Name: aqSeabiscuit Date: 2020/4/14
Second-order sucient optimality conditions
Problem:
Suppose that f : R
n
→ R is twice dierentiable at x. Please show that x is a strict local
minimum if ∇f(x) = 0 and the Hessian matrix H(x) is positive denite.
Solve:
First consider the situation where n = 2, then the function f : R
n
→ R can be written as
f(x
1
, x
2
), x
1
, x
2
∈ R
2
. Suppose that f is twice dierentiable at ˆx, where ˆx is ( ˆx
1
, ˆx
2
), then
we can expand the function f (x) near ˆx with the help of Taylor series up to second-order. [1]
It can be given as
f (x) = f (ˆx) +
∂f
∂x
1
ˆx
∆x
1
+
∂f
∂x
2
ˆx
∆x
2
+
1
2
∂
2
f
∂x
2
1
ˆx
∆x
2
1
+ 2
∂
2
f
∂x
1
∂x
2
ˆx
∆x
1
∆x
2
+
∂
2
f
∂x
2
2
ˆx
∆x
2
2
(1)
where ∆x
1
= x
1
− ˆx
1
, ∆x
2
= x
2
− ˆx
2
.
Turn the Taylor expansion above into a matrix form, then
f (x) = f (ˆx) +
∂f
∂x
1
∂f
∂x
1
ˆx
∆x
1
∆x
1
+
1
2
(∆x
1
∆x
2
)
∂
2
f
∂x
2
1
∂
2
f
∂x
1
∂x
2
∂
2
f
∂x
2
∂x
1
∂
2
f
∂x
2
2
ˆx
∆x
1
∆x
1
(2)
Equivalently, it’s vector form can also be given as
f (x) = f (ˆx) + ∇f ( ˆx) ∆x +
1
2
∆x
T
H (ˆx) ∆x (3)
where
∆x = [∆x
1
∆x
2
]
T
(4)
∇f (ˆx) =
h
∂f
∂x
1
∂f
∂x
2
i
(5)
1
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