Partial Differential Equations, nd Edition, L.C.Evans Chapter 5 Sobolev Spaces Shih-Hsin Chen, Yung-Hsiang Huang 7.8.3 Abstract In these exercises always denote an open set of with smooth boundary. As usual, all given functions are assumed smooth, unless otherwise stated.. Suppose k Z, < γ. Show that C,β () is a Banach space. Proof. We prove the case k = for simplicity. The general case is almost the same (except we need the standard convergence theorem between {f n } and {Df n }). Given a Cauchy sequence {f n } C,β (), then there is a f C() such that f n f uniformly and a constant M > such that for each n N and s t, f n (s) f n (t) s t β M. Then for each s t, there is some N = N(s, t) N such that f(s) f(t) f(s) f N (s) + f N (s) f N (t) + f N (t) f(t) s t β + M s t β. Therefore, f C,β (). It remains to show f n f in C,β (). For each s, t and ɛ >, by Cauchy s criteria, we can find a N = N(ɛ) such that for k, n N f k (s) f n (s) f k (t) + f n (t) ɛ s t β Department of Math., National Taiwan niversity. Email: d3@ntu.edu.tw Department of Math., National Taiwan niversity. Email: d4@ntu.edu.tw
For each η >, we can find a K = K(η, s, t) > N, such that f(x) f K (x) η for x = s, or x = t. Therefore, for every n N f(s) f n (s) f(t)+f n (t) f(s) f K (s) + f K (s) f n (s) f K (t)+f n (t) + f(t) f K (t) η+ɛ s t β. Letting η, we obtain that f(s) f n (s) f(t) + f n (t) ɛ s t β for all n N(ɛ).. Proof. u C,γ = sup u(x) + sup Set t s.t. ( t)β + t = γ. Then u(x) u(y) x y γ = u(x) u(y) x y γ. ( ) t ( ) t u(x) u(y) u(x) u(y). x y β x y u C,γ u t u t + [u] t [u] t C,β C, := a t b t + a t b t. Since y = x t, < t <, is a concave function, we have ( a t b t + a t b t a ( b ) t a ( b ) ) t = (a + a ) + a + a a a + a a Then (a + a ) ( b a + a + b a + a = (a + a ) t (b + b ). ) t u C,γ u t u t + [u] t C,β [u] t C, ( u + [u] C,β) t ( u + [u] C,) t = u t C,β u t C,. Remark. There is a general convexity theorem for Higher-order Hölder norm (due to Hörmander), see Helms [4, chapter 8]. 3. Omit. 4. Assume u W,p (, ) for some p <. (a) Show that u is equal a.e. to an absolutely continuous function and u belongs to L p (, ). (b) Prove that if < p <, then for a.e. x, y [, ], u(x) u(y) x y p u p.
Proof. (b) follows easily from (a) and Hölder inequality. For (a), denote the weak derivative of u by u w. Define v(t) = t u w(s) ds, then v is absolutely continuous. Given φ C c, [v u]φ dx = = x = t u w(t) dtφ (x) dx φ (x) dxu w(t) dt + φ(t)u w(t) dt + u(x)φ (x) dx u w(x)φ(x) dx u w(x)φ(x) dx =. (This argument avoids the use of the Lebesgue Differentiation theorem if we only want to show the -d weakly differentiable function F = H a.e., an absolutely continuous function.) We are done if we use Exercise. There are many other way to prove the second fundamental lemma of calculus of variations without using mollifications. We also note such statement is true for distributions. For example, see []. 5. Proof. For any V W, we can find a smooth function u such that u on V and supp(u) W. Take ε = 3 dist(v, W c ) >. Let φ ɛ (z) be the standard mollifier supported in { z ɛ}, V := {z : dist(z, V ) < ɛ} and u(x) := χ V φ ε (x) = χ V (x y)φ ε (y) dy = φ ε (y) dy. x V Then u is smooth (by LDCT), u = on V and supported in {dist(x, V ) ɛ} W. 6. Apply Exercise 5. We omit it since it can be found in many Differential Geometry textbooks. 7. Proof. u p ds = = u p α ν ds div ( u p α) dx u p div α + ( u p ) α dx 8. Proof. Consider u n (x) = n dist(x, ) +. Note that u n C(Ū), u n(x) and {u n (x)} is decreasing to for each x, MCT tells us u n p,. On the other hand, u n p, = p, is a positive constant. Hence T u n p, u n p, = u n p, u n p, which means T will not be bounded, if T exists. 3
9. Proof. Given u H () H (), there exists u k C (Ū) with u k u in H (), and v k Cc (Ū) with v k u in H (). Hence { u k } and { u k } are bounded in L. Note u k u k n ds = div (u k u k ) dx = u k dx + u k u k dx. The right-hand side tends to u dx + u u dx. Note u k u k n ds = = div ((u k v k ) u k ) dx (u k v k ) u k + (u k v k ) u k dx. which tends to by Hölder and (u k v k ) in H (). Therefore, for all u H () H(), u + u u dx = Then we apply Hölder s inequality to get the desired inequality.. Proof. (Sketch) (b) is similar to (a). To prove (a), we start with the formula in Hint, do integration by parts once, and then apply the generalized Hölder with exponents pair (p, p, p p ).. Proof. Given K, compact. Then u L (K). Note that the weak derivative D(u η ɛ ) = Du η ɛ =. Since is connected and u η ɛ is smooth, u η ε is a constant c ɛ which will converges to u in L (K) as ɛ and hence {c } n= forms a Cauchy sequence (note that K n has finite measure) that will converges to some constant c. Therefore, u = c a.e. in K and hence in.. Proof. Consider in R, let u(x) = χ (,) (x). 3. Proof. Consider = B(, ) {(x, y) x > } in R, let u(x, y) = rθ in polar coordinate. ( ) 4. Let n >. Show that log log + W,n (B x ()). Proof. u(x) n dx = ω n ( log log + ) n r n dr, let t = r r log log( + t) n = ω n dt tn t log log( + t) n log log( + t) n Since lim =, there is T > such that < for t T. Then t t t u(x) n log log( + t) n dx = ω n dt tn t T log log( + t) n ω n dt + ω tn n dt <. t tn 4
On the other hand, we omit the proof that the first weak derivative of u exists and u xi = log( + ) + x x x i x 3. So u xi n dx ω n log( + ) + n r r r r n dr, let t = r = ω n log( + t) n ( + t) n tn dt ω n n <. log s n e sn esn ds (s = log( + t)) 5. Proof. Applying Poincaré inequality, ( Applying Hölider inequality, ) ( ) ( ) u dx (u) dx + u (u) dx ( ) (u) + C Du dx. (u) ( {u } u dx dx {u } ( = ( α) ) ( {u } ) u dx. ) u dx Therefore, ( ) ( u α dx ) ( ) u dx + C Du dx, Since α <, we proved the desired inequality ( )( ) ( ) α u dx C Du dx. Remark. For further discussions on Poincare-Type inequality, see G.Leoni [5, Section.] and L.Tartar [7, Section -]. 5
6. Proof. For any u C c ( ), x u u dx = u div x = ( x u n R x n x dx. ) dx, by Divergence theorem The Hölder s inequality implies that n Rn u ( x dx u ) ( R x dx n Rn Given u H ( ), there exists a sequence u k C c u dx ). ( ) converging to u in H. By picking a further subsequence, we may assume u k converging to u pointwisely. Fatou s lemma implies u dx = liminf k u k dx liminf k (n ) 4 u k (n Rn ) u dx x 4 x dx. Remark 3. This proof is simpler than the one suggested in the hint. Also note that the same approximation argument should be applied at the end of the proof to Theorem 7 on page 96. Remark 4. A family of inequalities that interpolate between Hardy and Sobolev inequalities is given by the Hardy-Sobolev inequality, ( u p n n x dx p ) p C(p) u dx for any u Cc ( ), where p n/(n ), n 3. One huge extensions and improvements of the Hardy-Sobolev inequalities is the Caffarelli-Kohn-Nirenberg inequalities established in their well-know paper []. See also a further generalization by C.S.Lin [6]. 7. Proof. For p <, there is a sequence u m W,p C () such that u m u in W,p. Then F (u m ) F (u) p sup F p u m u p, as m. F (u m )Du m F (u)du p sup F Du m Du p + F (u m ) F (u) p Du p. p to a subsequence, we know u m converges to u a.e. Since F is continuous, F (u m ) converge to F (u) a.e. Hence the last integral tends to by the dominate convergence theorem. Consequently, {F (u m )} and {F (u m )Du m } converge to F (u) and F (u)du in L p. By the uniqueness of weak derivative, D[F (u)] = F (u)du. On the other hand, F (u) L p since F (u) F u + F (). 6
Remark 5. Note that the hypothesis is bounded is only used to show F (u) L p. Such restriction can be removed if we know F () =. (Still need F L.) A trivial counterexample for unbounded and F () is F. 8. Proof. (a) and (c) follow from (b), (b) is proved by the method given in the hint easily. Note that can be unbounded. (See the remark for problem 7.) 9. Proof. Let φ be a smooth, bounded, nondecreasing function, such that φ is bounded and φ(z) = z, if z. Let u ɛ (x) := ɛφ(u/ɛ). Since u L, u is finite a.e. Therefore u ɛ as ɛ a.e.. Moreover by the mean-value theorem, for a.e. x u ɛ (x) = ɛ φ(u(x)/ɛ) φ(/ɛ) sup(φ ) u(x) L Hence u ɛ in L by dominated convergence theorem. By exercise 7, {Du ɛ } = {φ (u/ɛ)du} is bounded on L. Given w Cc (), (Du ɛ )w = u ɛ Dw, since u ɛ L Apply the density argument with the help of boundedness of { Du ɛ L }, we know the above is true for any w L. Take w = (Du)χ {u=} L, we see Du = φ (u/ɛ)du Du = Du ɛ (Du)χ {u=}, as ɛ {u=} {u=} which is the desired result. Remark 6. In the next two problems, we define the norm in fractional Sobolev spaces by u H s ( ) = ( + ξ ) s û(ξ) dξ, for s R. which is different from Evans definition. This definition works for negative power s.. (Sobolev Lemma) Proof. Since u H s L, Fourier inversion formula on L implies that u( x) = û(x) in L sense. However ( û(ξ) dξ = u H s ( ) ] ) [( ( + ξ ) s [ ] ) û(ξ) dξ ( + ξ ) s dξ ( ) Rn ( + ξ ) s dξ C u H s (R ), since s > n, n we know û L, and the inversion formula holds pointwisely now, this tells us that for each x u(x) e ix ξ û(ξ) dξ C u (π) n/ H s ( ) 7
Remark 7. It s true that for s = k + α + n, < α <, Hs is continuously embedded in C k,α. Remark 8. The converse of Sobolev lemma is true: If H s is continuously embedded in C k, then s > k + n. The proof is sketched in Folland [3, page 95].. (H s is an algebra if s > n/) We need the following convolution identity for Fourier transform on L ( ) : Lemma 9. ûv = (π) n/ û ˆv, u, v L. Proof. Since s, ( + ξ ) s/ ( + ξ η + η ) ) s/ ( + ξ η + η ) s/ s[ ( + ξ η ) s/ + ( + η ) s/]. Hence (( uv H = ( + ξ ) s ûv(ξ) dξ (π) n + ξ ) s/ ( u v )(ξ) ) dξ s ( [( C(n, s) + ξ η ) s/ + ( + η ) s/] û(ξ η) ˆv(η) dη ) dξ ( [( = C + ξ η ) s/ û(ξ η) ˆv(η) dη + ( + η ) s/] û(η) ˆv(ξ η) dη ) dξ C ( s û ˆv (ξ)) + ( û sˆv(ξ)) dξ, where x := ( + x ) /. C ( < > s û L ˆv L + û L < ) >s ˆv L, by Young s inequality. C u H s v H s, since ˆv C v H s Acknowledge The authors would like to thank Tommaso Seneci for his/her reminder that the original proof for problem 9 is wrong. (7.8.) References [] Luis Caffarelli, Robert Kohn, and Louis Nirenberg. First order interpolation inequalities with weights. Compositio Mathematica, 53(3):59 75, 984. 8
[] Bernard Dacorogna. Direct methods in the calculus of variations, volume 78. Springer Science & Business Media, 7. [3] Gerald B Folland. Introduction to Partial Differential Equations. Princeton university press, nd edition, 995. [4] Lester L Helms. Potential Theory. Springer Science & Business Media, nd edition, 4. [5] Giovanni Leoni. A first course in Sobolev spaces, volume 5. American Mathematical Society Providence, RI, 9. [6] Chang Shou Lin. Interpolation inequalities with weights. Communications in Partial Differential Equations, (4):55 538, 986. [7] Luc Tartar. An introduction to Sobolev spaces and interpolation spaces, volume 3. Springer Science & Business Media, 7. 9