Strong convexity inequality implied by a second derivative which is positive definite outside a convex set

Question

I am interested in having a proof of the following result:

Let $f: \mathbb{R}^n \to \mathbb{R}$ be a $C^2$ function satisfying $$ \frac{\partial ^2 f}{\partial x\partial x}(x) \geq \underline{f} >0 \qquad \forall x: f(x) > 0 $$ There exists a strictly positive real number $b$ such that $$ f(x_1) \geq f(x_2) + \frac{\partial f}{\partial x}(x_2) (x_1-x_2) + b |x_1-x_2|^2 \qquad \forall (x_1,x_2) : f(x_i)> 0 $$

Remarks:

1- The difficulty is in the fact that the property on the second derivative holds only when $f(x) > 0$

2- I am expecting the result to hold with $b=\frac{\underline{f}}{2}$

3- An argument on why this result should hold is that I can prove it if the restriction $f(x) > 0$ is removed. Then, since the desired inequality does not involve anything in the set $\{x : f(x) \leq 0\}$, I can modify $f$ inside this set without altering the inequality. Going along these lines, I could use a theorem on extension of convex functions as in M. Yan, Extension of convex functions, Journal of Convex Analysis 21 (2014), No. 4, pp. 965-987 But this does not give me the bound $b=\frac{\underline{f}}{2}$

Answer 1

Partial answer:

2 is not true. That is, $b$ cannot be purely a function of ${\underline {f}}$.

Define $f_T(x) = \begin{cases} (x+T)^2-1, & x\le -T \\ (x-T)^2-1, & x \ge T \\ g_T(x),& \text{otherwise} \end{cases}$,

where $g_T$ is chosen so that $g_T<0$, and $f_T$ is smooth.

Note that $f_T(x) \ge 0$ iff $|x| \ge T+1$. and $f''_T(x) = \underline{f} = 2$ for $|x| \ge T$.

We have $f_T(-T-1) = 0 = f_T(T+1) = 0$ and $f'_T(-T-1) = -2$.

The above hold for all $T >0$.

Letting $x_2 = -T-1, x_1 = T+1$ gives $0 \ge -4(T+1)+ 4b (T+1)^2$.

If $b$ was purely a function of $\underline{f}$ then the above would have to hold for all $T>0$ which cannot be true.

Answer 2

The counter-example of copper.hat is excellent. Indeed it made me realize that the desired inequality imposes an upper bound for the diameter of the level set $S_y=\{x: f(x)=y\}$ for $y>0$. This bound is $\frac{1}{b} \sup_{x\in S_y} \left| \frac{\partial f}{\partial x}(x)\right|$. For the counter example, with $y=0$, the bound is $\frac{2}{b}$ while the diameter is $2(T+1)$. So a necessary condition is $b\leq \frac{1}{T+1}$, as already obtained by copper.hat.

The question of validity of the inequality remains, but the guess for $b$ is more complex, being apparantly, related to the curvature of the level sets of $f$