A convex function has a lower bound?

Question

Suppose that $f=f(x)$ is strictly convex for $x\in\mathbb{R}$, i.e. there exists $\epsilon>0$ such that $f''(x)\geq\epsilon>0$ for $x\in\mathbb{R}$. Does there exist $\delta>0$ such that $f(x)\geq \delta$ for $x\in\mathbb{R}$?

Answer 1

Use the mean value theorem. For any $x > 0$ you have $f'(x) \ge \epsilon x + f'(0)$ and for any $x < 0$ you have $f'(x) \le \epsilon x + f'(0)$. This means there exists $M > 0$ with the property that $f$ is increasing on $(M,\infty)$ and decreasing on $(-\infty,-M)$. It follows that $f$ attains a minimum in the interval $[-M,M]$.

Answer 2

Alternatively, you can integrate the inequality $f'' \ge \epsilon$ twice to get an explicit lower bound in terms of the function and its derivative evaluated at a point.

That is, by integrating once from some fixed point $x_0$, we get: $$ \begin{aligned} \int^{x}_{x_0} (f''(t) - \epsilon)dt &\ge 0\\ \Rightarrow (f'(x) - f'(x_0)- \epsilon(x-x_0))\operatorname{sign}(x-x_0) &\ge 0 \end{aligned} $$ Likewise, integrating again and completing the square, we get: $$ \left(f(x) - f(x_0) - f'(x_0)(x-x_0) - \frac{\epsilon}{2}(x-x_0)^2 \right) \operatorname{sign}(x-x_0)^2\ge 0 $$ $$ \begin{aligned} \Rightarrow f(x) &\ge f(x_0) + f'(x_0)(x-x_0) + \frac{\epsilon}{2}(x-x_0)^2\\ &=f(x_0) - \frac{f'(x_0)^2}{2\epsilon} + \frac{\epsilon}{2}\left(x-x_0 + \frac{f'(x_0)}{\epsilon}\right)^2\\ &\ge f(x_0) - \frac{f'(x_0)^2}{2\epsilon} \end{aligned} $$ So for any $x_0 \in \mathbb{R}$, you can take $\delta = f(x_0) - \frac{f'(x_0)^2}{2\epsilon}$.

Note, you specify in the question that $\delta > 0$, but in general this will not be true, as shown by the example $f(x)=x^2-1$.

Answer 3

A function satisfying the condition given in this question is called "strongly convex".

There is a more general definition of strong convexity which applies to functions that may not be differentiable: A function $f:\mathbb R^N \to \bar{\mathbb R}$ is strongly convex (with parameter $\epsilon > 0$) if and only if the function $h(x) = f(x) - \frac{\epsilon}{2} \|x\|_2^2$ is convex.

It would be nice to give a version of this result that doesn't require $f$ to be differentiable.

Assume that $f:\mathbb R^N \to \mathbb R \cup \{\infty\}$ is strongly convex (with parameter $\epsilon$) but not necessarily differentiable. The function $h(x) = f(x) - \frac{\epsilon}{2} \|x\|_2^2$ is convex and never equal to $-\infty$, therefore $h$ has an affine minorant: there exists $m \in \mathbb R^N$ and $b \in \mathbb R$ such that $h(x) \geq \langle m, x \rangle + b$ for all $x \in \mathbb R^N$. It follows that \begin{equation} f(x) \geq \frac{\epsilon}{2} \|x\|_2^2 + \langle m, x \rangle + b \end{equation} for all $x \in \mathbb R^N$. The quadratic function on the right has a lower bound, so $f$ has a lower bound.