## Show a function is strictly convex and coercive

So we know f is convex on R^n
If f(x) has continuous first partial derivatives and E>0, then g(x)=f(x)+E*||x||^2 is strictly convex and coercive.

So i started with my definition of f being convex, and I want to show g is convex.

so f(ax+by)<= af(x)+bf(y), where a,b>0 and a+b=1
then adding E*||ax+by||^2 to both sides, I get g(ax+by)<=af(x)+bf(y)+E*||ax+by||^2

(btw, I also know ||x|| is convex because ||ax+by||<=|a| ||x|| + |b| ||y||)

I think I need to show that E*||x||^2 is strictly convex, then g(x) would be strictly convex since it's the addition of a convex and strictly convex function, but I can't prove that ||x||^2 is strictly convex....or that my original g(ax+by)<af(x)+bf(y)+Ea||x||^2 + Eb||y||^2

For the second part, again I know ||x||^2 is coercive since it goes to infinity as ||x|| goes to infinity...but I don't know where f(x) goes or even if its bounded. I need g(x) to go to infinity as ||x|| goes to infinity

Any ideas would be greatly appreciated!! Thanks!