S chauder Theory. x 2. = log( x 1 + x 2 ) + 1 ( x 1 + x 2 ) 2. ( 5) x 1 + x 2 x 1 + x 2. 2 = 2 x 1. x 1 x 2. 1 x 1.

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1 Sep. 1 9 Intuitively, the solution u to the Poisson equation S chauder Theory u = f 1 should have better regularity than the right hand side f. In particular one expects u to be twice more differentiable than f. The validity of this conjecture depends on the function spaces we are looking at. Note. Schauder Theory in fact denotes the similar results for the general linear elliptic PDE u ai j x + u b i x + c x u x = 0. 2 x i x j x i Nevertheless we use it instead of C 2, α estimates as the title of this lecture to make it easy to display on the web. 1. C ounter-examples. The most natural conjecture one would make is f C u C 2. Anyway, it is indeed true in 1 D. However it cease to be true when the dimension is bigger than 1. Example 1. f L but u C 1, 1. We compute in x 1, x 2 > 0 2 x = [ 1 x 1 Thus u x 1, x 2 = x 1 x 2 log x 1 + x 2. 3 x 2 log x 1 + x 2 + x 1 x 2 = 2 x 2 x 1 + x 2 x 1 + x 2 x 2 2 = 2 x 1 x 1 + x 2 and the RHS is a bounded function. However, we compute x 1 x 2 = log x 1 + x x 1 x 2 x 1 + x 2 2 ; 4 x 1 x 2 x 1 + x u = 2 2 x 1 x 2 x 1 + x x 1 x 2 x 1 + x 2 2 L. 7 Example 2. f continuous but u C 1, 1. [ u = f x x x 1 n log + 1, x B 1 / 2 2 log 3/ 2 R R n. 8 f x is continuous after setting f 0 = 0. However, the solution has Therefore u C 1, 1. 1 u x = x 1 2 x 2 2 log x 1 / 2 9 x 1 2 x One can show that there is no classical solution to this problem. Assume otherwise a classical solution v exists, then the difference u v is a bounded harmonic function in B R \{ 0}. In the next lecture we will see that such functions can be extended as a harmonic function in the whole B R which means 2 u must be bounded, a contradiction.

2 2. C α regularity. The right space to work on are the Hölder spaces. Definition 3. Hölder continuity Let f: continuous at x 0 with exponent α if sup x R, x 0, 0 < α < 1. The function f is called Hölder f x f x 0 α <. 1 1 x x 0 f is called Hölder continuous in if it is Hölder continuous at each x 0 with the same? exponent α, denoted f C α. When α = 1, f is called Lipschitz continuous at x 0, denoted f Lip or f C 0, 1. C k, α contains f C k whose kth derivatives are uniformly Hölder continuous with exponent α over, that is sup x, y f x f y α <. 1 2 C k, α contains f C k whose kth derivatives are uniformly Hölder continuous with exponent α in every compact subset of. Example 4. The functions f x = α, 0 < α < 1, is Hölder continuous with exponent α at x = 0. It is Lipschitz continuous when α = 1. Remark 5. When k = 0, we usually use C α for C 0, α since there is no ambiguity for 0 < α < 1. We can define the seminorm and the norms where f C k, α = f C α sup x, y f x f y α, 1 3 f C α = f C 0 + f C α, 1 4 α k α f C 0 + α = k The following property is important. In short, C α is an algebra. Lemma 6. If f 1, f 2 C α, then f 1 f 2 C α G, and f 1 f 2 C α sup f 1 f 2 C + α α f C α. 1 5 f C 0 = sup f. 1 6 x sup f 2 f 1 C. α 1 7 Proof. Left as exercise. Theorem 7. Let R d be open and bounded, u x Φ x y f y dy, 1 8 where Φ is the fundamental solution. Then α a If f C 0, 0 < α < 1, then u C 2, α, and u C 2, α c f C α. 1 9 b If f L α = 0 case, then u C 1, α for any 0 < α < 1, and u C 1, α c f L. 20

3 c If f Lip α = 1 case with support contained in, then u C 2, α for any 0 < α < 1, and Proof. a Recall that Φ x y = C log x y for n = 2 and Φ x y = C 1. We first show u C 1. Formally differentiating we obtain x i u = x i Γ x, y f y dy = C u C 2, α c f L ip x y n 2. for n 3. x i y i n f y dy. 22 It is easy to check that the integrand is integrable. Therefore by the theorem regarding differentiating with respect to a parameter for Lebesgue integrals, we see that the formal relation indeed holds. x i u = C x i y i n f y dy Next we show u C 2, α. In the following we will omit the constant factor C. In this step we do some preparations. Again formally differentiating, we obtain x ix j u = δ i j n n x i y i x j y j n+ 2 f y dy. 24 But this time the integrand is not automatically integrable and therefore this equality is dubious. To overcome this difficulty, we first work in the weak sense. By extending f outside to be 0 resulting in a distribution with compact support, we can write x i u = x i n f 25 x in the sense of distributions. Thus we have [ x i x j u = x j xi n f 26 x in the sense of distributions. We compute the distributional derivative i x j n Take any φ C 0 R n, we know [ xi xi x j n φ = n x j φ x dx = lim ε 0 Now integrate by parts, we have x i n x R n \ B ε x j φ dx = φ x B ε = where = ε S i j x = δ i j = ε x i n φ x x i x j ε n+ 1 + R n \ B ε now. x i n x j φ dx. 27 x j + S i j x φ x dx x R n \ B ε R n \ B ε S i j x φ x dx. 28 n n x i y i x j y j n is the formal derivative. For the boundary term, we write φ x x i x j ε n+ 1 = φ 0 x i x j ε n+ 1 + x [ φ x i x j φ 0 εn = ε = ε

4 Note that since φ C 0, φ x φ 0 = O = O ε which makes the second term an O ε quantity. For the first term, a symmetry argument shows that the integral vanishes when i j. When i = j, we use symmetry and the fact that xi x i n+ 1 = 1 = ε ε n 1 = ω n 1, 31 = ε where ω n 1 is the surface area of the n 1 dimensional unit sphere, to conclude that the limit is c φ 0 for some constant c. Therefore we have shown that [ xi x j n φ = lim S i j x φ x dx + c δ 32 ε 0 R n \ B ε As a consequence, we have x ix j u x = lim ε 0 \ B ε S i j x y f y dy + c f x. 33 We now show directly that the second derivative x i x j u is Hölder continuous with power α. Since f x C α, we only need to show that [ S i j x 1 y f y dy S i j x 2 y f y dy x 1 x 2 α <. 34 lim ε 0 \ B ε x 1 uniformly for x 1, x 2. \ B ε x 2 3. x ix j u C α. Inspection of S i j reveals that for any 0 < R 1 < R 2 : R 1 y R 2 S i j x y dy = To make things simple, we extend f to be 0 outside. The resulting function is in C α 0 R n 2. We have S i j x y f y dy = R n \ B ε S i j x y [ f y f x dy. R n \ B ε 36 Note that since f C α, the integrand is integrable now, which means u C 2 has been proved. To show u C 2, α we need more refined analysis of the integral. First note that in writing the quantity as lim S i j x y [ f y f x dy 37 ε 0 R n \ B ε the singularity has been removed and we can take the limit and write x ix j u x = S i j x y [ f y f x dy. 38 R n For any x 1, x 2, setting δ = 2 x 1 x 2, we have x ix j u x 1 x ix j u x 2 = S i j x 1 y [ f y f x 1 S i j x 2 y [ f y f x 2 R n = + A + B. 39 B δ x 1 R n \ B δ x 1 2. Let f be the extended function. Then one notices that f x f y = f x f y when x, y, vanishes when x, y, and equals f x f y when x and y, where y is the intersection of and the line connecting x, y.

5 For A, we bound f y f x 1 f C α y x 1 α and f y f x 2 f C α y x 2 α, and get For B, we have B = = R n \ B δ x 1 R n \ B δ x 1 + A C f C α δ α = C f C α x 1 x 2 α. 40 R n \ B δ x 1 S i j x 1 y [ f y f x 1 S i j x 2 y [ f y f x 2 S i j x 1 y [ f x 2 f x 1 dy [ S i j x 1 y S i j x 2 y [ f y f x 2 dy B 1 + B It is easy to see that B 1 = 0. For B 2, we estimate 3 S i j x 1 y S i j x 2 y S i j x 3 y x 1 x 2 C x 1 x 2 n x 3 y for some x 3 lying on the line segment connecting x 1, x 2. We have x 1 x 2 B 2 C f C α n+ 1 R n \ B δ x 1 x 3 y y x α 2 α n+ 1 C f C α x 1 x 2 x 1 y dy R n \ B δ x 1 α 1 = C f C α x 1 x 2 x 1 x 2 = C f C α x 1 x α where we have used the fact that x i y are all comparable i = 1, 2, 3 for y B δ x 1. b We prove the stronger statement x i u is Log-Lipschitz, that is 1 x i u x 1 x i u x 2 C sup f x 1 x 2 log x 1 x It is easy to get x i u x 1 x i u x 2 sup f x 1 y i x 1 y n x 2 y i n x 2 y dy. 45 We extend f by 0 and break the integral to + with δ = 2 B δ x 1 R n \ B δ x 1 x 1 x 2. For the first term we obtain a bound C sup f x 1 x 2, for the second we use x 1 y i x 1 y n x 2 y i n x 2 y C x 1 x 2 n 46 x 3 y with a uniform C. Now note that for R big enough, =. The R n \ B δ B R \ B δ x 1 B R \ B δ / 2 x 3 integration can be carried out explicitly and yields the bound Thus ends the proof the details are left as exercise. C x 1 x 2 log R log x 1 x c This part is the same as b. Omitted. Remark 8. The techniques involved in the above proof is standard in the theory of singular integrals and are applied extensively in equations arising from fluid mechanics, mathematical biology, etc. 3. Note that the intermediate value theorem gives x 3 depending on y. But when we are working outside B δ x 1, ξ y are all comparable for any ξ between x 1 and x 2.

6 Remark 9. One may notice that when f L, one cannot reach x i u Lip that is x ix j u L. The reason is that the operator x i x j 1 does not map L into L. Details can be found in any textbook in real harmonic analysis. When f does not have compact support, we cannot obtain uniform bounds for u over the whole, but we can obtain estimates on any smaller set 0. 4 Theorem 1 0. Let R n be open and bounded, and 0. Let u solve u = f in. Here a If f C 0, then u C 1, α for any α 0, 1, and u c C 1, α 0 f C 0 + u. 48 L 2 b If f C α for 0 < α < 1, then u C 2, α, and u c C 2, α 0 f C α + u. 49 L 2 u L 2 = u 2 1 / Proof. We just give an outline of the proof here. Set η be a cut-off function and consider φ = η u. We have where the RHS has compact support. This gives and φ = F η f + 2 u η + u η 51 F L c η f L + C η u C 1 F C α c η f C α + C η u C 1, α Next we show that for any ε > 0, there is N ε > 0 such that and u C 1 N ε u L 2 + ε u C 1, α 54 u C 1, α N ε u L 2 + ε u C 2, α. 55 This is shown via reductio ad absurdum using the Arzela-Ascoli theorem. Thus we obtain [ u C η C 1, ε α 0 u + C 1, α u + L c η N ε f 2 C 56 0 and a similar estimate for u C 2, α 0 with the problem that the C1, α norm on the LHS is on 0 while that on the RHS is on a bigger set and therefore cannot be absorbed into the LHS. This difficulty is overcome by the following technical trick. Consider the case when 0 = B r, = B R2, we have [ u C η C 1, ε α B r u C 1, α + B R2 u L 2 + B R2 c η N ε f C B R2 Now set 5 A sup R r 3 u. 58 C 1, α B r 0 r R for some R > R Meaning: The closure 0 is a compact subset of. 5. Here it seems we need to assume the finiteness of this quantity.

7 Now choose R 1 such that This gives A 1 2 R R 1 3 u C 1, α B R1, 59 Now observe that C η A 1 2 R R 1 3 u C 1, α [ B R1 2 R R 1 3 ε C η u C 1, α + B R2 C η u L 2 B R2 A 1 C R R 1 3 R R 2 3 ε R 1 R 2 2 A 1 + C N ε + 2 R R 1 3 c η N ε f C 0 B R R 2 R 1 2 and c η 1, we have, using the definition of A 1, R R 1 3 R 2 R 1 2 u L 2 B R2 + C R R 1 3 f C 0 B R2. 61 Now for fixed R, R 1, one can choose R 2 and ε appropriately so that the coefficient of A 1 on the RHS is less than 1. Thus we obtain the desired estimate for u C 1, α B r 1 R r 3 A Now we can cover 0 by balls B r, and set R = r + d where d = dist 0,, and finish the proof. Corollary 1 1. If u solves u = f with f C k, α for k N and 0 < α < 1, then u C k + 2, α 0 for any 0 and u c C k + 2, f + u α 0 C k,. 63 α L 2 In particular, u C when f C. 3. Regularity and existence: method of continuity. We briefly discuss why the regularity estimates matter. Consider two bounded linear operators L, L from Banach spaces X to Y. 6 Assume that we know that L is surjective and wish to establish that L is also surjective, in other words the solvability of for arbitrary y Y. Define a family of operators L x = y. 64 L t = 1 t L + t L. 65 Thus L 0 = L and L 1 = L. Assumption. We have uniform that is, independent of t a priori that is, assuming the existence of solutions estimates Under this assumption, one has Theorem 1 2. If L 0 is surjective, so is L 1. u X c L t u Y. 66 Proof. The idea is to show that there is ε independent of t, such that if L τ is surjective, so is L t for all t τ, τ + ε. To see this, note that the estimate u X c L t u Y implies that all L t s are injective. Thus the inverse L τ 1 is well-defined and bounded. We write 6. For example, in the case L = i, j ai j 2 with a i j C α, X = C 2, α, Y = C α. x i x j L t u = f 67

8 into This gives L τ u = f + L τ L t u = f + t τ L 0 L 1 u. 68 u = L τ 1 f + t τ L τ 1 L 0 L 1 u. 69 Therefore all we need to do is to show the existence of a fixed point of the mapping from X to X : u Tu L τ 1 f + t τ L τ 1 L 0 L 1 u. 70 It is clear that if we take t τ small enough, we can find 0 < r < 1, such that Tu Tv X r u v X. 71 Now set v 0 = 0 and v n = Tv n 1, we see that { v n } is a Cauchy sequence in X and therefore has a limit v which is a fixed point. An application of this theorem is to show the existence of the solutions to L u = i, j a i j x x i x j + i b i x u x i + c x u x = f 72 for Hölder continuous a i j, b i, c starting from the existence of the Poisson equation which can be shown by explicitly construct the solutions. Further readings. D. Gilbarg, N. S. Trudinger, Elliptic Partial Differential Equations of Second Order, Chapter 4. J. Jost, Partial Differential Equations, GTM 21 4, Chapter 1 0. Exercises. Exercise 1. Prove the following: If f 1, f 2 C α, then f 1 f 2 C α G, and f 1 f 2 C α sup f 1 f 2 C + α Hint: Use the identity f 1 x f 2 x f 1 y f 2 y = f 1 x [ f 2 x f 2 y + f 2 y [ f 1 x f 1 y. Exercise 2. Write down the detailed proof of If f L, then u C 1, α for any 0 < α < 1, and where sup f 2 f 1 C. α 73 u C 1, α c f L. 74 u x Φ x y f y dy. 75