The concept of duality in Asymptotic Geometric
Analysis:
The Legendre Transform.
Shiri Artstein-Avidan and Vitali Milman
Tel Aviv University.
Adrien-Marie Legendre1752-1833
The Legendre transform :
This is an involution on the class of convex lower semi-continuous
functions on Rn
For a function f : Rn R let
(Lf()x = )sup { <x,y>- f)y( } y
U ±{∞}
Fix a scalar product <·, ·> on Rn
Define: Cvx(Rn) is the set of lower semi continuous convex functions on Rn with values
possibly infinite .
Example: For f(x) = /2 , we have (Lf)(x) = /2
2|||| Kx2|||| K
x
Most simple Examples: functions which are infinity everywhere but one point ; linear functions.
Abstract duality concept:
1 .For all f, T(T(f)) = f
2 .For all f < g one has T(f)> T(g)
We illustrate in this talk how the concrete formula for the Legendre transform can be obtained directly from
“abstract duality’’
(Notice that condition 1 implies that T is 1-1 and onto)
Theorem 1: 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) > T(g)must satisfy for some symmetric B in GLn , v0 in Rn, and C0 in R, that
( Tf()x = )C0 + <v0 , x> + )Lf( )Bx+v0(. Remark: <v0 , x> versus (Lf)(x+v0)…
Remark: <v0 , x> versus (Lf)(x+v0)…
Notice that:
(L(f+<•, v>)()x )
= sup <x, y> - f(y)-<y,v<
= sup <x-v, y> -f(y)( = Lf()x-v .)
Theorem 1’: Any 1-1 and onto T: Cvx(Rn) Cvx(Rn) satisfying that
( 1 )For all f < g one has T(f) > T(g)
( 2 )For all T(f) < T(g) one has f > gmust satisfy for some fixed B in GLn , v0 and v1 in Rn, C1 positive and C0 in R, that
( Tf()x = )C0+<v0 , x>+C1)Lf()Bx+v1(.
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
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 .
First step:
If T is 1-1 and onto and satisfies
1 .f < g implies T(f) > 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 .
The proof is quite simple :
Since min(f,g)< 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 .
In fact, we may show a more general fact:
If T is 1-1 and onto and satisfies
1 .f<g implies T(f)>T(g) 2 .T(f) < T(g) implies f >g
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 .
Notation: the delta functions Dx :
Dx(x)= 0, and infinity elsewhere
Claim: If we know what happens to Dx+c for all x in Rn and c in R, we know the form of the
transform .
Proof: f(x) = inf (Dy(x) + f(y))
and so (Tf)(x) = sup T(Dy+f(y))(x)
y
y
Notation: linear functions hx :
lu(y)= <u,y<Claim: If we know what happens to hx+c for all x in Rn, we know the form of the
transform .
Proof: f(x) = sup (hu(x) + c(u))
and so (Tf)(x) = inf T(hu+c(u))(x)
)here inf is regularized infimum(
u
u
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”
Delta functions
x
Dx
…if f> Dx , then f=Dx+c for some
c>0…
x Dx+c’…
…same true for the functions
Linear functions
hu=<x, u<
…if f<hu , then f=hu-c for some
c>0…
u
…same true for the functions
hu+c’…
u
Let T(hu)= w. (for hu = <·,u>) If f >w and g>w , then T(f)<
T(w) = hu , and T(g)<T(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)>T(g) or T(g)>T(f)Therefore: either f<g or g<f.
Any two functions f>w and g>w are comparable .
We see that: (letting w =T(hu))
(Notice: this is true for w=Dx). Second: If the “support” of w includes two points x and y, we may build two non-comparable
functions f>w , g>w :
x yw(x
)
w(y)
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 cT(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’ u(x
)c’(x,
c)
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 thatT(Dx+c) = hBx+v0+ <x,v0> — c + c0
u(x)
c’(x,c)
)Tf)(x = (sup T(Dy+f(y))(x)This would complete the proof since:
T(Dx+c) = hBx+v0 +<x,v0> — c + c0
)Tf)( x = (sup (<y,v0>+hBy+v0 (x)— f(y) + c0) = c0+ sup (<y,v0> +<x, By+v0>— f(y)) = c0+ <x, v0> + sup (<v0+Bx,y> — f(y))
= c0+ <x, v0>+(Lf )(v0+Bx)
And thus:
we know
The linearity of the correspondence
is established as follows:
T(Dx+c) = hBx+v0 +<x,v0> — 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) = Ax+vfor some fixed A in GLm and v in Rm.
Any (z,c) inside this interval satifies that
Dz+c ≥ “min” (Dx+c1, Dy+c2)
And that the same is not true for (Dz+c’) for any c’<c .
Define the mapping F : Rn+1 Rn+1 by
F((x,c)) = (u,c’)Where T(Dx+c) = hu+c ’
Consider the interval [(x,c1),(y,c2)] .
Consider the interval [(x,c1),(y,c2)] .
Any (z,c) inside this interval satisfies that
Dz+c ≥ “min” (Dx+c1, Dy+c2)
And that the same is not true for (Dz+c’) for any c’<c .
Letting F(x,c1) = (u,c’) and F(y,c2)=(v,c’’):
( hw+c)=’’’ T(Dz+c) ≤ max (hu+c’, hv+c’’)
And the same is not true for (Dz+c’) for any c’<c .
hu+c’
hv+c’’
(notice that T(Dz+c) are all parallel)
So, T(Dz+c) = hw+c’’’
with w in [u,v]
~
F(x,c) = A(x,c)+vfor some fixed A in GLn+1 and v in Rn+1.
Conclude: the mapping F given by F((x,c)) = (u,c’), where T(Dx+c) = hu+c’ is mapping intervals to intervals and so, by the fact stated before, must be affine linear:
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 = ( Bx+v0 , <v0 ,x> — c+c0)
u(x)
c’(x,c)
This is not difficult.. (show that: first coordinate doesn’t depend on c, second coordinate’s dependency on c is involutive, relation between A’s last row and v’s entries: again, from involution)
So that T(Dx+c) = hBx+v0 +<v0 , x> —c+c0
A(x,c)+v = ( Bx+v0 , <v0 ,x> —c+c0)
End of Proof .
We have sketched the proof of:Theorem 1: 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) > T(g)must satisfy for some symmetric B in GLn , v0 in Rn, and C0 in R, that
( Tf()x = )C0 + <v0 , x> + )Lf( )Bx+v0(.
Theorem 2:
Any T: Log-Conc(Rn) Log-Conc(Rn) satisfying that
( 1 )For all f, T(T(f))=f
( 2 )For all f < g one has T(f) > T(g)must satisfy for some symmetric B in GLn , v0 in Rn, and C0>0, that
( Tf()x = )C0 e-<v0,x> inf e-<Bx+v0,y>/f)y(. (first defined in A-Klartag-
M)
Theorem (Böröczky-Schneider) (using Grüber): For n ≥ 2, any T: K0(Rn) K0(Rn) satisfying that
( 1 )For all K, T(T(K))=K
( 2 )For all K1 K2 one has T(K2) T(K1)
must satisfy for some symmetric B in GLn that
T(K) = BK0
Theorem 3: Any T: K(Rn) K(Rn) satisfying that
( 1 )For all K, T(T(K))=K
( 2 )For all K1 K2 one has T(K2) T(K1)
must satisfy for some symmetric B in GLn that
T(K) = BK0
Theorem 4: Any T: Concs(Rn) Concs(Rn) satisfying that
( 1 )For all f, T(T(f))=f
( 2 )For all f < g one has T(f) > T(g)must satisfy for some symmetric B in GLn and C0>0, that
( Tf()x = )C0 inf )1-<x,y>(s/f)By(. y +
(also used in A-Klartag-M)
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.)
(f g()z = )inf (f(x) + g(y))x+y=z
(Part of a joint work with Semyon Alesker)
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