Chapter 4: Hilbert Spaces

Hilbert spaces, Basic Projections Riesz Representation Theorem Application of Riesz representation theorem

I-Liang Chern

Department of Applied Mathematics National Chiao Tung University and Department of Mathematics National Taiwan University

Fall, 2013

1 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Outline

1 Hilbert spaces, Basic Inner product structure Sobolev spaces

2 Projections Projections in Banach spaces Orthogonal projections

3 Riesz Representation Theorem

4 Application of Riesz representation theorem Solving Poisson equations Error estimates for ﬁnite element method

2 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Inner product structure

Deﬁnition Let X be a complex linear space. An inner product (·, ·) is a bilinear form: X × X → C which satisﬁes (a) (x, x) ≥ 0 and (x, x) = 0 if and only if x = 0, (b) (x, y) = (y, x), (c) (x, αy + βz) = α(x, y) + β(x, z). The linear space X equipped with the inner product (·, ·) is called an inner product space.

3 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Examples.

The space Cn with X (x, y) := xiyi i is an inner product space. Let A be a symmetric positive deﬁnite matrix in Rn. Deﬁne

hx, yiA := (x, Ay) n Then h·, ·iA is an inner product in R . The space C[0, 1] with the inner product Z 1 (f, g) := f(t)g(t) dt 0 is an inner product space. 4 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem

The space L2(0, 1) is the completion of C[0, 1] with the above inner product. In fact, it is the space of all functions whose squares are Lebesgue integrable. Let T be the unit circle and Z 2 2 L (T) := {f : T → C | |f(t)| dt < ∞} T It is the space of all square summable and periodic functions. The space `2(N) is deﬁned to be ∞ 2 X 2 ` (N) := {x | x = (x1, x2, ··· ), |xi| < ∞} i=1 equipped with the inner product ∞ X (x, y) := xiyi. i=1 5 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem

Let wn > 0 be a positive sequence. Deﬁne ∞ 2 X 2 `w := {x| x : N → C, wi|xi| < ∞} i=1 with the inner product: ∞ X (x, y) := wixiyi i=1 Let w :(a, b) → R+ be a positive continuous function. Consider the space Z b 2 2 Lw(a, b) := {f :(a, b) → C | |f(x)| w(x) dx < ∞} a equipped with the inner product Z b (f, g) := f(x)g(x) w(x) dx. a 6 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Cauchy-Schwarz inequality

We can deﬁne kxk = p(x, x). Theorem Let X be an inner product space. For any x, y ∈ X, we have

|(x, y)| ≤ kxkkyk.

Proof:

1 From non-negativity of (·, ·), we get

2 2 2 0 ≤ (x+ty, x+ty) = kxk +2Re(x, y)t+kyk t for all t ∈ R. From this, we obtain one form of Cauchy-Schwarz: |Re(x, y)|2 ≤ kxk2kyk2.

7 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem

2 We claim that

|Re(x, y)| ≤ kxk kyk for any x, y ∈ X

if and only if

|(x, y)| ≤ kxk kyk for any x, y ∈ X.

3 Suppose (x, y) is not real, we choose a phase φ such that eiφ(x, y) is real. Now we replace x by eiφx. Then

|Re(eiφx, y)| ≤ kxk kyk

But the left-hand side is |(x, y)|. This proves one direction. The other direction is trivial.

8 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Cauchy-Schwarz and Triangle inequality

From kx + yk2 = kxk2 + 2Re(x, y) + kyk2 and

(kxk + kyk)2 = kxk2 + 2kxk kyk + kyk2,

by comparing the two equations, we get that the triangle inequality is equivalent to the Cauchy-Schwarz inequality. In fact, the following statements are equivalent: (a) For any x, y ∈ H, Re(x, y) ≤ kxk kyk; (b) For any x, y ∈ H, |(x, y)| ≤ kxk kyk; (c) For any x, y ∈ H, kx + yk ≤ kxk + kyk.

9 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Remark

If we are care about the cosine law, that is kx − yk2 = kxk2 + kyk2 − 2kxk kyk cos θ, then we should deﬁne the angle between x and y by Re(x, y) cos θ := . kxk kyk However, this creates a problem, the orthogonality in this sense may not have (x, y) = 0. This is not what we want. So, we deﬁne the acuate angle between two vectors x and y by |(x, y)| cos θ := , kxk kyk and we give up the traditional cosine law. 10 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem

Proposition (Parallelogram law) A normed linear space is an inner product space if and only if

kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2, for all x, y ∈ X.

Proof. Suppose a norm satisﬁes the parallelogram law, we deﬁne 1 (x, y) := kx + yk2 − kx − yk2 − ikx + iyk2 + ikx − iyk2 . 4 I leave you to check the parallelogram law implies the bilinearity of the inner product.

11 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Sobolev spaces

1 1 The H space

2 1 The H0 space

3 Poincar´einequality

4 Optimal constant in Poincar´einequality

12 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem The H1 space.

Similarly, we deﬁne

Z b 1 2 0 2 H (a, b) = {u :(a, b) → C| |u(x)| + |u (x)| dx < ∞} a with the inner product

Z b (u, v) = uv¯ + u0v0 dx. a Why is H1(a, b) complete?1 Indeed, it is the completion of C1[a, b], or C∞[a, b] under the above inner product.

1 2 0 2 0 If un → u in L (a, b) and un → v in L (a, b), then prove u = v. 13 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem 1 The H0 spaces

1 1 Deﬁne H0 (a, b) = {u ∈ H (a, b)|u(a) = u(b) = 0}. u(a) and u(b) are well-deﬁned for u ∈ H1(a, b). In fact, for any two points x1 and x2 near a, we can express

Z x2 0 |u(x2) − u(x1)| = | u (x) dx| x1

Z x2 1/2 Z x2 1/2 ≤ 12dx |u0(x)|2dx x1 x1 1/2 0 ≤ (x2 − x1) ku k → 0 as x1, x2 → a.

1 ∞ Alternatively, H0 (a, b) is the completion of C0 [a, b] ∞ under the above inner product. Here, C0 [a, b] are those C∞ function on [a, b] satisfying zero boundary condition.

14 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Poincar´einequality

1 In H0 (a, b), we can deﬁne another inner product

Z b hu, vi := u0(x)v0(x) dx. a We check that hu, ui = 0 implies u ≡ 0. R b 0 2 0 From a |u (x)| dx = 0, we get that u (x) ≡ 0 on (a, b). This together with u(a) = u(b) = 0 lead to u ≡ 0. 1 Now, in H0 , we have two norms:

2 2 0 2 2 0 2 kuk1 ≡ kuk + ku k , kuk2 ≡ ku k .

1 We claim that they are equivalent in H0 .

15 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem

We recall that two norms k · k1, k · k2 are equivalent in a normed space X if there exist two positive constants

C1,C2 such that for any u ∈ X, we have

C1kuk2 ≤ kuk1 ≤ C2kuk2.

Clearly we have

ku0k2 ≤ kuk2 + ku0k2.

Theorem (Poincar´einequality) 1 There exists a constant C > 0 such that for any u ∈ H0 (a, b), we have kuk2 ≤ Cku0k2. (1.1)

16 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Proof of Poincar´einequality

x x 1 R 0 R 0 u(x) = u(a) + a u (y) dy = a u (y) dy.

Z x 2 2 0 |u(x)| = u (y) dy a Z x Z x ≤ 12 dy |u0(y)|2 dy a a Z b ≤ (x − a) |u0(y)|2 dy a

2 We integrate x over (a, b) to get

Z b (b − a)2 Z b |u(x)|2 dx ≤ |u0(y)|2 dy. a 2 a

17 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Remarks

1 The Poincar´einequality is valid by just assuming u(a) = 0. b 2 R It is also valid by assuming a u(x) dx = 0. 3 Dimension analysis for the Poincar´einequality: Denote the dimensions by [x] = L and [u] = U. We have

[kuk] = (U 2L)1/2 = UL1/2, [ku0k] = (UL−1)L1/2 = UL−1/2

0 4 Thus in kuk ≤ Cku k, the dimension [C] = L.

18 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Best constant in Poincar´einequality

To ﬁnd the best constant C in the Poincar´einequality, we look for the following minimum R b u0(x)2 dx min a u(a)=u(b)=0 R b 2 a u(x) dx This problem is equivalent to Z b Z b min u0(x)2 dx subject to u(x)2 dx = 1. u(a)=u(b)=0 a a By the method of Lagrange multiplier, there exists a λ such that Z b Z b δ u0(x)2 dx − λ u(x)2 dx = 0. a a

19 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem

The corresponding Euler-Lagrange equation is

−u00 − λu = 0

with the two boundary condition u(a) = u(b) = 0. This is a standard eigenvalue problem. The minimal value of λ is the ﬁrst eigenvalue of −D2 with the Dirichlet boundary condition. The corresponding eigenvector and eigenvalue are x − a π 2 u(x) = sin π , λ = . b − a b − a Thus, the best constant is 1 b − a C = √ = . λ π

20 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem

Exercise: The weighted Sobolev space. Let w(x) > 0 on [a, b]. Deﬁne the inner product

Z b 0 0 hu, viw := u (x)v (x)w(x) dx a

0 2 and the corresponding norm ku kw := hu, uiw. Let

1 0 Hw,0(a, b) := {u :(a, b) → C | ku kw < ∞, u(a) = u(b) = 0}

1 1 0 Then the space H0,w(a, b) = H0 and the norm ku kw is equivalent to ku0k.

21 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections Outline

1 Hilbert spaces, Basic Inner product structure Sobolev spaces

2 Projections Projections in Banach spaces Orthogonal projections

3 Riesz Representation Theorem

4 Application of Riesz representation theorem Solving Poisson equations Error estimates for ﬁnite element method

22 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections Projections in Banach spaces

Deﬁnition (a) A projection P in a Banach space X is a linear mapping from X to X satisfying P 2 = P . (b) The direct sum of two subspaces M and N in a Banach space X is deﬁned to be

M ⊕ N := {x + y | x ∈ M, y ∈ N }.

23 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections

Theorem If P is a projection on a linear space X, then X = Ran P ⊕ Ker P , and RanP ∩ KerP = {0}. Conversely, if X = M ⊕ N and M ∩ N = {0}, then any x ∈ X can be uniquely represented as x = y + z with y ∈ M and z ∈ N . Furthermore, the mapping P : x 7→ y is a projection.

Here, RanP : range of P , KerP : kernel of P .

24 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections Proof

(⇒)

1 We ﬁrst show that x ∈ Ran P ⇔ x = P x.(⇐) If x = P x clearly x ∈ RanP .(⇒)If x ∈ RanP , then x = P y for some y ∈ H. From P 2 = P , we get P x = P 2y = P y = x. 2 Next, if x ∈ RanP ∩ KerP , then x = P x = 0. Hence, RanP ∩ KerP = {0}. 3 Finally, we can decompose x ∈ X into x = P x + (x − P x). The part P x ∈ RanP . The other part x − P x ∈ Ker P because P (x − P x) = P x − P 2x = 0. 25 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections

(⇐)

1 If x = y1 + z1 = y2 + z2 with yi ∈ M and zi ∈ N , then

y1 − y2 = z2 − z1 and it is in M ∩ N . Thus, y1 = y2 and

z1 = z2.

2 For y ∈ M, P y = y. For any x, P x ∈ M, hence P (P x) = P x. Remarks.

1 If P is a projection, so is I − P .

2 We have RanP = Ker(I − P ), KerP = Ran(I − P ).

3 A projection in a Banach space needs not be continuous in general.

26 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections

Theorem Let X be a Banach space and P is a projection in X. (a) If P is continuous, then both KerP and RanP are closed. (b) On the other hand, if Y is closed subspace and there exists a closed subspace Z such that X = Y ⊕ Z. Then the projection P : x 7→ y is continuous, where x = y + z is the decomposition of x with y ∈ Y and z ∈ Z.

27 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections Proof

1 We show the graph of P is closed. That is, if xn → x

and yn := P xn → y, then y ∈ Y and x − y ∈ Z (thus,

P x = y). From the decomposition, we have yn ∈ Y and

zn := xn − yn ∈ Z. From the closeness of Y , we get

y ∈ Y . From xn → x and yn → y, we get xn − yn converges to x − y. From the closeness of Z, we get x − y ∈ Z. Thus, x = y + (x − y) with y ∈ Y and x − y ∈ Z.

2 The theorem follows from the closed graph theorem: A closed graph linear map A from Banach space X to Banach space Y is also continuous.

28 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections Orthogonal projections in Hilbert spaces

Theorem (Orthogonal Projection Theorem) Let H be a Hilbert space and let M ⊂ H be a closed linear subspace of H. Then (a) for any x ∈ H, there exists a unique y ∈ M such that

kx − yk = min kx − zk; z∈M

(b) (x − y) ⊥ M; (c) the mapping P : x 7→ y is a projection.

29 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections Proof

Existence: 2 1 Let ` = infz∈M kx − zk . and let {yn} be a minimal 2 sequence of kx − ·k in M. That is yn ∈ M and 2 limn→∞ kyn − xk = `.

2 Then from the parallelogram law 1 y + y ky −y k2 = ky −xk2 +ky −xk2 −2k m n −xk2. 2 m n m n 2 The ﬁrst two terms tend to 2` as n, m → ∞, while the last term is greater than 2` by the deﬁnition of `. This

implies {yn} is a Cauchy sequence in M hence it has a limit y in M.

30 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections

y yn ym z M yynm 2

Figure : Orthogonal projection of x onto a closed subspace M.

{yn} are minimal sequence. 31 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections

Uniqueness: 2 Suppose y1 and y2 are two minima, that is kyi − xk = `. By the parallelogram law, 1 y + y ky −y k2 = ky −xk2+ky −xk2−2k 1 2 −xk2 ≤ 2`−2` = 0. 2 1 2 1 2 2 Orthogonality: (x − y) ⊥ M. From kx−yk2 ≤ kx−y−tzk2 = kx−yk2−2Re(x−y, tz)+|t|2kzk2

for all t ∈ C and z ∈ M, we choose t = eiφ so that Re(x − y, tz) = |t||(x − y, z)| Then we get |(x − y, z)| ≤ 2|z|2. Taking → 0+, we get (x − y, z) = 0. 32 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections Orthogonal Projection

Let H be a Hilbert space. M ⊂ H be a subset. Deﬁne the orthogonal complement of M by

M⊥ := {x ∈ H|x ⊥ y for all y ∈ M}.

The orthogonal complement of a subset M in H is a closed linear subspace.

33 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections

Corollary If M is a closed subspace of a Hilbert space H, then H = M ⊕ M⊥ and M⊥⊥ = M.

Proof. We only prove M⊥⊥ ⊂ M. Suppose x ∈ M⊥⊥. That is, (x, w) = 0 for all w ∈ M⊥. By the orthogonal projection theorem, we can decompose x = y + z with y ∈ M and z ∈ M ⊥. Then 0 = (x, w) = (y + z, w) = (z, w) for all w ∈ M⊥. Since z ∈ M⊥, we can take w = z and get (z, z) = 0. Hence, x = y ∈ M. This proves M⊥⊥ ⊂ M.

34 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections

Theorem Let P : H → H be a projection. The following two statements are equivalent:

(a) (P x1, x2) = (x1, P x2) for all x1, x2 ∈ H; (b) H = Ran P ⊕ Ker P and Ran P ⊥ Ker P .

Deﬁnition A mapping P on H is called an orthogonal projection if it 2 satisﬁes (i) P = P , (ii) (P x1, x2) = (x1, P x2).

35 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections Proof

1 (a) ⇒ (b): For any x ∈ RanP , then x = P y for some y ∈ H. Then for any z ∈ Ker P ,

(x, z) = (P y, z) = (y, P z) = 0.

Hence, RanP ⊥ Ker P .

2 (b) ⇒ (a): For any x1, x2 ∈ H, they can be uniquely decomposed into

⊥ x1 = y1 + z1, x2 = y2 + z2, with yi ∈ M, ziM .

Thus,

(P x1, x2) = (y1, y2) = (x1, P x2).

36 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections Examples

Example 1. Given a vector y ∈ H. Deﬁne y ⊥ P : x 7→ (y, x) kyk2 . Then RanP = h{y}i and KerP = y . Example 2.

1 Given n independent vectors {v1, ··· , vn} in H. Let

M = h{v1, ··· , vn}i. Given any x ∈ H, the orthogonal projection y of x on M satisﬁes: 1 y = arg min { kx − zk2 | z ∈ M}. 2

2 The Euler-Lagrange equation is (x − y) ⊥ M.

37 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremProjections Application in of Banach Riesz representationspaces Orthogonal theorem projections

Pn 3 Since y ∈ M, we can express y as y = i=1 αivi. The condition (x − y) ⊥ M is equivalent to (x − y, vi) = 0, i = 1, ··· , n. This leads to the following n × n system of linear equations

n X (vi, vj)αj = (x, vi), i = 1, ··· , n. j=1

From the independence of {v1, ··· , vn}, we can get a unique solution of this equation.

38 / 60 Hilbert spaces, Basic Projections Riesz Representation Theorem Application of Riesz representation theorem Outline

1 Hilbert spaces, Basic Inner product structure Sobolev spaces

2 Projections Projections in Banach spaces Orthogonal projections

3 Riesz Representation Theorem

4 Application of Riesz representation theorem Solving Poisson equations Error estimates for ﬁnite element method

39 / 60 Hilbert spaces, Basic Projections Riesz Representation Theorem Application of Riesz representation theorem Dual space

Let H be a normed linear space. The set

∗ H := {` : H → C bounded linear functional} forms a linear space called the dual space of H. It is a Banach space equipped with the operator norm.

Given a y ∈ H, the mapping `y(x) := (y, x) is a bounded linear functional, by Cauchy-Schwarz inequality. Its norm

k`yk ≤ kyk. On the other hand, by choosing x = y/kyk,

we obtain k`yk ≥ |`y(y/kyk)| = kyk. Thus k`yk = kyk. Riesz representation theorem: every bounded linear functional on H must be in this form. In other word, H∗ is isometric to H. 40 / 60 Hilbert spaces, Basic Projections Riesz Representation Theorem Application of Riesz representation theorem Riesz representation theorem

Theorem (Riesz representation theorem) Let ` be a bounded linear functional on a Hilbert space H. Then there exists a unique y ∈ H such that `(x) = (y, x).

Proof.

1 We suppose ` 6= 0. Our goal is to ﬁnd y such that `(x) = (y, x). We ﬁrst notice that such y must be in (Ker `)⊥ and P : x 7→ (y, x)y/kyk2 is an orthogonal projection.

2 Let N = Ker `. Then N is closed and N 6= H. Hence

there exists a z1 6∈ N .

41 / 60 Hilbert spaces, Basic Projections Riesz Representation Theorem Application of Riesz representation theorem

3 By the orthogonal projection theorem, there exists a

y1 ∈ N and z := (z1 − y1) ⊥ N . From z1 6∈ N , we get z 6= 0. 4 Let `(x) P x := z. `(z) Then P is an orthogonal projection, i.e. P 2 = P and H = RanP ⊕ Ker P , and RanP ⊥ Ker P . Since RanP = {αz|α ∈ C} and Ker P = Ker ` = N . We thus have H = {αz|α ∈ C} ⊕ Ker `. 5 Thus, any x ∈ H can be represented uniquely by

x = αz + m, m ∈ N , α = (z, x)/kzk2.

42 / 60 Hilbert spaces, Basic Projections Riesz Representation Theorem Application of Riesz representation theorem

6 We have 1 `(x) = `(αz) = (z, x)`(z) = (y, x). kzk2

`(z) where, y := kzk2 z. We have shown the existence of y such that `(x) = (y, x).

7 For the uniqueness, suppose there are y1 and y2 such that

`y1 = `y2 . That is,

(y1, x) = (y2, x), for all x ∈ H.

Choose x = y1 − y2, we obtain ky1 − y2k = 0.

43 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method Outline

1 Hilbert spaces, Basic Inner product structure Sobolev spaces

2 Projections Projections in Banach spaces Orthogonal projections

3 Riesz Representation Theorem

4 Application of Riesz representation theorem Solving Poisson equations Error estimates for ﬁnite element method

44 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method Applications: Solving Poisson equation

We consider the Poisson equation on a bounded domain Ω ⊂ Rn: (P ): 4u = f in Ω, u = 0 on ∂Ω. This problem can be reformulated as the following weak form:

1 (WP ): Find u ∈ H0 (Ω) such that 1 (∇u, ∇v) = −(f, v), for all v ∈ C0 .

Theorem Let Ω ⊂ Rn be a smooth bounded domain. Let f ∈ L2(Ω). 1 Then (WP) has a unique solution in H0 (Ω).

45 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method

To show the existence for the Poisson equation, we need the following lemma. It was proven before. Lemma (Poincar´einequality) Let Ω ⊂ Rn be a smooth bounded domain. Then there exists 1 a constant C such that for u ∈ H0 (Ω), we have

kuk2 ≤ Ck∇uk2.

46 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method Proof of existence of solution of Poisson equation

1 From the Poincar´e’sinequality, we see that Z hu, vi1 := (∇u, ∇v) := ∇u(x) · ∇v(x) dx Ω 1 deﬁnes an inner product in H0 (Ω). ( 2 On the other hand, for f ∈ L Ω), `v := (f, v) is a 2 1 bounded linear map in both L and H0 : |(f, v)| ≤ kfkkvk ≤ Ckfk k∇vk

3 Thus, by the Riesz representation theorem, there exists a 1 unique u ∈ H0 (Ω) such that

hu, vi1 = (∇u, ∇v) = (−f, v)

1 for all v ∈ H0 (Ω). 47 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method Application: analysis for ﬁnite element method

We consider the Poisson equation in one dimension: −u00 = f on (a, b), u(a) = u(b) = 0. (4.2) We shall ﬁnd an approximate solution by ﬁnite element method. 1 We chosse an n > 0. Let h := (b − a)/n the mesh size,

xi = a + ih, i = 0, ··· , n the grid point.

2 Deﬁne the nodal function φi(x) to be φi(xj) = δij and φ(x) is continuous and piecewise linear. 3 Let

Vh = hφ1, ··· , φn−1i called the ﬁnite element space. 48 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method

4 An element v ∈ Vh is a continuous and piecewise linear function and is uniquely expressed by

n−1 X v(x) = v(xi)φi(x). i=1

5 The approximate solution uh ∈ Vh is expressed as

n−1 X uh(x) = Uiφi(x) i=1

6 We project the equation (4.2) onto Vh:

00 (−u − f, v) = 0, for all v ∈ Vh.

49 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method

This leads to the following equations for T U = (U1, ··· ,Un−1) :

huh, φii1 = (f, φi), i = 1, ··· , n − 1.

Or n−1 X 0 0 (φi, φj)Uj = (f, φi), i = 1, ··· , n − 1. j=1

We can compute (φi, φj) directly and obtain the matrix 0 0 A = (φi, φj)(n−1)×(n−1) as 1 A = diag(−1, 2, −1) h This matrix is invertible.

50 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method

Error of the approximate solution uh

Let u be the exact solution and eh := u − uh be the true error.

Since both u and uh satisfy

0 0 0 0 (u , v ) = (f, v), (uh, v ) = (f, v) for all v ∈ Vh,

we obtain 0 0 (eh, v ) = 0 for all v ∈ Vh.

That is, (u − uh) ⊥1 Vh. This is equivalent to say that

uh is the h·, ·i1-orthogonal projection of u on Vh.

51 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method

Thus,

0 0 0 0 ku − uhk2 ≤ ku − v k2 for all v ∈ Vh.

In particular, we can choose

n−1 X v = πhu := u(xi)φi, i=1 then 0 0 0 0 ku − uhk2 ≤ ku − (πhu) k2. (4.3) Thus, the true error is controlled by the approximation error.

52 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method Approximation error

The approximation error w(x) = u(x) − πhu(x) satisﬁes

w(xi) = w(xi+1) = 0. 0 00 We want to estimate kwk2 and kw k2 in terms of ku k∞ 00 and ku k2.

53 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method Approximation errors in terms of ku00k : √ ∞ h2 00 ku − πhuk2 ≤ b − a 8 ku k∞

From w(xi) = w(xi+1) = 0, we get for x ∈ (xi, xi+1), ∃ξi w00(ξ ) w(x) = i (x − x )(x − x ). 2 i i+1 Hence n−1 Z b X Z xi+1 |w(x)|2 dx = |w(x)|2 dx a i=0 xi h2 2 ≤ (b − a) max |w00(x)|2. 8 x∈[a,b] 00 00 From w (x) = u (x) on (xi, xi+1), we get √ 2 h 00 ku − πhuk2 ≤ b − a ku k∞ 8 54 / 60 Hilbert spaces, Basic Projections Riesz Representation√ TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method 0 0 00 ku − (πhu) k2 ≤ b − ahku k∞

First, there exists a ζ1 ∈ (xi, xi+1) such that 0 u (ζ1) = (u(xi+1 − u(xi))/h.

For any x ∈ (xi, xi+1), there exists ζ2 ∈ (xi, xi+1) such 0 0 00 that u (x) − u (ζ1) = u (ζ2)(x − ζ1).

Therefore, we get for x ∈ (xi, xi+1) u(x ) − u(x ) u0(x)−(π u)0(x) = u0(x)− i+1 i = u00(ζ )(x−ζ ). h h 2 1

n−1 Z b Z xi+1 0 0 2 X 0 0 2 |u − (πhu) | dx = |u − (πhu) | dx a i=0 xi n−1 X 2 00 2 2 00 2 ≤ h h max |u (x)| = (b − a)h ku k∞ x∈[a,b] i=0 55 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method 00 Approximation error in terms of ku k2

00 It is desirable to estimate ku − πhuk2 in terms of ku k2. We should use the integral representation for the interpolation error w.

Recall that for w(xi) = w(xi+1) = 0, w has the representation: Z xi+1 x − x y − x w(x) = h2 g i , i w00(y) dy xi h h Z xi+1 0 x − xi y − xi 00 w (x) = h gx , w (y) dy xi h h 2 2 where g is the Green’s function of d /dx on (xi, xi+1). 0 00 Thus, we can estimate kwk2 and kw k2 in terms of kw k2

on (xi, xi+1). 56 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method

For x ∈ (xi, xi+1),

Z xi+1 x − x y − x Z xi+1 |w(x)|2 ≤ h4 |g i , i |2 dy |w00(y)|2 dy . xi h h xi

Z xi+1 Z xi+1Z xi+1 x − x y − x Z xi+1 |w(x)|2 dx ≤ h4 |g i , i |2 dy dx |w00(y)|2 dy xi xi xi h h xi 1 Z xi+1 ≤ h4 |w00(y)|2 dy. 90 xi As we sum over i = 1, ··· , n − 1, we get

1 2 00 kwk2 ≤ √ h kw k2. 90 Similarly, we get

0 1 00 kw k2 ≤ √ hkw k2. 6

57 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method Approximation error in Sobolev space

Theorem 2 1 For u ∈ H (a, b) ∩ H0 [a, b], the interpolation error has the following estimates

1 2 00 ku − πhuk2 ≤ √ h ku k2, 90

0 0 1 00 ku − (πhu) k2 ≤ √ hku k2. 6

58 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method True error of the ﬁnite element method

Theorem For the ﬁnite element method for problem (4.2), the true error

u − uh has the following estimate

0 0 1 00 ku − u k2 ≤ √ hku k2, h 6 1 ku − u k ≤ h2ku00k . h 2 6 2

59 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremSolving Application Poisson of equations Riesz representation Error estimates theorem for ﬁnite element method Proof by duality argument

1 Let eh = u − uh. We ﬁnd the function φh such that 00 φh = −eh and φ(a) = φ(b) = 0. 2 Then

00 0 0 0 0 0 (eh, eh) = −(eh, φh) = (eh, φh) = (eh,φ h − (πhφh) ). 0 0 Here, I have used (eh, v ) = 0 for all v ∈ Ran(πh).

3 Applying interpolation estimate to φh, we get

2 0 0 h 0 00 h 0 kehk ≤ ke kk(φh−πhφh) k ≤ √ ke k kφ k = √ ke k kehk h 6 h h 6 h Hence, we get

1 0 1 2 00 kehk2 ≤ √ hke k2 ≤ h ku k2. 6 h 6

60 / 60