Useful Inequalities and Existence-Uniqueness Theorems for Finite
Element Analysis
Spencer Patty
January 26, 2015
Contents
1 Notations
1
2 Statement of Main Inequalities and Theorems without Proof
2
3 Inequalities with Proofs
3.1 Cauchy-Schwarz Inequality
3.2 Young’s Inequality . . . . .
3.3 Hölder’s Inequality . . . . .
3.4 Poincaré Inequality . . . . .
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4 Existence-Uniqueness Theorems with Proofs
4.1 A Riesz Representation Theorem for Separable Hilbert
4.2 Symmetric Lax-Milgram Theorem . . . . . . . . . . .
4.3 Contraction Mapping Principle on a Banach Space . .
4.4 Non-Symmetric Lax-Milgram Theorem . . . . . . . . .
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Notations
We will use V to refer to an inner product space (V, (·, ·))V with inner product (·, ·)V . (Note that in these
notes I will consistently use the parenthesis notation for inner product). The norm given by the inner product
1/2
is then kukV = (u, u)V for all u ∈ V . We will drop the subscript V notation when it is clear which space
we are talking about: kuk and (u, v) for u, v ∈ V .
Given a Banach Space (or Inner Product Space), V , we denote by V 0 the dual space to V consisting of
bounded linear functionals on V . The action of a dual element, f ∈ V 0 , on an element u ∈ V will be denoted
by f (u) or by angled brackets < f, u >.
When we are talking about function spaces for instance given p ≥ 1,
Z
p
L (Ω, dµ) := {f : Ω → C(orR) | f is µ-measurable and kf k
Lp (Ω,dµ)
:=
1/p
|f (x)| dµ
< ∞}
p
(1)
Ω
we will simplify notations when it is clear what space we are talking about. For instance we could say
Lp = Lp (Ω) = Lp (Ω, dµ) and kf kLp (Ω,dµ) = kf kLp (Ω) = kf kLp = kf kp .
It turns out that almost all of the spaces we will be working with are separable Banach spaces. The only
non-separable examples worth mentioning are L∞ (Ω), the set of bounded functions on a set Ω, and BV (Ω)
the set of functions of bounded variation on a set Ω. And even these two spaces will hardly be used in our
class. All the continuous, and Lp and W k,p spaces are separable when considered on Ω ⊂ Rn . And only
L2 (Ω) and the sets H k (Ω) = W k,2 (Ω) = {f ∈ L2 (Ω) | |Dj f | ∈ L2 (Ω) for 1 ≤ j ≤ k} for k ≥ 0 are separable
1
Hilbert spaces. We often denote L2 (Ω) = H ( 0)(Ω). Most of the time in this class, our pde’s will naturally
support weak solutions in L2 or H k for some k.
2
Statement of Main Inequalities and Theorems without Proof
Cauchy-Schwarz
Theorem 2.1. Given u, v ∈ V an inner product space with inner product (·, ·), then
|(u, v)| ≤ kukkvk.
(2)
Young’s Inequality
Theorem 2.2. Let a, b be non negative real numbers and p, q ∈ (0, ∞] such that
ab ≤
ap
bq
+
p
q
1
p
+
1
q
= 1. Then
(3)
For p = q = 2, we have
a2
b2
+
2
2
and for an ε > 0 we have the so called ”Peter-Paul Inequality”
ab ≤
ab ≤
a2
εb2
+
.
2ε
2
(4)
(5)
Hölder’s Inequality
Theorem 2.3. Let p, q ∈ [1, ∞] with
1
p
+
1
q
= 1. Then for f, g : Ω → C(R)
kf ḡkL1 (Ω) ≤ kf kLp (Ω) kgkLq (Ω)
ie
Z
Z
|f ḡ|dµ ≤
Ω
1/p Z
1/q
q
|f | dµ
|g| dµ
p
Ω
(6)
(7)
Ω
Poincaré Inequality
Theorem 2.4. Assume 1 ≤ p ≤ ∞, Ω is a bounded open subset of Rn with Lipschitz boundary. Then
there exists a finite constant c = c(Ω, p) depending only on the space Ω and on p such that for all
u ∈ W 1,p (Ω),
ku − uΩ kLp (Ω) ≤ ck∇ukLp (Ω)
(8)
where uΩ is the average value of u on Ω
uΩ :=
1
|Ω|
Z
u(y)dy
(9)
Ω
Lax-Milgram Theorem
Theorem 2.5. Given a Hilbert space (H, (·, ·)) and a continuous(bounded) bilinear form a(·, ·) which
is coercive on V ⊂ H, a closed subset of H, and a continous(bounded) linear functional F ∈ V 0 , then
there exists a unique u ∈ V such that for all v ∈ V ,
a(u, v) = F (v)
2
(10)
3
Inequalities with Proofs
3.1
Cauchy-Schwarz Inequality
Theorem 3.1. Given u, v ∈ V an inner product space with inner product (·, ·), then
|(u, v)| ≤ kukkvk.
(11)
Proof. If u = 0 or v = 0 or (u, v) = 0 then the statement is true, so suppose otherwise. Let
(u, v)
v.
(v, v)
(12)
(u, v)
(u, v)
u−
v, v = (u, v) −
(v, v) = 0.
(v, v)
(v, v)
(13)
z =u−
Then z⊥v since
(z, v) =
Rearranging, we have u =
(u,v)
(v,v) v
+ z which are orthogonal directions, so apply Pythagorean Theorem to get
2
2
(u, v) 2
kvk2 + kzk2 = |(u, v)| + kzk2 ≥ |(u, v)| .
kuk = (v, v) kvk2
kvk2
(14)
|(u, v)|2 ≤ kuk2 kvk2
(15)
2
Hence,
and so our result holds by taking the square root of both sides.
3.2
Young’s Inequality
Theorem 3.2. Let a, b be non negative real numbers and p, q ∈ (0, ∞] such that
ab ≤
1
p
+
1
q
= 1. Then
bq
ap
+
p
q
(16)
For p = q = 2, we have
a2
b2
+
(17)
2
2
and for an ε > 0 we have the so called ”Peter-Paul Inequality” (You have to rob Peter to pay Paul ie. for
small ε you can obtain a tighter control on the second part at expense of less control on the first)
ab ≤
ab ≤
a2
εb2
+
.
2ε
2
(18)
Proof. (General Young’s Inequality) We use the concavity of log(x), ie for θ ∈ [0, 1], x, y ∈ R
log ((1 − θ)x + θy) ≥ (1 − θ) log(x) + θ log(y).
Let θ = p1 , then 1 − θ = 1 −
1
p
= 1q and
1 p 1 q
log
a + b = log (θap + (1 − θ)bq )
p
q
≥ θ log (ap ) + (1 − θ) log (bq )
1
1
= log (ap ) + log (bq )
p
q
= log(a) + log(b)
= log(ab)
The result follows from exponentiating both sides.
3
(19)
(p=q=2)
0 ≤ (a − b)2 = a2 + b2 − 2ab
so
ab ≤
b2
a2
+ .
2
2
(20)
(21)
(Peter-Paul Inequality) Using the above result,
ab =
3.3
a
√
ε
√ 1
εb ≤
2
a
√
ε
2
+
a2
εb2
1 √ 2
+
.
εb =
2
2ε
2
(22)
Hölder’s Inequality
Theorem 3.3. Let p, q ∈ [1, ∞] with p1 + 1q = 1. Then for f, g functions on a measure space (Ω, Σ, µ)
mapping into C (or in most cases R), we have
kf ḡkL1 (Ω) ≤ kf kLp (Ω) kgkLq (Ω)
ie
Z
Z
|f ḡ|dµ ≤
Ω
|f |p dµ
1/p Z
Ω
1/q
|g|q dµ
(23)
(24)
Ω
Proof. If kf kp = ∞, kgkq = ∞, f = 0 or g = 0 then the inequality holds so assume otherwise. Apply
|g(x)|
Young’s Inequality with a = |fkf(x)|
kp and b = kgkq (using |g(x)| = |ḡ(x)|)
f (x)ḡ(x) 1 |f (x)|p
1 |g(x)|q
≤
+
.
kf kp kgkq p kf kpp
q kgkqq
(25)
1
1 kf kpp
1 kgkqq
1 1
kf ḡk1 ≤
+ =1
p +
q =
kf kp kgkq
p kf kp
q kgkq
p q
(26)
kf ḡk1 ≤ kf kp kgkq
(27)
Integrate both sides over Ω to get
so that
as desired. Note that if f and g map into R and not C, then ḡ = g.
3.4
Poincaré Inequality
Assume 1 ≤ p ≤ ∞, Ω is a bounded open subset of Rn with Lipschitz boundary. Then there exists a finite
constant c = c(Ω, p) depending only on the space Ω and on p such that for all u ∈ W 1,p (Ω),
ku − uΩ kLp (Ω) ≤ ck∇ukLp (Ω)
(28)
where uΩ is the average value of u on Ω
uΩ :=
1
|Ω|
Z
u(y)dy
Ω
Proof. May be given as exercise in class for Ω ⊂ R2 , Ω a rectangle.
4
(29)
4
4.1
Existence-Uniqueness Theorems with Proofs
A Riesz Representation Theorem for Separable Hilbert Spaces
Theorem 4.1. Let H be a separable Hilbert space and H 0 its dual space consisting of all continuous (bounded)
linear functionals from H to R(C). Then given F ∈ H 0 , there exists x ∈ H such that for all y ∈ H,
F (y) = (y, x)
(30)
and kF kH 0 = kxkH
Proof. Let ϕi i = 1, 2, . . . be an orthonormal basis for H (one exists since H is separable) and F a bounded
linear functional. Set ai := F (ϕi ).
n
n
X
X
Given any y ∈ H let ci = (y, ϕi ) and define yn =
ci ϕi =
(y, ϕi )ϕi . Thus since {ϕi } is an
i=1
i=1
orthonormal basis for H, we have ky − yn kH → 0 as n → ∞.
n
X
Now since F is linear, we have F (yn ) =
ai ci and since F is bounded (kF kH 0 ≤ ∞) we have |F (y) −
i=1
F (yn )| ≤ kF kH 0 ky − yn kH → 0 as n → ∞. Thus
F (y) = lim F (yn ) =
∞
X
n→∞
ai ci .
(31)
i=1
In fact, since |F (y)| ≤ kF kH 0 kykH , we have
∞
X
ai ci ≤ kF kH 0
n
X
!1/2
c2i
.
(32)
i=1
i=1
Now choose y =
∞
X
ai ϕi ie ci = ai for i = 1, . . . , n and 0 beyond. Then
i=1
n
X
a2i ≤ kF kH 0
n
X
a2i
(33)
≤ kF kH 0
(34)
i=1
or
n
X
!1/2
i=1
!1/2
a2i
i=1
for any n. Thus if we take the limit, we have
∞
X
ai ϕi i=1
Thus x =
∞
X
=
∞
X
!1/2
a2i
≤ kF kH 0
(35)
i=1
H
ai ϕi is a well defined element of H and
i=1
F (y) =
∞
X
ai ci = (x, y).
(36)
i=1
Now, kxkH =
P
a2i
1/2
≤ kF kH 0 and for all y ∈ H
|F (y)| = |(x, y)| ≤ kxkH kykH
5
(37)
So that
kF kH 0 := max
y6=0
|F (y)|
≤ kxkH
kykH
(38)
Thus kF kH 0 = kxkH .
There is a more simple proof for the general Hilbert space that takes advantage of the Orthogonal
Decomposition Theorems that come out of the geometry of a Hilbert space. However I decided go this
route because it is common for people to have seen the fourier type decompositions like above we are often
already comfortable talking about sum of scalars squared etc. I also didn’t want to have to prove a bunch of
geometry conditions on Hilbert spaces. They are interesting in their own right but it would have been more
work than I wanted to give.
4.2
Symmetric Lax-Milgram Theorem
1/2
A bilinear form A(·, ·) on (H, (·, ·)H ), a Hilbert space with norm kukH = (u, u)H for u ∈ H, is said to be
bounded or continuous if there exists a constant C < ∞ such that for all v, w ∈ H
|a(v, w)| ≤ CkvkH kwkH .
(39)
We say a(·, ·) is coercive on V ⊂ H a closed subset of H (often we use V = H) if there exists a constant
α > 0 such that for all v ∈ V ⊂ H
a(v, v) ≥ αkvk2H .
(40)
Lemma 4.2. Let (H, (·, ·)H ) be a Hilbert space with a symmetric bilinear form, a(·, ·) which is continuous
on H and coercive on V ⊂ H, V closed in H. Then (V, a(·, ·)) is also a Hilbert space.
Proof. Since a(·, ·) is coercive, bilinear and symmetric on V , it is an inner product on V . (You can prove
1/2
all the tenets of an inner product using these properties) Let kvkE := (a(v, v))
be the induced norm for
v ∈ V . (Sometimes it is called the energy norm, hence the subscript E). We must show that V is closed
under this energy norm and then it is a Banach Space and a closed subset of a Hilbert space and so is a
Hilbert space itself with inner product
√ a(·, ·). To this end, consider a Cauchy sequence {vn } in (V, k · kE ).
By coercivity, kvkE = a(v, v)1/2 ≥ αkvkH , so {vn } is also Cauchy in (V, k · kH ). Since V is closed in H
and H is complete, Cauchy sequences in (V, k · kH ) converge. In other words, there is a v ∈ V such that
kv − vn kH → 0. Finally by boundedness of a(·, ·) in H, we have
kv − vn k2E = a(v − vn , v − vn ) ≤ Ckv − vn kH → 0.
(41)
So that vn → v in k · kE . So (V, k · kE ) is complete and thus (V, a(·, ·)) is a Hilbert space.
Theorem 4.3. Symmetric Lax-Milgram Theorem Given a(·, ·) a symmetric bounded bilinear form on
H coercive on V ⊂ H a closed subspace of H and F ∈ V 0 , a bounded linear functional on V , then there
exists a unique u ∈ V such that a(u, v) = F (v) for all v ∈ V and we have kukE = kF kV 0 .
Proof. Using the previous lemma, (V, a(·, ·)) is a Hilbert space on V . Apply the Riesz representation theorem
to the linear functional F to get a unique u ∈ V such that F (v) = (u, v)E := a(u, v). In particular we also
have kF kV 0 = kukE .
4.3
Contraction Mapping Principle on a Banach Space
Theorem 4.4. Given a Banach Space (V, k · kV ) and a mapping T : V → V satisfying for all v1 , v2 ∈ V
kT v1 − T v2 kV ≤ M kv1 − v2 kV
for some fixed M ∈ [0, 1). Then there exists a unique fixed point u ∈ V such that u = T u.
6
(42)
Proof. We will drop the subscript V notation on the norm as it is clear which space we are talking about.
(uniqueness) Suppose T v1 = v1 and T v2 = v2 , then by the contraction mapping, kv1 −v2 k = kT v1 −T v2 k ≤
M kv1 − v2 for some 0 ≤ M < 1. So that kv1 − v2 k ≤ M kv1 − v2 k but since M < 1 this can only be
true if kv1 − v2 k = 0. Thus v1 = v2 .
(existence) We will construct a Cauchy sequence in V . Given v0 ∈ V then for i = 1, 2, . . . , we define
vi := T vi−1 . Thus we have
kvk − vk−1 k = kT vk−1 − T vk−2 k ≤ M kvk−1 − vk−2 k ≤ · · · ≤ M k−1 kv1 − v0 k
so that for all m > n we have (using 0 ≤ M < 1 so that for any r ∈ N,
r
X
k=0
Mk ≤
∞
X
(43)
M k = 1/(1 − M )
k=0
makes sense)
m
X
kvm − vn k = (vk − vk−1 )
≤
≤
k=n+1
m
X
k=n+1
m
X
(44)
kvk − vk−1 k
(45)
M k−1 kv1 − v0 k
(46)
k=n+1
n
M
kv1 − v0 k
1−M
Mn
kT v0 − v0 k.
=
1−M
≤
(47)
(48)
Thus kvm − vn k → 0 as m, n → ∞. So {vn } is a Cauchy sequence in (V, k · k). Now since V is complete
(Banach), there is a v ∈ V such that vn → v. Then since T is continuous,
v = lim vn+1
n→∞
(49)
= lim T vn
n→∞
= T lim vn
(50)
= Tv
(52)
n→∞
(51)
and v is a fixed point of the contraction mapping T .
4.4
Non-Symmetric Lax-Milgram Theorem
Theorem 4.5. Given a Hilbert space (H, (·, ·)) and a continuous(bounded) bilinear form a(·, ·) which is
coercive on V ⊂ H, a closed subset of H, and a continous(bounded) linear functional F ∈ V 0 , then there
exists a unique u ∈ V such that for all v ∈ V ,
a(u, v) = F (v)
(53)
Proof. Given a fixed u ∈ V , we can define a functional Au(v) := a(u, v) for all v ∈ V . Au is linear since
a(·, cdot) is linear in the second term. Au is bounded (or continuous) since for all v ∈ V
|Au(v)| = |a(u, v)| ≤ Ckukkvk ≤ (Ckuk)kvk.
7
(54)
Thus Au ∈ V 0 , is a bounded linear functional on V, since
kAukV 0 := sup
v6=0
|Au(v)|
≤ Ckuk < ∞.
kvk
(55)
We can also show that the mapping u → Au is a linear map from V → V 0 . We must find a unique u ∈ V
such that Au(v) = F (v) are the same for all v ∈ V . In other words, we want to find u ∈ V such that Au = F
in V 0 . By the Riesz-representation theorem, we can find an element τ Au ∈ V such that Au(v) = (τ Au, v)
for all v ∈ V and τ F ∈ V such that F (v) = (τ F, v). Note that since the mapping τ : V 0 → V is a 1-1
mapping, it suffices to find u such that τ Au = τ F in V .
To find u, we will construct a contraction mapping which will have as fixed point u ∈ V such that
τ Au = τ F . Let ρ be a constant (we will specify it later) such that T : V → V is a contraction mapping
where for v ∈ V
T v := v − ρ (τ Au − τ F ) .
(56)
We must show there is some ρ 6= 0 such that this is a contraction mapping since then there exists a unique
u ∈ V such that u = T u = u − ρ (τ Au − τ F ) which is equivalent to 0 = ρ (τ Au − τ F ) which is equvalent to
τ Au = τ F .
To choose ρ 6= 0 we take any v1 , v2 ∈ V and define v = v1 − v2 . Then,using coercivity and boundedness
of a(·, ·) and since kτ Avk ≤ kAvk ≤ Ckvk and kτ Avk2 = (τ Av, τ Av) = Av(τ Av) = a(v, τ Av), we have
kT v1 − T v2 k2 = kv1 − v2 − ρ (τ Av1 − τ Av2 ) k2
2
= kv − ρτ Avk
(58)
2
2
= kvk − 2ρ(τ Av, v) + ρ kτ Avk
2
= kvk − 2ρ(τ Av, v) + ρ a(v, τ Av)
2
= M kv1 − v2 k .
(60)
(61)
2
(62)
≤ kvk − 2ραkvk + ρ Ckvkτ Avk
≤ 1 − 2ρα + ρ2 C 2 kvk2
2
(59)
2
= kvk − 2ρa(v, v) + ρ a(v, τ Av)
2
(57)
(63)
(64)
2α
We want 0 < M 2 = 1 − 2ρα + ρ2 C 2 < 1, so if we choose ρ ∈ (0, C
2 ) then we have that T is a contraction
mapping and so has a unique fixed point u ∈ V . Thus τ Au = τ F in V or Au = F in V 0 or a(u, v) = F (v)
for all v ∈ V as desired.
8
Index
Banach space, 1, 7
< f, x >, action of dual f on x, 1
bounded bilinear form, 6
BV (Ω), 1
Cauchy sequence, 7
Cauchy-Schwarz Inequality, 2, 3
coercive bilinear form on V, 2, 6
continuous bilinear form, 6
contraction mapping, 6, 8
Dual Space, V 0 , 1
f (x) action of dual f on x, 1
Hilbert space, 2, 7
H k (Ω), 1
Hölder’s Inequality, 2, 4
(·, ·), inner product, 1
Inner product space, 1
Lax-Milgram Theorem, 2
Non-Symmetric Form, 7
Symmetric Form, 6
L∞ (Ω), 1
Lipschitz boundary, 2
Lp (Ω, dµ), 1
Peter-Paul Inequality, 2, 3
Poincaré Inequality, 2, 4
Riesz Representation theorem, 5, 8
separable, 1, 5
Young’s Inequality, 2, 3
9