Download Convex Analysis

Institut für Mathematik
Universität Würzburg
Dr. Tim Hoheisel
May 24, 2016
Convex Analysis
Problem Set No. 6 - hints for solution
6.1
a) First, we show that V := K ∩ (−K) is actually a subspace. As V is a convex cone
it is closed with respect to addition and multiplication with nonnegative scalars.
To see that it is also closed w.r.t. multiplication with negative scalars first notice
that a ∈ V if and only if −a ∈ V . As λa = (−λ)(−a) this property is fulfilled.
Altogether, V is a subspace and since every subspace contained in K must contain
negative mulitples of its elements, there cannot be a bigger one than V .
b) Put V := K − K. Then V is a convex cone and (analogous to a)) the only thing
lacking to be a subspace is closedness w.r.t. multiplication with negative scalars.
Hence, let x = a − b ∈ V and λ < 0. Then λx = (−λ)b − (−λa) ∈ K − K ∈ V .
Thus, V is a subspace which obviously (0 ∈ K!) contains K. Since, clearly, V ⊂
span K, we must have V = span K.
c) Follows immediately from a).
6.2
a) First, let v ∈ TK (0), i.e. there exist {xk ∈ K} → 0 and {tk } ↓ 0 such that
But as K is a cone, xtkk ∈ K, and as K is closed, v ∈ K.
xk
tk
→ v.
If, in turn, v ∈ K, take {tk } ↓ 0 and put xk := tk v (k ∈ N). Then {xk ∈ K} → 0
and xktk−0 → v, hence v ∈ TK (0).
b) Let y ∈ R{x}, i.e. y = λx with λ ∈ R. Now, take {tk } ↓ 0 and put xk := tk y + x.
Then xk → x, xkt−x
→ y. Moreover, we have
k
xk = (1 + λtk )x ∈ K
for k sufficiently large, which proves that y ∈ TK (x).
6.3 We first show that T (C ∞ ) ⊂ T (C ∞ ) (which holds without the assumption that ker T ∩
C ∞ = {0}): To this end, let v ∈ T (C ∞ ), i.e. v = T (w) and w ∈ C ∞ . Hence, there
exists {xk ∈ C} and {tk } ↓ 0 such that tk xk → w. By linearity and continuity we have
tk T (xk ) = T (tk xk ) → T (w) = v,
hence v ∈ T (C)∞ .
In order to see the converse inclusion let v ∈ T (C)∞ . i.e. there exists {xk ∈ C} and
{tk } ↓ 0 such that tk T (xk ) → v. By linearity and continuity this implies T (tk xk ) → v.
In particular, {T (tk xk )} is bounded, hence, using the assumption that ker T ∩C ∞ = {0}
and the reasoning from the proof of Theorem 2.4.26, we see that {tk xk } is bounded, hence
w.l.o.g. tk xk → w ∈ C ∞ . Thus, v = T (w) ∈ T (C ∞ ), which completes the proof.
6.4 We first prove the generic inclusion
p
X Ci
i=1
∞
p
⊂
X Ci∞ .
i=1
(1)
∞
p
For these purposese, let (v1 , . . . , vp ) ∈ Xi=1 Ci , i.e. there exist {(x1 , . . . , xp ) ∈
Xpi=1 Ci } and {tk } ↓ 0 such that tk (x1 , . . . , xp ) → (v1 , . . . , vp ). But then λk xi → vi , i.e.
p
vi ∈ Ci∞ for all i = 1, . . . , p, i.e. (v1 , . . . , vp ) ∈ Xi=1 Ci∞ , which proves (1).
We now show that the converse inclusion of (1) holds, in particular, in the following
two scenarios:
• Ci is convex for all i = 1, . . . , p: By Theorem 2.4.23 we have
p
X Ci
i=1
∞
+
=0
cl
p
X Ci
i=1
+
=0
p
X Di
i=1
p
and
p
X Ci∞ = i=1
X 0+ (Di ),
i=1
where Di := cl Ci (i = 1, . . . , p). Hence, it suffices to prove that
p
p
+
0
X Di ⊃ X 0+ (Di ).
i=1
i=1
with Di closed and convex for all i = 1, . . . , p. To this end, let (v1 , . . . , vp ) ∈
Xpi=1 0+ (Di ) and take (x1 , . . . , xp ) ∈ Xpi=1 Di and λ ≥ 0. For all i = 1, . . . , p we
have vi ∈ 0+ (Di ), i.e. λvi +xi ∈ Di . Hence, we have λ(v1 , . . . , vp )+(x1 , . . . , xp ) ∈
Xpi=1 Di , which shows that (v1 , . . . , vp ) ∈ 0+ Xpi=1 Di .
• At most one set Ci is unbounded: Assume w.l.o.g. that at most C1 is unbounded
p
and take (v1 , . . . , vp ) ∈ Xi=1 Ci∞ . As, by Proposition 2.4.21, we have
p
p
X Ci∞ = C1∞ × i=2
X {0},
i=1
this means (v1 , . . . , vp ) = (v1 , 0, . . . , 0). As v1 ∈ C1∞ , there exist {x1k ∈ C1 }
and tk ↓ 0 such that tk x1k → v1 . Now take ci ∈ Ci ; (i = 2, . . . , p) and put
p
xk := (x1k , c2 , . . . , cp ). Then xk ∈ Xi=1 Ci and λk xk → (v1 , . . . , vp ), hence
∞
p
(v1 , . . . , vp ) ∈ Xi=1 Ci .
6.5 It holds that u ∈ U , x − u = u0 ∈ U ⊥ = U ◦ and hx − u, ui = hu0 , ui = 0. By
Proposition 2.5.5 we have PU (x) = u.
The linearity is a standard result from Linear Algebra.