These notes are based on Boyd and Vandenberghe. Here, I have suppressed the domains of functions everywhere for simplicity.

**Definition**: is convex if

when and .

Or, informally, a function is convex if a “chord” between two points on the function always lies above the function.

A function is convex if any only if the restriction to a line is always convex. This follows pretty easily from the definition:

is convex if and only if is convex for any and .

A *differentiable* function, it is convex if any only if a first-order approximation is a global underestimator:

A *twice differentiable* function is convex if any only if the Hessian is positive definite at all points.

.

(This is often the easiest way to check for convexity.)

The *epigraph* of is roughly the set of all points “above the function”.

.

is convex if any only if is a convex *set*.

**Operations that preserve convexity**:

* If is convex and , then is convex.

* If and are convex, so is . (This also works for infinite sums or integrals.)

* If is convex, so is .

* If are convex, then is convex.

* If is convex in (for all ), then is convex.

* Suppose . (So is just a scalar function.) Then, is convex if either A) is convex and is convex and nondecreasing. or B) is concave and is convex and nonincreasing. (Be careful if h has a limited domain.)

* Suppose . (So is a function with a vector domain.) Then, is convex if either A) is convex and is convex and nondecreasing in each argument. or B) is concave and is convex and nonincreasing in each argument. (Be careful if h has a limited domain.)

* If is convex, and is a convex set, then is convex.

* Suppose . If is convex, then so is (restricted to the domain ).

A function is convex if any only if the restriction to a line is always convex. This follows pretty easily from the definition:

is convex if and only if is convex for any and

Would you please explain why this definition is formed to me?

I want to know origin of this definition

please send e-mail to me..

I highly recommend Boyd and Vandenberghe’s book:

http://www.stanford.edu/~boyd/cvxbook/

I read the book.

Also I attended the lecture.

But he had assigned proving the upper definition to their students so I coudln’t find the reason I wanted to know.

Could you please prove the upper definition again for me?

Thank you.

Well… OK. First, assume is convex, and let’s prove is convex. We have that

.

Now, instead assume is convex for all choices of . Then,

(where in the definition of ).

In all honesty that wasn’t as trivial as I remembered it being…