛ x) [u p, x 1 )] D ij ϕ ɛ x) = ɛ n 2 [uy) py, x 2 )]D ij ϕx y)/ɛ) dy = I II y x <ɛ ɛ n 2 y x <ɛ [uy) py, x 1 )]D ij ϕx y)/ɛ) dy If we let x = x 1 + x 2 )/2, ɛ = x 1 x 2 /2, we get that B ɛ x) B 2ɛ x i ) for i = 1, 2, so from 11) we get I ɛ n 2 C 1 y x 2 2+α D ij ϕx y)/ɛ) dy y x 2 <2ɛ ɛ n 2 C 1 D ij ϕ II ɛ n 2 C 1 13) follows y x 1 <2ɛ ɛ n 2 C 1 D ij ϕ y x 2 <2ɛ y x 2 2+α dy = C n C 1 ɛ α = C x 1 x 2 α, y x 1 2+α D ij ϕx y)/ɛ) dy y x 1 <2ɛ y x 1 2+α dy = C n C 1 ɛ α = C x 1 x 2 α, REFERENCES [CZ61] A P Calderón A Zygmund, Local properties of solutions of elliptic partial differential equations, Studia Math ), DEPARTMENT OF MATHEMATICS, TEMPLE UNIVERSITY, PHILADELPHIA, PA address: gutierrez@mathtempleedu

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