Answers HW2 (Partial)
A4.d
f (x) =
x1
x2
Applying Jensens Inequality
?
f (x+ (1 ) y) f (x) + (1 ) f (y)
Let x =
f
x1
y
and y = 1 where x, y R2 , then
x2
y2
x1 + (1 ) y1
x2 + (1 ) y2
x1
x2
?
f
+ (1 ) f
y1
y2
y1
x1 + (1 ) y1 ? x1
+ (1 )
x2 + (1 ) y2
x2
y2
x1 + (1 ) y1
x1
y1 ?
(1 )
0
x2 + (1 ) y2
x2
y2
?
(x1 + (1 ) y1 ) x2 y2 x1 y2 (x2 + (1 ) y2 ) (1 ) x2 y1 (x2 + (1 ) y2 ) 0
?
2
(1 ) x1 x2 y2 + x2 y1 y2 x1 y2 x2 y1 0
2
?
(1 ) (x1 y2 x2 y1 ) (x2 y2 ) 0
(1 ) is always non-negative, moreover there is no guarantee that (x1 y2 x2 y1 ) (x2 y2 )
is negative denite, e.g. let x1 = 1, x2 = 1, y1 =,3 y2 = 2, then (x1 y2 x2 y1 ) (x2 y2 ) =
1, hence inequality test fails.
f (x) =
x1
is not convex!.
x2
A4.e
f (x) =
x2
1
x2
Applying Jensens Inequality
?
f (1 (x+ ) y) f (x) + (1 ) f (y)
Let x =
f
x1
y
and y = 1 where x, y R2 , then
x2
y2
x1 + (1 ) y1
x2 + (1 ) y2
?
f
1
x1
x2
+ (1 ) f
y1
y2
2
(x1 + (1 ) y1 ) ? x2
y2
1 + (1 ) 1
x2 + (1 ) y2
x2
y2
?
2
2
(x1 + (1 ) y1 ) x2 y2 x2 y2 (x2 + (1 ) y2 ) (1 ) x2 y1 (x2 + (1 ) y2 ) 0
1
?
2
2
2
2
x2 y2 + 2x1 x2 y1 y2 x2 y1 + x2 y2 2x1 x2 y1 y2 + x2 y1 0
1
2
1
2
2?
2
(x1 y2 x2 y1 ) + (x1 y2 x2 y1 ) 0
2?
( 1) (x1 y2 x2 y1 ) 0
2
: (x1 y2 x2 y1 ) is always non-negative, ( 1) is always non-positive, hence
inequality holds.
x2
f (x) = 1 is convex!.
x2
You can repeat the same procedure for other parts also to nd
4.a is convex,
4.b is neither convex not concave,
4.c is convex,
4.d is neither convex nor concave,
4.e is convex, and
4.f is concave.
2