1. No category

EE5138R: Simplified Proof of Slater’s Theorem for Strong Duality
Vincent Y. F. Tan
March 6, 2015
In this document, we provided a simplified proof of Slater’s theorem that ensures strong duality holds
for convex problems.
Consider the convex primal problem
min f0 (x)
x∈Rn
s.t.
fi (x) ≤ 0, i ∈ [m].
(1)
We consider no equality constraints because if there were, each equality constraint can be cast as two separate
inequality constraints. We assume that
1. The objective f0 and all the constraint functions fi , i ∈ [m] are convex;
2. The optimal value p∗ = inf x {f0 (x) : fi (x) ≤ 0, i ∈ [m]} is finite.
Slater’s condition states that there exists a vector x
¯ ∈ dom f0 (called a Slater vector) such that
∀ i ∈ [m]
fi (¯
x) < 0,
(2)
Theorem 1. Let Assumptions 1 and 2 as well as Slater’s condition’s hold. Then:
1. There is no duality gap, i.e., d∗ = p∗ ;
2. The set of dual optimal solutions is nonempty and bounded.
Recall that the Lagrangian L : Rn × Rp → R is
L(x, λ) := f0 (x) +
m
X
λi fi (x)
(3)
i=1
and the Lagrange dual function is
g(λ) := inf L(x, λ).
x
(4)
The Lagrange dual problem is
max g(λ)
λ
s.t.
λ 0.
(5)
Slater’s theorem says that under Assumptions 1 and 2 and the existence of a Slater vector that the optimal
values of the primal in (1) and the dual in (5) are equal. By weak duality, we always have that d∗ ≤ p∗ .
Proof. Consider the set V ⊂ Rm × R given by
V := {(u, w) ∈ Rm × R : f0 (x) ≤ w, fi (x) ≤ ui , ∀ i ∈ [m], ∀ x}.
(6)
This set has several properties including: (i) it is convex; (ii) if (u, w) ∈ V, then (u0 , w0 ) ∈ V for any
(u0 , w0 ) (u, w). Convexity of V follows from the convexity of the fi , i ∈ {0} ∪ [m]. The second property
follows directly from the definition of V.
1
We first claim that the vector (0, p∗ ) is not in the interior of the set V . Suppose it is, i.e., (0, p∗ ) ∈ int(V).
Then, there exists an ε > 0 such that (0, p∗ − ε) ∈ int(V), thus clearly contradicting the optimality of p∗ .
Thus, either (0, p∗ ) ∈ bd(V) or (0, p∗ ) ∈
/ V. By the Supporting Hyperplane Theorem, there exists a
hyperplane passing through (0, p∗ ) and supporting the set V. In other words, there exists (λ, λ0 ) ∈ Rm × R
with (λ, λ0 ) 6= 0 such that
(λ, λ0 )T (u, w) = λT u + λ0 w ≥ λ0 p∗ ,
∀ (u, w) ∈ V.
(7)
This relation means that λ 0 and λ0 ≥ 0 because if there were one negative component in (λ, λ0 ), we could
make the corresponding component of (u, w) arbitrarily large and still in V (property (ii) of V) and hence
contradicting (7). Now we consider two different cases: (i) λ0 = 0; and (ii) λ0 > 0.
ˆ Case (i): The relation in (7) and λ 6= 0 implies that
λT u = 0.
inf
(8)
(u,w)∈V
On the other hand, by the definition of the set V, since λ 0 and λ 6= 0, we have
λT u = inf
inf
x
(u,w)∈V
m
X
λi fi (x) ≤
i=1
m
X
λi fi (¯
x) < 0
(9)
i=1
where x
¯ is the Slater vector and the last inequality is due to Slater’s condition. This contradicts (8)
so λ0 = 0 is not possible.
ˆ Case (ii): Hence, the only possibility is λ0 > 0. Now we may divide (7) by λ0 yielding
n
o
˜ T u + w ≥ p∗
inf
λ
(10)
(u,w)∈V
˜ := λ/λ0 0. Therefore,
with λ
(
˜ = inf
g(λ)
x
f0 (x) +
n
X
)
˜
λi fi (x) ≥ p∗ .
(11)
i=1
˜ 0, we obtain
Now if we maximize the LHS over all λ
d∗ ≥ p∗
(12)
which completes the proof of strong duality by weak duality d∗ ≤ p∗ .
˜ 0, we have
Now we show that the set of dual optimal solutions is bounded. For any dual optimal λ
(
)
m
X
˜ = inf f0 (x) +
˜ i fi (x)
d∗ = g(λ)
λ
(13)
x
i=1
≤ f0 (¯
x) +
m
X
˜ i fi (¯
λ
x)
(14)
i=1
≤ f0 (¯
x) + max {fi (¯
x)}
1≤i≤m
Therefore
"
min {−fi (¯
x)}
1≤i≤m
m
X
"m
X
#
˜i
λ
(15)
i=1
#
˜
λi ≤ f0 (¯
x) − d∗
(16)
i=1
implying that
˜ ≤
kλk
m
X
i=1
˜i ≤
λ
f0 (¯
x) − d∗
<∞
min1≤i≤m {−fi (¯
x)}
where the final equality is again due Slater’s condition.
2
(17)