Basic question: why and how smooth convex function $f(x)$ with domain $R^n$ is equivalent to $g(x):= \frac{L}{2} x^T x - f(x)$?

Question

I am so sorry to ask probably the most trivial and fundamental question. But it is just bothering me and not able to understand,

why and how smooth convex function $f(x)$ with domain $R^n$ is equivalent to another convex function $g(x):= \frac{L}{2} x^T x - f(x)$?

(reference: page 1-11 of http://www.seas.ucla.edu/~vandenbe/236C/lectures/gradient.pdf)

I am so confused. If I minimize the function $f(x)$, then I need to maximize the function $g(x)$, right? or I have confused myself more.

score 1 · Accepted Answer · answered Jan 24 '19 at 17:55

No wonder you're confused, since you're reading it all wrong.

The claim that they make is that “this” (which refers to the inequality at the top of the page, not to the function $f$) is equivalent to the condition that $\frac{L}{2} x^T x - f(x)$ is convex, provided that $f$ is convex with domain $\mathbf{R}^n$.

(What would it even mean for a function $f$ to be equivalent to another function $g$ in this context?)

And it also says “we will see that”, meaning (I suppose) that you will get an explanation of why this is true later in the text, or in a later lecture perhaps (since this seems to be lecture notes). So just keep reading on and see if it gets clearer.

Oh yes, thank you so much for the clarification. I started to think it is some kind of convex conjugate :|.. As you said, I must continue and try to understand ... — learning, Jan 24 '19 at 18:30

Basic question: why and how smooth convex function $f(x)$ with domain $R^n$ is equivalent to $g(x):= \frac{L}{2} x^T x - f(x)$?

1 Answers1