Would you like to elaborate a bit? Condition (3) says the Hessian is positive definite at the origin, so by smoothness, the Hessian is positive definite in some neighborhood of the origin. On that neighborhood the function is strictly convex. What do we infer outside that neighborhood given conditions (1)-(3) ?

This is brilliant. But the inequality $f(x)>\delta^{-1} f(\delta x)$ applies to $x$ that lies in convex set that contains $0$ on which $f$ is convex. The question remains : Is the claim ''$f$ is convex '' true and what is the best way to argue that?

I edited the question. Do you find my answer correct ? Thanks

@ Davidi Cone If you find an answer to your question correct and useful, you should accept it by clicking on the "check " sign. I think your questions are interesting. Good Luck!

It is useful to recall the definition of the space $D^{1,2}$ here.

Constantin-Nicolae Beli gave a simple correct answer.