Lecture 4 The Sobolev space H1, and applications In Section 4.1 we present the definition and some basic properties of the Sobolev space H1. This treatment is prepared by several important tools from analysis. The main objective of this lecture is the Hilbert space treatment of the Laplace operator in Section 4.2. In particular, the Dirichlet Laplacian will be presented as our first (non-trivial) example of a generator of a contractive holomorphic C0-semigroup. 4.1 The Sobolev space H1 4.1.1 Convolution We recall the definition of locally integrable functions on an open subset Ω of Rn, L1,loc(Ω) := f : Ω K; for all x Ω there exists r > 0 such that → ∈ B(x, r) Ω and f L (B(x, r)) . ⊆ B(x,r) ∈ 1 Moreover, Cc∞(Ω) := C∞(Ω) Cc(Ω) is the space of infinitely differentiable functions with compact support. ∩ n n 4.1 Lemma. Let u L1,loc(R ), ρ Cc∞(R ). We define the convolution of ρ and u, ∈ ∈ ρ u(x) := ρ(x y)u(y) dy = ρ(y)u(x y) dy (x Rn). ∗ n − n − ∈ ZR ZR n n Then ρ u C∞(R ), and for all α N0 one has ∗ ∈ ∈ ∂α(ρ u) = (∂αρ) u. ∗ ∗ Proof. (i) The integral exists because ρ is bounded and has compact support. (ii) Continuity of ρ u: There exists R > 0 such that spt ρ B(0,R). Let R0 > 0, ∗ ⊆ δ > 0. For x, x0 B(0,R0), x x0 < δ, one obtains ∈ | − | ρ u(x) ρ u(x0) = ρ(x y) ρ(x0 y) u(y) dy | ∗ − ∗ | − − − ZB(0,R+R0) sup ρ(z) ρ(z0) ; z z0 < δ u(y) dy. 6 | − | | − | | | ZB(0,R+R0) The second factor in the last expression is finite because u is locally integrable, and the first factor becomes small for small δ because ρ is uniformly continuous. 38 (iii) Let 1 6 j 6 n. The existence of the partial derivative of ρ u with respect to the j-th variable and the equality ∂ (ρ u) = (∂ ρ) u are a consequence∗ of the differentiability j ∗ j ∗ of integrals with respect to a parameter. The function (∂jρ) u is continuous, by step (ii) above, and therefore ρ u is continuously differentiable with∗ respect to the j-th variable. ∗ n (iv) Induction shows the assertion for all α N0 . ∈ n A sequence (ρk)k N in Cc(R ) is called a δ-sequence if ρk 0, ρk(x) dx = 1 and ∈ > spt ρk B[0, 1/k] for all k N. (The notation ‘δ-sequence’ is motivated by the fact that ⊆ ∈ n R the sequence approximates the ‘δ-distribution’ Cc∞(R ) ϕ ϕ(0).) 3 7→ 4.2 Remarks. (a) We recall the standard example of a Cc∞-function ϕ. The source of this function is the well-known function ψ C∞(R), ∈ 0 if t 0, : 6 ψ(t) = 1/t (e− if t > 0. 2 n n Then ϕ(x) := ψ(1 x )(x R ) defines a function 0 6 ϕ Cc∞(R ), with the property that ϕ(x) = 0 if and− | only| if ∈x < 1. ∈ 6 n | | (b) If 0 ρ Cc(R ), ρ(x) dx = 1, spt ρ B[0, 1], and we define 6 ∈ ⊆ R n n ρk(x) := k ρ(kx)(x R , k N), ∈ ∈ then (ρk) is a δ-sequence. n 4.3 Proposition. Let (ρk) be a δ-sequence in Cc(R ). n n (a) Let f C(R ). Then ρk f f uniformly on compact subsets of R as k . ∈ ∗n → n → ∞ (b) Let 1 p , f Lp(R ). Then ρk f Lp(R ), 6 6 ∞ ∈ ∗ ∈ ρk f p f p for all k N. k ∗ k 6 k k ∈ If 1 p < , then 6 ∞ ρ f f 0 (k ). k k ∗ − kp → → ∞ Proof. (a) The assertion is an easy consequence of the uniform continuity of f on compact subsets of Rn. (b) (i) For p = the estimate is straightforward. ∞ 1 1 If 1 < p < and p + q = 1, then we estimate, using H¨older’sinequality in the second step, ∞ 1 + 1 ρ f(x) = ρ (x y) q p f(y) dy | k ∗ | k − Z 1/q 1/p p ρk(x y) dy ρk(x y) f(y) dy 6 − − | | Z Z 1/p = ρ (x y) f(y) p dy . k − | | Z 39 This estimate also holds (trivially) for p = 1. Then, with Fubini’s theorem in the second step, ρ f(x) p dx ρ (x y) f(y) p dy dx | k ∗ | 6 k − | | Z Zx Zy = ρ (x y) dx f(y) p dy = f p. k − | | k kp Zy Zx n (ii) Now let 1 6 p < . For k N we define Tk (Lp(R )) by Tkg := ρk g for n ∞ ∈ ∈ L ∗ g Lp(R ); then step (i) shows that Tk 6 1. ∈ n k nk If g Cc(R ), then Tkg g in Lp(R ) as k . Indeed, there exists R > 0 such that spt∈g B[0,R]. It is easy→ to check that this→ implies ∞ that spt(ρ g) B[0,R + 1] ⊆ k ∗ ⊆ for all k N. Also, ρk g g uniformly on B[0,R + 1], by part (a). This shows that ∈ ∗ →n ρk g g (k ) in Lp(R ). ∗ → → ∞ n n Now, the denseness of Cc(R ) in Lp(R ) (which we will accept as a fundamental fact from measure and integration theory) together with Proposition 1.6 implies that Tkg g n n → (k ) in Lp(R ), for all g Lp(R ). → ∞ ∈ n 4.4 Corollary. Let Ω R be open, 1 p < . Then Cc∞(Ω) is dense in Lp(Ω). ⊆ 6 ∞ n Proof. Let (ρk) be a δ-sequence in Cc∞(R ). n n Let g Cc(Ω), and extend g by zero to a function in Cc(R ). Then ρk g C∞(R ) for ∈ 1 n ∗ ∈ all k N, by Lemma 4.1. If < dist(spt g, R Ω), then spt(ρk g) spt g+B[0, 1/k] Ω ∈ k \ ∗ ⊆ ⊆ (see Exercise 4.1(a)), and therefore ρk g Cc∞(Ω). From Lemma 4.3 we know that n ∗ ∈ ρk g g (k ) in Lp(R ). So, we have shown that Cc∞(Ω) is dense in Cc(Ω) with ∗ → → ∞ respect to the Lp-norm. Now the denseness of Cc(Ω) in Lp(Ω) yields the assertion. 4.1.2 Distributional derivatives Let P (∂) = a ∂α be a partial differential operator with constant coefficients α N0, α 6k α ∈ | | n k aα K ( α k), with k N. Let Ω R be open, f C (Ω). Then for all “test ∈ | |P6 ∈ ⊆ ∈ functions” ϕ C∞(Ω) one has ∈ c α α P (∂)f ϕ dx = f ( 1)| |aα∂ ϕ dx Ω Ω − Z Z α 6k |X| (integration by parts!). Now, let f L1,loc(Ω). Then we say that P (∂)f L1,loc(Ω) if there exists g L1,loc(Ω) such that ∈ ∈ ∈ α α gϕ dx = f ( 1)| |a ∂ ϕ dx − α α k Z Z |X|6 for all ϕ Cc∞(Ω), and we say that P (∂)f = g holds in the distributional sense. In ∈ α α particular, if ∂ f L1,loc(Ω), then we call ∂ f the distributional (or generalised or weak) derivative∈ of f. In order to justify this definition we have to show that g = P (∂)f is unique. This is the content of the following “fundamental lemma of the calculus of variations”. 40 n 4.5 Lemma. Let Ω R be open, f L1,loc(Ω), ⊆ ∈ fϕ dx = 0 (ϕ C∞(Ω)). ∈ c Z Then f = 0. The statement ‘f = 0’ means that f is the zero element of L1,loc(Ω), i.e., if f is a representative, then f = 0 a.e. Proof of Lemma 4.5. Let ϕ C∞(Ω). We show that ϕf = 0 a.e. Defining ∈ c ϕ(x)f(x) if x Ω, g(x) := ∈ 0 if x Rn Ω, ( ∈ \ n we have g L1(R ). ∈ n Let (ρk) be a δ-sequence in Cc∞(R ). From Lemma 4.3 we know hat ρk g g in n n ∗ → L1(R ). For x R , k N we have ∈ ∈ ρ g(x) = ρ (x y)ϕ(y)f(y) dy = 0, k ∗ k − ZΩ because ρ (x )ϕ C∞(Ω). This shows that ρ g = 0, and hence g = 0 (as an k − · ∈ c k ∗ L1-function). From ϕf = 0 a.e. for all ϕ C∞(Ω) we conclude that f = 0 a.e.; see Exercise 4.5. ∈ c In the remainder of this subsection we give more information on the one-dimensional case. The aim is to show that an L1,loc-function with L1,loc-derivative can be written as the integral of its derivative. 4.6 Proposition. Let a < x < b , f, g L (a, b). Then f 0 = g in the −∞ 6 0 6 ∞ ∈ 1,loc distributional sense if and only if there exists c K such that x ∈ f(x) = c + g(y) dy (a.e. x (a, b)). ∈ Zx0 Note that the right hand side of the previous equality is continuous as a function of x, and that therefore f has a continuous representative. For the proof we need a preparation. 4.7 Lemma. Let a < b , h L (a, b), and assume that h0 = 0 in the −∞ 6 6 ∞ ∈ 1,loc distributional sense. Then there exists c K such that h = c. ∈ Proof. (i) We start with the observation that a function ψ C∞(a, b) is the derivative of ∈ c a function in Cc∞(a, b) if and only if ψ dx = 0.