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University of Washington
Lecturer: James R. Lee
CSE 599I: Geometric embeddings and highdimensional pheneomena
February 1, 2007
Scribe: DangTrinh HuynhNgoc
Lecture 6
Our goal in this lecture is to prove the following theorem.
Theorem 1.
For arbitrary large
n
∈
N
, there exists
n
point subsets
X
⊆
‘
1
such that for any
D
embedding of
X
into
‘
d
1
,
d
≥
n
Ω(1
/D
2
)
.
Note that the best known upper bound for dimension reduction in
‘
1
is due to Talagrand. It says
that every
n
dimensional subspace of
‘
1
embeds into
‘
O
(
n
log
n
)
1
with
O
(1) distortion. In particular,
this holds for every
n
point subset of
‘
1
.
1 Uniform convexity
Our proof of Theorem 1 relies on the following
uniform convexity
inequality for points in
L
p
, for
p
∈
(1
,
2].
Lemma 2.
Fix
1
< p
≤
2
and
a,b
∈
L
p
. Then,
k
a
+
b
k
2
p
+ (
p

1)
k
a

b
k
2
p
≤
2(
k
a
k
2
p
+
k
b
k
2
p
)
.
Proof.
See course web page.
Note that when
p
= 2, Lemma 2 is the wellknown Parallelogram law for Euclidean spaces.
The inequality expresses a property of the
L
p
spaces (for
p
∈
(1
,
2]) called uniform convexity. The
geometry of every norm space is reﬂected in its unit ball and we know that the unit ball of every
norm space is convex. For some spaces the ball is even strictly convex, in the following sense. If
one ﬁxes
X
and
Y
on the boundary of the unit ball, then the midpoint
M
not only lies inside the
ball, but at some nonzero distance inside the boundary. As we can see from Figure 1, the unit ball
of
L
1
is not strictly convex, while for
p
∈
(1
,
2], the unit ball of
L
p
is strictly convex. Furthermore,
the
L
p
balls for
p
∈
(1
,
2] are not just strictly convex, they are uniformly convex. In other words, if
the length of the segment
XY
is
²
, then there is a certain distance
δ
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