Variational methods - prove f is convex in R->R

[braindead101]

Suppose f:R^N -> R is twice differentiable. Prove that f is convex if and only if its Hessian gradiant^2 f(x) is nonnegative.

How do I go about proving this? and my professor said I only need to consider when N=1. so R->R.
any help would be greatly appreciated.

For proving it backwards, this is what i have, but i am not sure if it is correct.
If the Hessian of F is nonnegative definite, then the function is locally strictly convex. A function that is locally strictly convex everywhere is strictly convex.

and I am not sure how to prove it other way

 

[mighty2000]

around.

To prove that f is convex if its Hessian is nonnegative, we can use the definition of convexity. A function f is convex if for any two points x and y in its domain, the line segment connecting f(x) and f(y) lies above or on the graph of f. In other words, f(tx + (1-t)y) <= tf(x) + (1-t)f(y) for all t in [0,1].

Now, let's consider the Hessian matrix of f at some point x. This matrix is defined as the matrix of second partial derivatives of f at x, denoted as H(x). Since f is twice differentiable, H(x) exists and is a symmetric matrix.

If H(x) is nonnegative, it means that all its eigenvalues are nonnegative. This implies that H(x) is positive semidefinite, which in turn implies that f is locally convex at x. This means that for any two points x and y in the domain of f, the line segment connecting f(x) and f(y) lies above or on the graph of f in a small neighborhood around x. Since this holds for all x, we can conclude that f is globally convex.

To prove that f is strictly convex, we can use the fact that a function is strictly convex if and only if its Hessian matrix is positive definite. So, if the Hessian of f is nonnegative, it means that all its eigenvalues are positive or zero. This implies that f is strictly convex at every point in its domain, and therefore, globally strictly convex.

In conclusion, we have shown that if the Hessian of f is nonnegative, then f is convex and strictly convex, which proves the statement in both directions.