Lecture Four: The Poincare Inequalities
In this lecture we introduce two inequalities relating the integral of a function to the
integral of it’s gradient. They are the Dirichlet­Poincare and the Neumann­Poincare in­
equalities.
1
The Dirichlet­Poincare Inequality
Theorem 1.1 If u : Br → R is a C 1 function with u = 0 on ∂Br then
�
�
2
2
u ≤ C(n)r
|�u|2 .
Br
(1)
Br
We will prove this for the case n = 1. Here the statement becomes
� r
� r
2
2
f ≤ kr
(f � )2
−r
(2)
−r
where f is a C 1 function satisfying f (−r) = f (r) = 0. By the Fundamental Theorem of
Calculus
� s
f (s) =
f � (x).
(3)
−r
Therefore
s
�
|f (s)| ≤
|f � (x)|.
(4)
−r
Recall the Cauchy­Schwarz inequality
h = 1, g =
|f � |
��
hg ≤
��
h2
��
r
�1/2 ��
g2
�1/2 �
. Apply this with
to get
��
s
|f (s)| ≤
� 2
(f )
�1/2
1/2
(r + s)
≤
� 2
(f )
�1/2
(2r)1/2 .
(5)
−r
−r
Squaring both sides gives
2
�
r
|f (s)| ≤ 2r
−r
1
(f � (s))2 ,
(6)
and finally we integrate over [−r, r] to give
� r
�
|f (s)|2 ≤ 4r2
−r
r
|f � (s)|2
(7)
−r
as required.
2
The Nueman­Poincare Inequality
�
u then
Theorem 2.1 If u is C 1 on Br , and we define A = 1
volBr Br
�
�
(u − A)2 ≤ C(n)r2
|�u|2 .
Br
(8)
Br
Again we give the proof in the case n = 1.
�r
1
Take a differentiable function f with A = 2r
−r f . Note by the intermediate value
theorem that there is a point c in [−r, r] with f (c) = A. We have
� s
f (s) = f (c) +
f � (s).
c
From this we get |f (s) − A| ≤
�s
c
|f � (t)| ≤
�r
−r
|f � (t)|. Apply Cauchy­Schwarz again to give
��
r
|f (s) − A| ≤ 2r
� 2
(f )
�1/2
.
(9)
−r
Squaring and integrating then gives our result.
� r
�
2
2
(f (s) − A) ≤ (2r)
−r
r
|f � (s)|2 .
(10)
−r
It is not difficult to extend these proofs to higher dimensional cubes.
There is another
interesting fact related to the Neumann­Poincare
�
�
� inequality. If we
define g(x) = Br (u − x)2 we can rearrange to get g(x) = Br u2 − 2x Br u − x2 . We then
R
u
find the minimum by taking the derrivative, and see that it occurs at x = Br , so the
volBr
mean really is the best constant approximation to a function.
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