Show that f is strictly convex

Question

For $\lambda >0$, I need to prove that $f(b) = ||Y-Xb||^2 + \lambda ||b||^2$ ( with $ b \in R^p $) is strictly convex and has a unique minimum I know that a norm is convex. But how can I use it to prove that f is a strictly convex?

Answer 1

It is a continuous function and it is coercive, since as $\|b\|\rightarrow \infty$ your function goes to infinity too. Thus, since it is convex, it is lower bounded and it has hence a minimizer. The uniqueness comes from the strict convexity. To prove it, you can notice that the function is twice continuously differentiable and its hessian is $$ \nabla^2 f(b) = X^TX+\lambda. $$ Since $\lambda >0$, this hessian is positive definite, and hence $f$ is strictly convex.

You can now conclude the existence and uniqueness of the minimizer.