I assume that your question concerns convex functions only; without convexity much of it would be false.
Question 2: strictly speaking, being Lipschitz smooth ($C^{1,1}$) does not imply $\nabla^2 f$ exists. But the statement is true if we interpret $\nabla^2 f\preceq LI$ as holding almost everywhere. Indeed, $\nabla^2 f$ is a positive semidefinite matrix, so having $\nabla^2 f\preceq LI$ a.e. is equivalent to $\nabla^2 f\in L^\infty$. And it is well known that having $L^\infty$ derivative is equivalent to being Lipschitz; thus $$\nabla^2 f\in L^\infty \iff \nabla f\in C^{0,1} \iff f\in C^{1,1}$$
Question 3: You misremembered. The correct inequality characterizing $\alpha$-strong convexity is
$$f(x+y) \ge f(x) + y^\top\nabla f(x) + \frac{\alpha}{2} \| x - y \|^2 \tag{1}$$
Indeed, (1) is equivalent to saying that the function $g(x)=f(x)-\frac{\alpha}{2} \| x \|^2$ is convex. The latter is equivalent to $\nabla^2 g\succeq 0$, which is $\nabla^2 f\succeq \alpha\, I$.
Question 4. Yes, there is a direct and important relation: a function is strongly convex if and only if its convex conjugate (a.k.a. Legendre-Fenchel transform) is Lipschitz smooth. Indeed, the gradients maps are inverses of each other, which implies that the Hessian of convex conjugate of $f$ is the inverse of the Hessian of $f$ (at an appropriate point). So, a uniform upper bound on $\nabla^2 f$ is equivalent to a uniform lower bound on $\nabla^2 (f^{*}) $, and vice versa. One can also argue without referring to the Hessian (which may fail to exist at some points): the Lipschitz smoothness of $f$, by your item 1, gives us at every $x_0$ a quadratic function $q$ so that $q(x_0)=f(x_0)$ and $f \le q$ everywhere. Taking convex conjugate reverses the order: $q^*\le f^*$; and this means that $f^*$ is strongly convex.
Question 1. The converse is true, but the only proof I see goes through the convex conjugate as described in Q4. Since strong convexity is characterized by the comparison property (1), taking the conjugate gives a matching characterization of Lipschitz smoothness.
Reference: Chapter 5 of Convex functions by Jonathan M. Borwein and Jon D. Vanderwerff.
Let’s take $(1)$ and write $g_1=f-\frac{\alpha}{2}\|\cdot\|_2^2$. Then $(1)$ is equivalent to $\langle \nabla g_1(y)-\nabla g_1(x),\,y-x \rangle \geq 0$, ie $g_1$ convex.
Do the same for $(2)$ with $g_2= f-\frac{\beta}{2}\|\cdot\|_2^2$.
Best Answer
Sure, you can just add the inequalities. That is, if $g$ is $m$-strongly convex then, for some $a \in \partial g(x)$ $$ g(y) \ge g(x) + a^T(y-x) + \frac{m}{2}\|y-x\|^2, $$ And if $h$ is convex, then for some $b \in \partial h(x)$: $$ h(y) \ge h(x) + b^T(y-x), $$ Then by adding we have: $$ g(y) +h(y) \ge g(x) +h(x) + (a+b)^T(y-x) + \frac{m}{2}\|y-x\|^2 $$ $$ f(y) \ge f(x) + (a+b)^T(y-x) + \frac{m}{2}\|y-x\|^2 $$ So $f$ is strongly $m$-convex and $a+b \in \partial f(x)$: