LECTURE 4, 18.155, 20 SEPTEMBER 2011
So now for some ‘hard analysis’ rather than softer measure theory.
(1) I will define the spaces Lp (Rn ) by completion and then show
that they have concrete realizations as spaces of (equivalence
classes of) functions. We can start with S(Rn ), or Cc∞ (Rn ) or
Cc0 (Rn ) – it doesn’t make much difference. Let me work with
Cc∞ (Rn ). On this space we have norms, one for each 1 ≤ p < ∞
(and there is a similar norm for p = ∞ but the present approach
does not work directly in that case – can you see why?)
1/p
Z
p
kφkLp =
|φ| )
.
Rn
Here the integral really is a Riemann, or iterated Riemann if
you only know the 1-D case, integral since there exists R > 0
such that φ(x) = 0 if |xj | ≥ R for any j.
(2) The first two properties of a norm – that kφkLp ≥ 0 with equality iff φ ≡ 0 follow from basic properties of the Riemann integral, as does kcφkLp = |c|kφkLp . The triangle inequality is called
Minkowkski’s inequality:kφ + ψkLp ≤ kφkLp + kψkLp ∀ φ, ψ ∈ Cc∞ (Rn )
needs proving! If you don’t know it you should look it up. The
usual chain of reasoning is to start with Young’s inequality (for
non-negative real numbers)
ap b q 1 1
ab ≤
+ , + =1
p
q p q
(which follows from calculus) then integrate this to get
Z
Z
Z
1 p
1 q
p
|af (x)||bg(x)|dx ≤ a
|f (x)| + b
|g(x)|q .
p
q
R
R
In non-trivial cases take ap |f (x)|p = 1, bq |g(x)|q = 1 in
which case we see that
Z
Z
| f (x)g(x)dx| ≤ |f (x)g(x)|dx ≤ kf kLp kgkLq
which is Hölder’s inequality. Then Minkowski’s inequality follows by observing that
|φ(x) + ψ(x)|p ≤ |φ(x)||φ(x) + ψ(x)|p−1 + |ψ(x)||φ(x) + ψ(x)|p−1
1
2
LECTURE 4, 18.155, 20 SEPTEMBER 2011
and applying Hölder’s inequality to the two products on the
right.
(3) It is easy to see that these norms are not complete on Cc∞ (Rn )
(or even on continuous functions with compact support). We
can always ‘complete’ a normed space to a Banach space abstractly. The standard methods are to embed it into its bidual
(the dual of its dual) which uses the Hahn-Banach theorem or
to embed it in a complete space – such as the space of continuous bounded functions on the space itself or, as I will quickly
outline, to take the linear space of all Cauchy sequences, give it
a seminorm and pass to the quotient Banach space.
Let V be a normed space. The space of maps N −→ N has a
linear structure given by addition in N. The Cauchy sequences
in N form a subspace, which we can temporarily denote c(N ).
Since the norm is continuous on N it follows that if {aj } ∈ c(N )
then kaj kN is Cauchy as a sequence in R and hence converges,
with limit necessarily non-negative. Set k{aj }kN = limj kaj kN .
This is a seminorm on c(N ) but not a norm. However, for any
seminorm on a linear space the null elements, those on which
the seminorm vanishes, form a linear subspace S and the quotient, in this case c(N )/S is a normed space – the properties
of the norm are inherited from the seminorm. For c(N ), S is
precisely the subspace of c(N ) consisting of the sequences which
converge to 0 and B(N ) = c(N )/S is a Banach space. This is
the usual sort of diagonalization argument. If {bj } is a Cauchy
sequence in B(N ) then, by definition, each bj = {aj,k }k is represented by a Cauchy sequence in N, i.e. an element of c(N ).
If k(j) is a sufficiently rapidly increasing sequence of integers
then cj = aj,k(j) is Cauchy in N, i.e. is an element of c(N ) and
bj −→ [c] converges in B(N ) which is therefore complete, i.e. a
Banach space. Now, one should embed N −→ B(N ) isometrically by mapping each a ∈ N to the constant sequence cj = a
to see that N can be realized as a dense subspace. You should
also know/check that any two Banach spaces in which a given
normed space N embed as isometric dense subspaces are naturally isometrically isomorphic – are ‘the same’ from this point
of view.
(4) So, in this way we can define spaces Lp (Rn ) by completion of
Cc∞ (Rn ) with respect to k · kLp . Of course the result is only an
abstract Banach space since we are essentially just ‘declaring’
that Cauchy sequences converge. The Lebesgue integral realizes
the new ‘ideal’ elements in terms of actual functions on Rn .
LECTURE 4, 18.155, 20 SEPTEMBER 2011
3
However, this is just a little tricky since the assumption that
φj is Cauchy with respect to k · kLp does not imply that φj (x)
converges (in C) for any one x ∈ Rn ! This pointwise sequence
may not converge at all.
(5) This statement notwithstanding, we can get some convergence
by ‘speeding up the Cauchy condition’. In a normed space (it
works even in a metric space) we can demand that a sequence
φj be absolutely summable, meaning just that
X
kφj+1 − φj k < ∞.
j≥1
Note that this condition does imply that the sequence is Cauchy,
since by the triangle inequality, if k > l then
kφk − φl k ≤
k−l
X
kφl+p − φl+p−1 k ≤
p=1
X
kφj+1 − φj k
j≥l
and the convergence above implies that this last sum tends to
zero as l → ∞.
Now, if one replaces ‘Cauchy’ by ‘absolutely summable’ in
the discussion above, nothing significant changes and one gets
the ‘same’ Banach space.
(6) So, the point is that for absolutely summable sequences with
respect to the Lp norms we do get some pointwise convergence.
In the homework you are supposed to think a bit about rectangles, meaning sets of the form
{x ∈ Rn ; aj ≤ xj < bj }, aj ≤ bj ,
and sets of measure zero.
Q The volume (Lebesgue measure) of a
rectangle is vol(R) = j (bj − aj ).
A set E is said to be of measure zero if for each > 0 it has
a countable coverging by rectangles
X
X
E⊂
Rk s.t.
vol(Rk ) < .
k
k
(7) Then we get a glimmer of the concrete realization by observing
that
Theorem 1. If φj ∈ Cc∞ (Rn ) is absolutely summable with respect to k · kLp for some 1 ≤ p < ∞, then
X
E = {x ∈ Rn ;
|φj+1 (x) − φj (x)| = ∞} has measure zero;
j≥1
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LECTURE 4, 18.155, 20 SEPTEMBER 2011
conversely, for any set of measure zero there exists
P an absolutely summable sequence φj ∈ Cc∞ (Rn ) such that
|φj+1 (x) −
j≥1
φj (x)| = ∞ at each point of the set.
(8) Proof. For the given sequence, set ψ1 = |φ1 | and ψj = |φj −φj−1 |
for j ≥ 2. Thus, our assumption is that
X
kψj kLp = C < ∞.
j
Let T > 0 be chosen – eventually we will choose it large to get
the in the definition of measurePzero. Now, consider x ∈ E.
There must exist M such that
ψj (x) ≥ T + 1. Since this
j≤M
sum is finite, it is a continuous function so there is P
a rectangle
containing x, R(x), (hence non-empty) on which
ψj ≥ T.
j≤M
We can choose this rectangle to have rational ‘end-points’ aj
and bj . The choices for all x ∈ E (where M may depend on x)
therefore gives a countable collection of rectangles which cover
E. Order them and denote them, without repeats now, as Rl0 .
Now a finite union of such rectangles can be written as a finite
union of disjoint rectangles and adding a new rectangle can be
done by adding a finite number of disjoint rectanges (disjoint
from each other and all the earlier ones). Thus we can replace
Rl0 by a new sequence of rectangles, Rl , with Rl ∩ Rm = ∅ if
l 6= m, still with the same total union and such that each Rl is
a subset of one of the R(x) so for each k there exists some M
for which
X
ψj ≥ T on Rl ∀ l ≤ k.
j≤M
From this it follows that
!p
Z
X
X
X
Tp
vol(Rl ) ≤
ψj
≤(
kψj kLp )p ≤ C p ,
l≤k
∪l≤k Rl
j≤M
j
using the triangle inequality. Passing to the limit shows that
X
vol(Rl ) ≤ C p /T p .
l
Given > 0 taking T p = C p / gives a countable open cover of
E by rectangles with total volume less than so E has measure
zero.
For the converse (only for p = 1), let F be the set of measure
zero and for each k choose a countable covering Rk,j of F by
LECTURE 4, 18.155, 20 SEPTEMBER 2011
rectangles with
P
5
vol(Rk,j ) ≤ 2−k−1 . We can enlarge each of
j
0
, replacing the upper and lower
the rectangles a little, to Rk,j
−k−j−1
limits by aj − 2
(bj − aj )/4n and bj + 2−k−j−1 (bj − aj )/4n
so that the volume is increased by at most 2−k−j vol(Rn,j ). Then
for each k and j we can choose a cut-off function ψk,j ∈ Cc∞ (Rn )
0
and is equal
which satisfies 0 ≤ ψk,j ≤ 1, vanishes outside Rk,j
to 1 on Rk,j . Now, let φl be anPordering
of the ψk,j into a seR
|φl | < ∞ since it is bounded
quence. It follows directly that
l
P
|φl (x)| = ∞ at each point of
by the sum of the volumes and
l
F, since for each k one of the φk,j = 1 at x.