determining convexity

Question 1

Convex or not, why?
a) f(x,y)=exp(3x+2y2)

b) f(x,y)=21x2+exp(−x)

Answer 1

f(x,y)=exp(3x+2y2)
Note that exp(x) is an increasing convex function, as its first and second derivatives both equal exp(x), which is positive on the entire real axis. Moreover, x, y2 are both convex functions (check their second derivatives), and 2, 3 > 0, which implies that 3x + 2y2 is a convex function by T10.1(1) and (2). I conclude that f (x, y) is also convex by T10.1(4).

−x is convex, exp(x) is convex and increasing, so exp(−x) is convex by T10.1(4). x2 is convex, and
so too is x2/2 by T10.1(1). Thus, by T10.1(2), f(x,y) is convex.

Question 2

c) f(x) = |x|
…
d) f(x) = 3x5n for even n
…

Answer 2

The absolute value function can be written, |x| = max{x, −x}. Both x, −x are convex functions, so the convexity of the absolute value function follows from T10.1(3).

Since n is even, I may write f(x) = 3x10m = 3 · (x2)5m for some positive number m. Now, x2 is a positive valued convex function, and x5m is convex on the domain [0, ∞) and increasing. Therefore, by T10.1(4) x10m is convex, and since 3 > 0 so too is 3x10m by T10.1(1).