P OINCARE I NEQUALITIES
Poincare inequalities are a simple way to obtain lower bounds on the distortion of mappings
X into Y. These are shown below to be sharp when we consider the L p spaces.
A Poincare inequality is one of the following type: suppose Ψ : [0, ∞) → [0, ∞) is a nondecreasing function and that au,v , bu,v are finite arrays of real numbers (for u, v ∈ X, and not
all of the numbers 0). We say that functions from X to Y obey a Poincare inequality if for all
f : X → Y we have that ∑u,v∈X au,v Ψ(d( f (u), f (v))) ≥ ∑u,v∈X bu,v Ψ(d( f (u), f (v))). Note that
the space X here only comes into play when we look at the indices of these sums, all that X
does is label.
∑u,v∈X bu,v Ψ(d X (u, v))
the Poincare ratio. Note that the
∑u,v∈X au,v Ψ(d X (u, v))
structure of X does, indeed, play a massive part in the value of this ratio. This means that an
‘optimal’ Poincare inequality (eg the one with the biggest ratio) for one n point space may be
different to the optimal Poincare inequality for another space.
We call the quotient Pa,b,Ψ(t) ( X ) =
We give an example: Suppose we take the case of a 4-cycle graph with distances all 1, and
consider f : X → `2 . Then k f (v1 ) − f (v3 )k2 + k f (v2 ) − f (v4 )k2 ≤ k f (v1 ) − f (v2 )k2 + k f (v2 ) −
f (v3 )k2 + k f (v3 ) − f (v4 )k2 + k f (v4 ) − f (v1 )k2 . This is true because we expand everything in
sight, rearrange, and then note that the resultant mess
√ is positive. This is a Poincare inequality
2
with Ψ(t) = t , and this gives us the Poincare ratio 2.
This brings us to thinking (well?) that particularly simple choices of Ψ would be t p . We are
also interested in ` p spaces (and L p spaces), specifically, we would like to bound the distortion
of embeddings into L p and ` p .
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Poincare Inequalities for L p
In fact, such ponderings are well founded. The Poincare ratio forms a bound for the distortion
of embeddings into the ` p spaces, and (moreover) the distortion of embeddings into ` p spaces
provides Poincare inequalities (within e). The latter is much harder than the former, so we
begin with the former.
Theorem 1.1. If Y valued functions on X satisfy the Poincare inequality with Ψ(t) = t p then cY ( X ) ≥
( Pa,b,t p ( X ))1/p
Proof. Suppose not. Then there is some embedding f : X → Y with D = dist ( f ) < ( Pa,b,t p ( X ))1/p ,
eg, there is some r > 0 such that rd X (u, v) ≤ dY ( f (u), f (v)) ≤ rDd x (u, v).
Consider ∑ bu,v (d X (u, v)) p ≤ r1p ∑ bu,v (dY ( f (u), f (v))) p by the distortion factor, which is ≤
r p ∑ au,v (dY ( f (u), f (v))) p by the Poincare inequality ≤ D p ∑ au,v (d X (u, v)) p by the distortion
factor. But this is in contradiction with definition of the Poincare inequality, and thus we are
done.
This allows us to, eg, conclude that for the 4-cycle we have that c2 ( X ) ≥
The converse is much more involved.
1
√
2.
1.1
The Converse
Theorem 1.2. Let ( X, d X ) be a finite metric space. Then the distortion c p ( X ) is equal to the supremum of constants C for which there exist positive and not all zero arrays au,v and bu,v for which
∑ bu,v (d X (u, v)) p ≥ C p ∑ au,v (d X (u, v)) p (which we denote PC1) and for any f : X → L p we have
that ∑ au,v k f (u) − f (v)k p ≥ ∑ bu,v k f (u) − f (v)k p (which we denote PC2).
Eg, the distortion of the embedding is equal to the largest Poincare ratio (up to an e).
Proving this will require us to go a little afield. The second part of the proof is, essentially, a
Hahn Banach argument.
1.1.1
Hahn Banach for Cones
We will be interested in using the Hahn Banach theorem for Cones. A Cone is a subset of a
linear space X that is closed under addition and multiplication by positive scalars. The reasoning
behind the name is apparent, they really do look like cones. We do not require that a cone must
be closed under multiplication by non-negative scalars (otherwise every two cones would meet
at zero.)
We will carefully state the Hahn-Banach separation theorem for cones here.
Theorem 1.3. Suppose that A and B are convex subsets of Rn , with B closed. Then there is a linear
functional s on Rn1 and α ∈ R for which s( x ) ≤ α on A and ≥ α on B.
Moreover, if A and BB are cones, then we can choose α = 0.
The proof of this is typical for Hahn-Banach.
1.1.2
Concatenating Embeddings
We mention here another prerequisite for the proof. Any separable subspace of L p (Ω, σ, µ) is
isometric to a subspace of L p [0, 1]. The proof of this statement is not trivial: however we will
S
only be using it in a particularly restricted case here. We will be mapping L p [ [αi , β i ]] (where
these intervals are disjoint) onto L p [0, 1]. To do this we consider a sequence of intervals of
S
[0, 1], say, [0,1/2], [3/4,7/8], [15/16,31/32], ..... Then the mapping f : [αi , β i ] → R maps to
the function f˜ can be broken up into the intervals, eg, f˜ restricted to the i’th interval [α, β] is
2i
Ki f ( βi2−αi ( x − α)), where Ki is a constant that makes the constant 1 function map correctly.
A similar trick can be done to remove atoms from the measure, eg, ` p can be embedded into L p
by choosing disjoint measurable sets σi with non zero Lesbegue measure. Then the mapping
that takes a sequence x to the function f (t) = |σ |x1/p if t ∈ σi and 0 otherwise suffices.
i
1.1.3
Proof of Result 1.2
To prove this requires us to show that if c p ( X ) ≥ C for each C satisfying the conditions in the
theorem, and that for each C < c p ( X ) we have that there exists au,v and bu,v satisfying both
conditions.
1 Linear
functionals on Rn correspond to inner products.
2
For the first part, assume that c p ( X ) < C and the conditions in the theorem are satisfied.
Then there exists an e > 0 and an f : X → L p such that for all u and v in X we have
that d X (u, v) ≤ k f (u) − f (v)k ≤ (C − e)d X (u, v). We use this map and put it in PC1 to
give that C p ∑ au,v (d X (u, v)) p ≤ ∑u,v bu,v (d X (u, v)) p ≤ ∑ bu,v k f (u) − f (v)k p . By PC2 this is
≤ ∑ au,v k f (u) − f (v)k p ≤ (C − e) p ∑ au,v (d X (u, v)) p , a contradiction.
For the second part let C < c p ( X ). Let P be the set of all unordered pairs of distinct eleemnts
of X and consider the linear space of functions on P by RP . Every semi-metric on X can be
regarded an element of RP , namely if d is a semi metric consider the function {u, v} 7→ d(u, v)
(where symmetricity of semi-norms means this is fine.) Moreover, note that for any function
f : X → L p , this induces a semi-metric, namely, {u, v} 7→ k f (u) − f (v)k p . Consider L p ,
the family of p−th power semi-metrics on X, and we know that L p contains all functions
f : X → L p in the way described.
We wish to observe that L p is a cone: positive multiplication is clear. To demonstrate that it is
closed under addition, we note that we can ‘concatenate’ embeddings, eg, if we have f 1 , f 2 we
can view f 1 as living on [0, 1] and f 2 living on [0,2]. Then the procedure described in the last
subsubsection means we can map correctly to produce another element of L p .2
Since C < c p ( X ) we have that the space ( X, d X ) does not admit a bi-Lipschitz embedding
into L p . Introduce the set K = {( xuv ) ∈ R p : ∃r > 0 such that ∀u, v r (d X (u, v)) p ≤ xu,v ≤
rC p (d X (u, v)) p }. The set K includes the p−th power of all semi-metrics arising by C biLipschitz embeddings of ( X, d X ). On the other hand, we do not have the triangle inequality, so K
can contain other things. Since there are no C biLipschitz embeddings of ( X, d X ) into L p we
have that K ∩ L p is empty. Since K and L p are cones we can separate them with a hyperplane,
eg, there is some s ∈ R P such that hs, x i ≥ 0 on K and ≤ 0 on L p .
Set bu,v = su,v if su,v is positive and 0 otherwise. Set au,v to be −su,v if su,v is negative and 0
otherwise. Then the condition hs, x i ≤ 0 on L p is PC2, and the condition hs, x i ≥ 0 on K is
PC1.
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Fourier Analytic Poincare Inequalities
We will now look at the Poincare inequalities that we can obtain for the following space: consider the space X consisting of 0, 1 sequences of length n. We define the Hamming distance
between two points to be the number of points at which the sequences differ. What can we say
about c2 ( X )?
To do this we introduce Fourier analysis. Consider the fields F2n , these are the fields of 0, 1
sequences. Given a subset A ⊂ {1, . . . , n} consider the Walsh function WA ( x ) = (−1)∑i∈ A xi .
Our claim is that the Walsh functions WA form an orthonormal basius of L2 (F2n , µ), where µ is
the uniform measure.
R
R
To show this, first note that hWA , Wb i = Fn WA ( x )WB ( x ) dx = Fn (−1)∑i∈ A4B xi dx. So our goal
2
2
R
is to now show that WA ( x ) dx = 0 if A 6= ∅ and 1 otherwise. It is clear that when A = ∅ the
answer is 1.
For A 6= ∅. Suppose e j is chosen such that e j = (R0, . . . , 0, 1, 0, . . . , 0) is the
R vector with the 1
in the j’th co-ordinate place, where j ∈ A. Then Fn WA ( x + e j ) dx = Fn WA ( x ) dx, by the
2
2I
2
have little to no idea why this is done here: I would have guessed it just followed from that fact that k.k
followed the triangle inequality.
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translation invariance of µ. However, WA ( x + e j ) = −WA ( x ), thus the thign integrates to zero
necessarily.
Since the functions WA ( x ) form a basis for L2 (F2n , µ) we get various properties
R that one expects
of them. We can express any function f ( x ) = ∑ A fˆ( A)WA where fˆ( A) = Fn WA ( x ) f ( x ) dx.
2
R
This gives us an equivalent of the Parsevals identity, eg, n ( f ( x ))2 dµ = ∑ A fˆ( A)2 .
F2
The same can be done with functions f : F2n → L2 , but we instead get
∑ A k fˆ( A)k2 , where the coefficients fˆ( A) become elements of L2 .
R
F2n
k f ( x )k22 dµ =
2
2.1
Partial Differentiation and Poincare Inequalities
We now define partial differentiation for functions f : F2n → X for X a Banach space. We define
f ( x +e )− f ( x )
j
∂j f =
. We note that when X = R and f ∈ L2 (F2n , µ) we get thatR ∂ j WA = −WA if
2
j ∈ A and 0 otherwise. The Parseval identity for each ∂ j f gives us that ∑nj=1 Fn (∂ j f ( x ))2 dµ =
2
∑ A∈{1,...,n} | A| fˆ( A)2 .
Observe that WA ( x + (1, 1, ..., 1)) = WA ( x )Rif | A| is even and −WA ( x ) Rif | A| is odd. Consider
f an arbitrary function f : F2n → R. Then Fn | f ( x ) − f ( x + e)|2 dµ = Fn | ∑ A fˆ( A)(WA ( x ) −
2
2
R
WA ( x + e))| dµ = Fn | ∑| A| is odd 2 fˆ( A)WA ( x )|2 dµ = ∑| A| is odd 4 fˆ( A) ≤ 4 ∑ A | A| fˆ( A)2 =
2
R
4 ∑nj=1 Fn |∂ j f ( x )|2 dµ. This is a Poincare inequality.
2
We now claim that a Poincare inequality of the type above, eg, ∑u,v∈X au,v | f (u) − f (v)|2 ≥
∑u,v∈X bu,v | f (u) − f (v)|2 gives rise to the inequality ∑u,v∈X au,v k F (u) − F (v)k22 ≥ ∑u,v∈X bu,v k F (u) −
F (v)k22 . To see this, consider the inequality ∑u,v∈X au,v ∑α | f α (u) − f α (v)|2 ≥ ∑u,v∈X bu,v ∑α | f α (u) −
f α (v)|2 , which is true by the first inequality. However, if f α = h F, gα i where gα is an orthonormal basis of L2 , this is exactly as required.
√
Now all we need to do is plug in numbers to see that c2 (F2n , ρ) ≥ n, and ≤ follows from the
obvious embedding.
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