The concept of duality in Asymptotic Geometric Analysis: The Legendre Transform.

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The concept of duality in Asymptotic Geometric
Analysis The Legendre Transform.
Shiri Artstein-Avidan and Vitali Milman Tel Aviv
University.
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The Legendre transform
Fix a scalar product lt, gt on Rn
This is an involution on the class of convex
lower semi-continuous functions on Rn
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Define Cvx(Rn) is the set of lower semi
continuous convex functions on Rn with values
possibly infinite.
Most simple Examples functions which are
infinity everywhere but one point linear
functions.
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We illustrate in this talk how the concrete
formula for the Legendre transform can be
obtained directly from abstract duality
Abstract duality concept 1. For all f,
T(T(f)) f 2. For all f lt g one has
T(f)gt T(g)
(Notice that condition 1 implies that T is 1-1
and onto)
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Theorem 1 Any T Cvx(Rn) Cvx(Rn) satisfying
that (1) For all f, T(T(f))f (2) For
all f lt g one has T(f) gt T(g) must satisfy for
some symmetric B in GLn , v0 in Rn, and C0 in R,
that (Tf)(x) C0 ltv0 , xgt (Lf)
(Bxv0).
Remark ltv0 , xgt versus (Lf)(xv0)
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Remark ltv0 , xgt versus (Lf)(xv0)
Notice that (L(flt, vgt))(x) sup ltx,
ygt - f(y)-lty,vgt sup ltx-v, ygt -f(y)
(Lf)(x-v).
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Remark The requirement of onto is important
Consider the transformation T given by T(f) L
(f x2) It is order reversing (not an
involution, of course) 1-1 but not onto.
Remark compare with Böröczky-Schneider
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The sketch of the proof we will see consists of
three parts
(1) Min and Max are interchanged
(2) It is enough to know what happens to delta
functions, or linear functions
(3) Because of order-reversion, and the special
properties of these extreme functions, we can
determine their behavior.
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First step If T is 1-1 and onto and satisfies
1. f lt g implies T(f) gt T(g) 2. T(T(f))
f then, T(min(f,g)) max(T(f), T(g)) and,
T(max(f,g)) min (T(f), T(g)) Where min should
be understood as regularized minimum.
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The proof is quite simple Since min(f,g)lt f, g
we have T(min(f,g)) T(f) , T(g) , so that
T(min(f,g)) max(T(f), T(g)).
Secondly, max(T(f), T(g)) T(h) for some h, in
fact, for h T(max(T(f), T(g))). Thus T(h)
T(min(f,g)) so h min (f,g). But then again h
f , and h g , so h min(f,g)
and we get equality.
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In fact, we may show a more general fact If T
is 1-1 and onto and satisfies 1. fltg
implies T(f)gtT(g) 2. T(f) lt T(g) implies f
gtg then, T(min(f,g)) max(T(f), T(g)) and,
T(max(f,g)) min (T(f), T(g)) where min should
be understood as regularized minimum.
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Notation the delta functions Dx
Dx(x) 0, and infinity elsewhere
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Notation linear functions hx
lu(y) ltu,ygt
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To find out what is the image of the delta
functions (and/or the linear functions)
Let us notice a few facts about these elementary
functions
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Delta functions
Dx
x
if fgt Dx , then fDxc for some cgt0
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Linear functions
hultx, ugt
u
if flthu , then fhu-c for some cgt0
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Let T(hu) w. (for hu lt,ugt)
If f gtw and ggtw , then T(f)lt T(w) hu , and
T(g)ltT(w) hu.
This means T(f) and T(g) are both linear T(f)
hu-c and T(g) hu-c.
In particular either T(f)gtT(g) or T(g)gtT(f)
Therefore either fltg or gltf.
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We see that (letting w T(hu))
Any two functions fgtw and ggtw are comparable.
(Notice this is true for wDx .)
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Conclude The support of w is just one point,
and so for every vector u we have that T(hu)
Dx c for some vector x and some constant c.
(In fact, for every vector u and constant c T(hu
c) Dx c for some vector x and some constant
c.)
We may similarly show that for every x and c
there are u and c with T(Dx c) hu c
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Next we establish the linearity of this relation,
namely we show There exists some symmetric B in
GLn , a vector v0 in Rn and a constant c0 such
that for every vector x and constant c we have
that T(Dxc) hBxv0 ltx,v0gt c c0
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This would complete the proof since
(Tf)(x) sup T(Dyf(y))(x)
we know
T(Dxc) hBxv0 ltx,v0gt c c0
And thus
(Tf )(x) sup (lty,v0gthByv0 (x) f(y) c0)
c0 sup (lty,v0gt ltx, Byv0gt f(y))
c0 ltx, v0gt sup (ltv0Bx,ygt f(y)) c0
ltx, v0gt(Lf )(v0Bx)
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The linearity of the correspondence is
established as follows
T(Dxc) hBxv0 ltx,v0gt c c0
Fact if F Rm Rm, 1-1 and onto, satisfies
that for every interval x,y we have that
F(x,y) is also an interval, then F is affine
linear, namely F(x) Axv for some fixed A in
GLm and v in Rm.
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Define the mapping F Rn1 Rn1 by
F((x,c)) (u,c) Where T(Dxc) huc
Consider the interval (x,c1),(y,c2).
Any (z,c) inside this interval satifies that
Dzc min (Dxc1, Dyc2) And that the same is
not true for (Dzc) for any cltc.
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Consider the interval (x,c1),(y,c2). Any (z,c)
inside this interval satisfies that Dzc min
(Dxc1, Dyc2) And that the same is not true for
(Dzc) for any cltc.
Letting F(x,c1) (u,c) and F(y,c2)(v,c)
(hwc) T(Dzc) max (huc, hvc) And
the same is not true for (Dzc) for any cltc.
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hvc
So, T(Dzc) hwc with w in u,v
huc

(notice that T(Dzc) are all parallel)
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Conclude the mapping F given by F((x,c))
(u,c), where T(Dxc) huc is mapping
intervals to intervals and so, by the fact stated
before, must be affine linear
F(x,c) A(x,c)v for some fixed A in GLn1 and v
in Rn1.
Finally, we should show that A and v are in fact
composed of a symmetric B in GLn and a vector v0
in Rn as follows A(x,c)v ( Bxv0 , ltv0
,xgt cc0)
u(x)
c(x,c)
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A(x,c)v ( Bxv0 , ltv0 ,xgt cc0)
This is not difficult.. (show that first
coordinate doesnt depend on c, second
coordinates dependency on c is involutive,
relation between As last row and vs entries
again, from involution) So that T(Dxc)
hBxv0 ltv0 , xgt cc0
End of Proof.
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We have sketched the proof of
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(first defined in A-Klartag-M)
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(also used in A-Klartag-M)
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Theorem 5 Any T Cvx(Rn) Cvx(Rn) satisfying
that (1) For all f, T(T(f))f (2) For
all f,g one has T(f g)T(f)T(g) must satisfy
for some symmetric B in GLn that (Tf)(x)
(Lf)(Bx).
(Part of a joint work with Semyon Alesker)
(f g)(z) inf (f(x) g(y))
xyz