How do I find out when a function is strongly convex?

Question

My professor said( i might have misunderstood) That if I have a function,and i take the second derivative,and somehow I get 0,hen it is not strongly comvex!

Example x^2

strongly comvex because its second derivative = 2

example x^4

not strongly convex becasue second derivative 2*x^2, and when x=0, the equation is 0

Now what about 1/2*x^2+x^4??

after the second derivative I get 1+12*x^2, if I set x=0 I get 1,if i set x=0,its not 0,i dont know is ters anyway this expression could be 0,does it mean it is srongly convex???

This question appears to be off-topic because it is about calculus, not programming, and would probably be better asked at http://math.stackexchange.com/ — Mureinik, Aug 16 '14 at 10:35

score 2 · Accepted Answer · answered Aug 16 '14 at 10:53

if I set x=0 I get 1,if i set x=0,its not 0

That's not what you're looking for. You need to check whether the expression is 0 for any x (which for the 2x² just happened to be x=0).

i dont know is ters anyway this expression could be 0

The typical way to solve this is to try:

1 + 12 x^2 = 0
12 x^2     = -1
x^2        = -1/12
x          = ↯

Since you can't take the square root of a negative number, there is no way this expression can yield 0.

does it mean it is srongly convex???

…which means that 1/2*x^2+x^4 is strongly convex.

How do I find out when a function is strongly convex?

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