Some lecture notes for Math 65E: PDEs, Fall 216 Tianling Jin December 1, 216 1 Variational methods We discuss an example of the use of variational methods in obtaining existence of solutions. Theorem 1.1. Let R n be a smooth bounded open set, with n 3. Let 1 < p < n+2 n 2. Then there exists a positive function u C 3 () satisfies u = u p in (1) u = on Proof. For u H 1 (), define the energy functional I[u] = u 2 ( u p+1) 2 p+1 and m = inf I[u]. u H 1() Clearly, m. We will show in the below that m is attained by some function in H 1 (), which will be called a minimizer. Moreover, we will show that this minimizer will be a desired solution of (1). Let u k } be a sequence in H 1 () such that By a normalization, we may assume that I[u k ] m as k. u k p+1 = 1. Consequently, u k is a bounded sequence in H 1(). Therefore, there exists u H1 () such that u k u weakly in H 1 (). Consequently, u 2 lim inf k u k 2 m. 1
Meanwhile, since p + 1 < 2n n 2, it follows from the Rellich compactness theorem that u k u strongly in L p+1 (). So, Therefore, and thus, u p+1 = 1. I[u] m, I[u] = m. Since I[ u ] I[u], we have that u or u. So we assume that u. Since u p+1 = 1, u. So we have found a minimizer u for m. This implies that for every ϕ H 1(), = d dt I[u + tϕ]. t= A calculation yield = 2 u ϕ 2m u p ϕ, that is u ϕ m u p ϕ = So u is weak solution of (1) (in the distribution sense) after a scaling (ũ = cu by some proper positive constant c). Two regularity theory: 1. W 2,p theory, that is for q (1, ), if u = f, where f L q, then u W 2,q loc. 2. Schauder theory, that is for α (, 1), if f C α, then u C 2,α loc. These two theories plus Sobolev embedding implies that the solutions of (1) u C 3. This is called bootstrap arguments. Note that such bootstrap arguments will NOT work when p = n+2 n 2. Finally, u is positive in by the strong maximum principle. Next, we will show that the equation (1) does not have non-trivial solutions when p n+2 n 2 and when the domain is strictly star-shaped with respect to zero, that is x ν > everywhere on. Here, ν is the outer normal of Theorem 1.2 (Pohozaev). Let p > n+2 n 2 and is strictly star-shaped with respect to zero. Suppose u C 2 () is a solution of u = u p 1 u in (2) u = on. Then u in. 2
Proof. Multiplying x u on the both sides of (2) and doing some calculations, we have ( ) n 2 u 2 + 1 u 2 (ν x) = n u p+1. 2 2 p + 1 This is usually called the Pohozaev identity. Multiplying u on the both sides of (2) and integrating by parts, we have u 2 = u p+1. Thus, 1 2 u 2 (ν x) = ( n p + 1 n 2 ) u p+1, (3) 2 where we used the fact that p > n+2 n 2 in the last inequality. Therefore, u. Remark 1.3. Theorem 1.2 also holds for p = n+2 n 2. In this case, it follows from (3) that u = on, since x ν >. Then by a unique continuation property, u, which is slightly complicated. However, if one additionally assumes that u in, then = u = u p, from which u follows. Remark 1.4. The assumption that is star shaped is necessary in Theorem 1.2. For example, if is an annulus, then there exists a positive radial solution of u = u n+2 n 2 u = on in 2 Method of subsolutions and supersolutions We will investigate the boundary -value problem for the nonlinear Poisson equation u = f(u) in u = on, (4) where f : R R is smooth and f L (R) C for some constant C. Definition 2.1. (i). We say that ū H 1 () is a weak supersolution of (4) if ū vdx f(ū)vdx for every v H 1 (), v a.e. (ii). Similarly, we say that u H 1 () is a weak subsolution of (4) if ū vdx f(ū)vdx for every v H 1 (), v a.e. 3
(ii). We say that u H 1 () is a weak solution of (4) if ū vdx = f(ū)vdx for every v H 1 (). Note that if ū, u C 2 (), then we have ū f(ū), u f(u). Theorem 2.2. Assume there exist a weak supersolution ū and a weak subsolution u of (4) satisfying u, ū on in the trace sense, u ū a.e. in. Then there exists a weak solution u of (4) such that u u ū a.e. in. Proof. Since f L (R) C for some constant C, we can choose a large λ > such that g(z) := f(z) + λz is an increasing function. Now we denote u = u. Given u k, k =, 1,,, we are going to inductively define u k+1 be the unique weak solution of the following liner boundary-value problem u k+1 + λu k+1 = f(u k ) + λu k in u = on. We claim that u = u u 1 u 2 u k u k+1 ū. (5) To prove this claim, we first note that for k =, (u u 1 ) v + λ(u u 1 )v for every v H 1 (), v a.e. Choose v = (u u 1 ) +, we obtain (u u 1 ) (u u 1 ) + + λ(u u 1 )(u u 1 ) + This implies that (u u 1 ) + =, that is, u u 1. Now we assume inductively that Then we have u k 1 u k. (u k u k+1 ) v + λ(u k u k+1 )v = (f(u k 1 + λu k 1 f(u k ) λu k ))v for every v H 1 (), v a.e. 4
Choosing v = (u k u k+1 ) + will lead to that (u k u k+1 ) + =, that is u k u k+1. This proves the claim on the monotonicity of the sequence u k }. Then we show that u k ū for all k. This is true for k =. Assume the induction that u k ū, then we have (u k+1 ū) v + λ(u k+1 ū)v = (f(u k + λu k f(ū) λū))v for every v H 1 (), v a.e. and let v = (u k+1 ū) +, we have (u k+1 ū) + =, that is u k+1 ū. Now, let u = lim k u k. By dominated convergence theorem, u k u in L 2 (). Since f(u k ) L 2 () C( u k L 2 + 1) C( ū L 2 + 1), we have that sup u k H 1 () <. k Therefore, subject to a subsequence which is still denoted as u k },we have u k u weakly in H 1 (). To verify that u is a weak solution of (4), we notice that ū k+1 v + λu k+1 vdx = (f(u k ) + λu k )vdx for every v H 1 (). Sending k and cancelling the term with λ, we have the confirm ū vdx = f(u)vdx for every v H 1 (). Note that the monotonicity (5) can also be proved by maximum principle if ū, u are smooth. Then by the Schauder estimate, we have that every u k is smooth. Then (u k u k+1 ) + λ(u k u k+1 ) = f(u k 1 ) + λu k 1 f(u k ) λu k in the classical sense. Since u k u k+1 on, we have u k u k+1 in. 3 The Dirichlet problem: Perron s method In this section, we will discuss the solvability of u = in, u = ϕ on. (6) Theorem 3.1. Let R n be a bounded open set. Suppose C 2. Let ϕ be a continuous function on. Then there exists a solution u C 2 () C () of (6). Remark: That satisfies the exterior ball condition at every point on would be sufficient. That is, for every ξ, there exists a ball B r (y) such that B r (y) = ξ}. We need some generalized subharmonic functions first. 5
Definition 3.2. A C () function u is called subharmonic (or superharmonic) in if for every ball B and every harmonic function h in B satisfying u ( )h on B, we also have u ( )h in B. Such subharmonic functions have several useful properties: 1. If u is subharmonic in a connected domain in, then it satisfies the strong maximum principle in. That is, if v is super harmonic in with v u on, then either u > v in or v u. To prove this, we suppose the contrary that at some point x, we have (u v)(x ) = sup(u v) = M, but there is a ball B = B r (x ) such that u v M on B. Let ū, v be the harmonic functions respectively equal to u, v on B (this is can be achieved by Green s representation). Then one sees that M sup(ū v) (ū v)(x ) (u v)(x ) = M. B Therefore, every inequality in the above has to be an equality. By the strong maximum principle for harmonic functions, it follows that ū v M. Thus u v M, which is a contradiction. 2. Let u be subharmonic in and B is a ball strictly contained in. Denote ū as the harmonic function in B satisfying ū = u on B. We define in he harmonic lifting of u by ū(x), x B, U(x) = u(x), x \ B. Then the function U is also subharmonic in. This can be proved as follows. Let B be an arbitrary ball. Let h be harmonic in B such that h = U on B. Then h u on B, and thus, h u in B. So h U in B \ B. Since U is harmonic in B, by maximum principle, we have h U in B B. Hence, U h in B, and thus, U is subharmonic in. 3. Let u 1, u 2,, u k be subharmonic in. Then the function u(x) = max(u 1,, u k ) is also subharmonic which is a trivial consequence of the definition of subharmonic functions. Now let us prove Theorem 3.1. A C () function u is called a subsolution of (6) if u is subharmonic, and u ϕ on. Similarly, a C () function v is called a supersolution of (6) if v is subharmonic, and v ϕ on. Denote S be the set of all subsolutions of (6). S since the constant function inf ϕ is a subsolution. Proposition 3.3. The function u(x) = sup v S y(x) is harmonic in. Proof. Since sup ϕ is a super solution, we know that v sup ϕ for every v S. Thus, u is well-defined. Let y be a fixed point. There exists v k S such that v k (y) u(y). By replacing v k with max(v k, inf ϕ), we may assume that the sequence v k } is bounded. Choose R > such that B = B R (y), and define V k as the harmonic lifting of v k in B. Then V k 6
S. By Harnack inequality and gradient estimate, there exists a subsequence V nk } converging uniformly in every ball B ρ (y) with ρ < R to a harmonic function in B. Clearly, v u in B and v(y) = u(y). We are going to show that v = u in B. Suppose v(z) < u(z) for some z B. Then there exists w S such that v(z) < w(z). Let w k = max(v nk, w) and W k be its harmonic lifting in B. As before, by Harnack and gradient estimate, a subsequence of W k converges to a harmonic function w in B satisfying v w u. Since v(y) = u(y), we have v(y) = w(y), and by strong maximum principle. v w, which contradicts with v(z) < w(z) w(z). Therefore, the u obtained in the above proposition is a candidate solution of (6). In the below, we will show that this u indeed satisfies the boundary condition u = ϕ on, which will finish the proof of Theorem 3.1. As along as the boundary satisfies the exterior boundary, we can construct some barrier functions for your purpose. Proof of Theorem 3.1. For ξ, there exists a ball B = B R (y) such that B = ξ}. Define R 2 n x y 2 n, n 3 w(x) := log R + log x y, n = 2. Note that w(ξ) =, w(x) > in \ ξ}, and w is harmonic. Let M = sup ϕ. Since ϕ is a continuous function, for ε >, there exists δ > such that Choose κ large enough such that ϕ(x) ϕ(ξ) < ε, x ξ < δ. κw(x) 2M x ξ δ. Then ϕ(ξ) + ε + κw and ϕ(ξ) ε κw are respectively supersolution and subsolutions of (6). Therefore, ϕ(ξ) ε κw(x) u(x) ϕ(ξ) + ε + κw(x). Hence Since w(ξ) =, we obtain that u(x) ϕ(ξ) ε + κw(x). u(x) ϕ(ξ) as x ξ. 4 Schauder estimates We first prove the following Schauder estimates for Possion s equation. 7
Theorem 4.1. Let α (, 1), f C α ( ), and u C 2 ( ) be a solution of u = f in. Then there exists C > depending only on α and the dimension n such that u C 2,α (/2 ) C( u L ( ) + f C α ( )) Proof. First of all, by multiplying some proper constant, we can assume that u L ( )+ f C α ( ) 1. Let ρ = 1/2. For k = 1, 2, 3,, let v k be the solution of v k = f() in B k, v k = u on B k, where B k = B ρ k centered at the origin. Claim 1: v k v L (B k ) Cρ (2+α)k. This claim can be proved as follows. Let ṽ k (x) = ρ 2k (v k u)(ρ k x), where x. Then ṽ k = f() f(ρ k ) in, ṽ k = on, By the maximum principle (one of the homework problem), we have ṽ k L ( ) C( u L ( ) + f() f(ρ k x) L ( )) Cρ αk, from which the claim follows. Claim 2: v k v k+1 L (B k+1 ) Cρ (2+α)k. This is because v k v k+1 is harmonic in B k+1, and therefore, v k v k+1 L (B k+1 ) = v k v k+1 L ( B k+1 ) v k u L ( B k+1 ) Cρ (2+α)k, where in the last inequality, we use Claim 1. Let w k = v k+1 v k, w = v 1. Then we know that from Claim 2 that, for every x B k+2 we have j w k (x) Cρ (2+α j)k. for ρ i+3 x < ρ i+2, 1 u(x) w l () Dw l () x 2 xt D 2 w l ()x l= u(x) + l=i+1 l= i w l (x) + l= w l () + ρ (2+α)(i+1) + 2c 2 x 3 + x 2 l=i+1 C 3 x 2+α. c 2 ρ αl l=i+1 i w l (x) l= l= i w l () l= Dw l () x + 1 2 i ρ (α 1)l + l= l=i+1 8 l=i+1 i Dw l () x l= ρ (2+α)l + x x T D 2 w l ()x l=i+1 i l= c 2 ρ (1+α)l 1 2 xt D 2 w l ()x
So we have proved that there exists a second order polynomial P such that u(x) P (x) C x 2+α where all the coefficients of the polynomial are universally bounded. This leads to the conclusion of the theorem. Theorem 4.2. Let α (, 1), f C α ( ), a ij (x) C α ( ), and u C 2 ( ) be a solution of a ij (x)u ij (x) = f in, where λi (a ij (x)) λ 1 I in. Then there exists C > depending only on α, λ and the dimension n such that u C 2,α (/2 ) C( u L ( ) + f C α ( )). Proof. Let ρ = 1/2. For k = 1, 2, 3,, let v k be the solution of a ij () ij v k = f() in B k, v k = u on B k, where B k = B ρ k centered at the origin. Then for ṽ k = v k u, we have a ij () ij ṽ k = f() f(x) + (a ij () a ij (x))u ij (x) in B k, Following the proof in Theorem 4.1, we can show that [ 2 u] C α (/4 ) C( f C α (/2 ) + u L (/2 ) + 2 u L (/2 )), (7) where the constant C depends only on α, n, λ. Now that the domain on the two sides of the inequality are DIFFERENT. Then the conclusion follows from the next iteration lemma. M. Giaquinta and E. Giusti, On the regularity of the minima of variational integrals. Acta Math. 148 (1982), 31 46. Lemma 1.1. Lemma 4.3. Let h(t) be a nonnegative bounded function defined for T t T 1. Suppose that for T t < s T 1 we have h(t) A(s t) α + B + θh(s) where A, B, α, θ are nonnegative constants, and θ < 1. Then there exists a constant C >, depending only on α, θ such that for every ρ, R, T ρ < R T 1, we have Proof. Consider the sequence t j } defined by h(ρ) C(A(R ρ) α + B). t = ρ, t j+1 t j = (1 τ)τ j (R ρ) with τ (, 1). By iteration ( h(t ) θ k A h(t k ) + (1 τ) α (R ρ) α + B ) k 1 θ j τ jα. We choose now τ such that τ α θ < 1 and let k. Then the conclusion follows. 9 i=
Proof of Theorem 4.2 continued. Let h(t) = [ 2 u] C α (B t). We will show that this h will satisfies Lemma 4.3. Let t < s 1. For every z B t, choose r = s t. For x, let Then u r (x) = u(z + rx), ã ij (x) = a ij (z + rx), f r (x) = r 2 f(z + rx). Therefore, the estimate (7) holds for u r, i.e., Scaling back, we have ã ij ij u r (x) = f r (x) in. [ 2 u r ] C α (/4 ) C( f r C α (/2 ) + u r L (/2 ) + 2 u r L (/2 )), [ 2 u] C α (B r/4 (z)) Cr 2 α ( f C α ( ) + u L ( )) + Cr α 2 u L (B r/2 (z)). By a covering, we have [ 2 u] C α (B t) Cr 3 ( f C α ( ) + u L ( )) + Cr 1 2 u L (B (s+t)/2 ). (From Gilbarg-Trudinger s book)we have the following interpolation lemma (the following is not sharp): there exists C > independent of s, t such that for all ε >, 2 u L (B (s+t)/2 ) εr 2 [ 2 u] C α (B s) + Cε 2+α α r 2 u L (B s). Choose ε = r 3 /2C, we have for some β > [ 2 u] C α (B t) 1 2 [ 2 u] C α (B s) + Cr β ( f C α ( ) + u L ( )). Then the conclusion from from Lemma 4.3. 5 De Giorgi estimates Let a ij L ( ) and uniformly elliptic, that is λi a ij (x) ΛI in for some λ, Λ >. Theorem 5.1 (De Giorgi, 1958). Let u H 1 ( ) be a weak solution of j (a ij i u) = in (8) Then there exists α (, 1) and C >, both of which depends only on n, λ and Λ, such that u C α (/2 ) C u L 2 ( ). 1
The proof of this theorem consists of two steps: from L 2 to L, and from L to C α. Proposition 5.2 (from L 2 to L ). Let u H 1 ( ) be a weak subsolution of (8), that is, j (a ij i u) in. Then there exists a constant δ = δ(n, λ, Λ) such that if u + L 2 ( ) < δ, then we have u + L (/2 ) 1. Applying this proposition to ( δ/ u + L 2 ( ))u, we obtain the following theorem: Theorem 5.3 (from L 2 to L ). Let u H 1 ( ) be a weak subsolution of (8), that is, Then j (a ij i u) in. u + L (/2 ) C(n, λ, Λ) u + L 2 ( ) Consequently, if u H 1 ( ) is a weak solution of (8), then u L (/2 ) C(n, λ, Λ) u L 2 ( ) To prove Proposition 5.2, we shall use the following Caccioppoli inequality (or energy estimate) Lemma 5.4 (Caccioppoli inequality). Let u H 1 ( ) be a weak subsolution of (8) and ϕ Cc ( ), we have ( (ϕu + )) 2 C(n, λ, Λ) (u + ) 2 ( ϕ) 2. Proof. Since u is a subsolution, we have a ij i u j (ϕ 2 u + ). That is a ij ϕ 2 i u + j u + + 2 a ij ϕu + i u + j ϕ. Then 1 ( (ϕu + )) 2 a ij i (ϕu + ) j (ϕu + ) λ B 1 C ϕ 2 (u + ) 2 + B 1 a ij ϕ 2 i u + j u + + a ij ϕu + i u + j ϕ + a ij ϕu + i u j ϕ B 1 C ϕ 2 (u + ) 2 + 3(a ij + a ji )ϕu + i u + j ϕ B 1 C ϕ 2 (u + ) 2 + u + i (u + ϕ) j ϕ B 1 C ϕ 2 (u + ) 2 + 1 2λ ( (ϕu+ )) 2, where in the last inequality we have used Holder s inequality. 11
Lemma 5.5. Let u H 1 ( ) be a weak solution of (8). Then u + is a weak subsolution of (8). Proof. Exercise. Now we can prove Proposition 5.2. Proof of Proposition 5.2. We will work on a family of ball and the family of truncated functions Define B k = x 1/2 + 2 k } u k = (u (1 2 k )) +. U k = u 2 k. B k Let ϕ k be a sequence of nonnegative shrinking cut-off functions: ϕ k = 1 in B k and ϕ = in B c k 1. Also, ϕ k C2 k. Note that where u k+1 >, then u > 1 2 k 1 = 1 2 k + 2 k 1, and thus, u k > 2 k 1. Therefore, x : ϕ k+1 u k+1 > } x B k : u k > 2 k 1 }. We have from the Sobolev inequality with p = 2n n 2 (or any p if n = 2) ( ) 2/p ϕ k+1 u k+1 p C (ϕ k+1 u k+1 ) 2. From Holder inequality ( ϕ k+1 u k+1 2 ϕ k+1 u k+1 p ) 2/p x : ϕ k+1 u k+1 > } 2 n. Therefore, U k+1 C (ϕ k+1 u k+1 ) 2 x : ϕ k+1 u k+1 > } 2 n. Using Lemma 5.4, we have U k+1 C2 2k (u k+1 ) 2 x : ϕ k+1 u k+1 > } 2 n. supp(ϕ k+1 ) The support of ϕ k+1 in contained in B k and u k+1 u k, we have U k+1 C2 B 2k u 2 k x : ϕ k+1u k+1 > } 2 n. k Since x : ϕ k+1 u k+1 > } x B k : u k > 2 k 1 }. 12
we have x : ϕ k+1 u k+1 > } 2 n x Bk : u k > 2 k 1 } 2 4k n ( B k u 2 k ) 2 N = 2 4k 2 n U n k. where the Chebyshev inequality was used in the last inequality. Therefore, we have U k+1 C2 4k (U k ) 1+ 2 n. The as long as U = δ is small enough, we will have U k as k. This proves that u + 1 in /2. In fact, U k decays faster than geometric sequences. The next step is to prove the regularit from L to C α. The kep step is this oscillation lemma. Proposition 5.6. Let u H 1 (B 2 ) be a weak subsolution of (8) in B 2. Suppose that u 1 in B 2. Assume that v } µ(> ). Then u 1 γ in /2, where γ depends only on λ, Λ, n, µ. In other words, if u is a subsolution and smaller than one, and is far from 1 in a set of non-trivial measure, then u cannot get too close to 1 in /2. Let us postpone the proof to the end, we shall first use this proposition to prove the following Holder estimate. Theorem 5.7. Let u H 1 (B 3 ) be a weak solution of (8) in B 3. Then u C α (/2 ). Moreover, u C α (/2 ) C u L 2 (B 3 ). Here, each of the constants α and C depend only on n, λ, Λ. Proof. First of all, we know from Theorem 5.3 that u L (B 2 ) < C u L 2 (B 3 ). Denote Consider the function v(x) = osc u = sup u inf u. ( 2 u(x) sup ) B 2 u + inf B2 u, osc B2 u 2 We have that 1 v 1 in B 2. Assume that v } /2, then we can apply Proposition 5.6 on v to obtain osc B1/2 u 2 γ. Hence, osc B1/2 u (1 γ/2)osc B2 u. If v } /2, then the same result holds by working with v. Therefore, we have proved that with γ = 1 γ/2 < 1 depends only on n, λ, Λ. osc B1/2 u γosc B2 u (9) 13
Take any x in /2, and introduce the rescaled functions u k (y) = u(x + y/2 n ), a (k) ij (y) = a ij(x + y/2 n ). Then a (k) ij satisfies the same assumptions as a ij, and u k is a weak solution of (8) with a ij replaced by a (k) ij. We apply recursively (9) to u k to obtain sup u(x ) u(x) 2 u L (B 2 ) γ k. x x 2 k Then for y such that 2 k 1 x x < 2 k, we have u(x ) u(x) 2 u L (B 2 ) γ k u L (B 2 )2 2 (k+1)α 2 (k+1)α γ k C u L (B 2 ) x x α, as long as the α is chosen by 2 α γ = 1, that is, log γ α = log 2 >. Note that this estimate is independent of the choice of x. Then the conclusion follows. then Now we are going to prove Proposition 5.6. We first note that if the set and thus, by Proposition 5.2, we have u } δ/4 u + L 2 ( ) δ/2, u 1/2. So we much bridge the gap between knowing that u } µ and knowing u } δ/4. The main tool is the so-called De Giorgi isoperimetric inequality, which rough says that for an H 1 function u, it must pay in measure to increase from u = to u = 1. Lemma 5.8. Consider a function w such that w + 2 C. Define A = w }, E = w 1}, D = < w < 1}. Then there exists a constant C 1 depends only on n such that C D C 1 ( A E 1 1 n ) 2. 14
Proof. Consider w = sup(, inf(w, 1)). Note that w = w + χ w 1}. For x E we integrate over lines in to x A 1 = w(x ) = Integrating over A, we obtain that A 1 1 1 2 2 C C dt D dt A 2 dt 2 r n 1 dr r n 1 dr = 1 1 d dt w(tx + (1 t)x ) w(tx + (1 t)x ) (x x ) w(tx + (1 t)x ) (x x ) dx r n 1 dr w(tx + (1 t)x ) rds(x) B r(x ) r n dr r r w(z) dz z x n 1 w(z) dz z x n 1 Integrating x E, we have A E C Note that E w(z) ds(z) B tr(x ) dt w(z) ds(z) B t(x ) t n 1 dt D B t(x ) w(z) ds(z) z x n 1 1 w(z) dz E z x n 1 dx 1 z x n 1 dx 1 B x n 1 dx = C E 1/n, where B is the ball centered at the origin such that B = E. Therefore, by Holder inequality, A E C E 1/n D 1/2 ( w(z) 2 dz) 1/2. D Proof of Proposition 5.6. We consider the following new sequence of truncation w k = 2 k (u (1 2 k )) 15
Note that w k 1 in B 2 and is also a subsolution. From the Caccippolli inequality, we have w + 2 C. We also have w k } µ. We will apply Lemma 5.8 recursevely on 2w k as long as (w + k+1 )2 δ. Note that 2w k 1} = w k+1 } (w + k+1 )2 δ. From Lemma 5.8, there exists a positive constant β independent of k such that Therefore, < w k < 1/2} β. w k+1 } = 2w k 1} w k } + β µ + kα. This clearly fails after a finite number of k. At this k, we have for sure that (w + k +1 )2 δ. Then Proposition 5.2 implies that Rescaling back, we have w k +1 1/2 in /2. u 1 2 k +1 + 2 k +2 = 1 2 k +2 in /2. 16