About strong convexity and beyond

Question

Let $\Phi:\mathbb R^{d}\to\mathbb R$ be a $C^2$ function. Consider the following conditions:

1. $\Phi=\Phi_1+\Phi_0\,$, where $\Phi_1\in C^2$ is strongly convex, namely there exists $c>0$ such that for all $x\in\mathbb R^d$ and all $u\in\mathbb R^d$ $$ u^T\,Hess\Phi_1(x)\;u \,\geq\, c \,|u|^2\;,$$ and $\Phi_0$ is bounded.

2. There exists $c>0$ such that $$ x^T\, \nabla\Phi (x) \,\geq\, c\,|x|^2 \,+\, \omega(x)$$ where $\omega(x)/|x|^2\to0\,$ as $|x|\to\infty\,$.

I am trying to figure out if there is a relation among conditions $1.$ and $2.$, such as $1.\Rightarrow 2.\,$ or $2.\Rightarrow 1.\,$ or maybe $1.+\textrm{"some extra condition"}\Rightarrow 2.\,$

I have seen many examples of functions that satisfy both conditions, so this seems a natural question to me.

Example. $\Phi(x) = |x|^2 + \Phi_0(x)$ where $\Phi_0(x)$ is bounded with bounded gradient.

Example. $\Phi(x) = |x|^4 - |x|^2 + \tilde\Phi_0(x)$ where $\tilde\Phi_0(x)$ is bounded with bounded gradient. Notice that in this case one needs to mollify $|x|^4$ around $0$ is order to obtain a strongly convex function.

Answer 1

Property 1 does not imply property 2. Intuitively, you cannot expect any bound on first derivatives for $\Phi_0 + \Phi_1$ to follow from mere boundedness of $\Phi_1$ plus some conditions for $\Phi_0$.

To show concretely that 1. does not imply 2., let $d = 1$ and consider $\Phi_0(x) = x^2/2$ and $\Phi_1(x) = \cos x^2$. Then $x\Phi'(x)$ is not bounded from below which contradicts 2.

Regarding the implication 2 $\Rightarrow$ 1: This has the flavor of an approximation problem, namely "given a function $\Phi$ satisfying 2, find a strongly convex function $\Phi_1$ such that $\Phi - \Phi_1$ is bounded". I doubt that that is always possible, but it's a harder question, since $\Phi_0, \, \Phi_1$ are unspecified.