I've revised a previous question that was ill-formed. Consider the following two definitions.
Def'n 1 (Lipschitz continuity of Hessian): A function $f:\mathbb{R}^n\to\mathbb{R}$ is said to have a Lipschitz continuous Hessian if there exists $L>0$ such that $\|\nabla^2f(x) - > \nabla^2f(y)\|\le L\|x-y\|$.
Def'n 2 (Self-concordance): Let $\Omega\subseteq\mathbb{R}^n$ be a convex open set. Then $f:\Omega\to\mathbb{R}^n$ is said to be self-concordant with parameter $\sigma>0$ at $x\in\Omega$ if \begin{align} \nabla^2f(x)(h,h) \ge 0 \end{align} and \begin{align} [\nabla^3f(x)(h,h,h)]^2 \le 4\sigma(h^\top\nabla^2f(x)h)^3 \end{align} for all $h\in\mathbb{R}^n$; $f$ is self-concordant on $\Omega$ if the above two inequalities hold for all $x\in\Omega$.
Note: The term $\nabla^3f(x)(h,h,h)$ means $\sum_{i,j,k=1}^n\frac{\partial^3 f(x)}{\partial x_i\partial x_j\partial x_k}h_ih_jh_k$.
The condition for self-concordance of a function is quite difficult to check, motivating a search for an alternate or sufficient condition.
My attempt:
Consider a three times continuously differentiable function $f$ that is strongly convex and has a Lipschitz continuous Hessian. I imagine a condition that looks something like the following may be enough for self-concordance but I'm not sure if it makes sense:
$f$ is self-concordant with parameter $\sigma>0$ if for all $x,y$ \begin{align} \|\nabla^2f(x)-\nabla^2f(y)\|\le2\sigma(h^\top\nabla^2f(x)h)^{3/2} \end{align} for all $h\in\mathbb{R}^n$.
That is, the Hessian of $f$ does not change too much (bounded by some function of the Hessian itself). Does anyone have any further insight?