Does convexity at a point imply existence of one-sided derivatives?

Question

Let $\phi:\mathbb (0,\infty) \to [0,\infty)$ be a continuous function, and let $c \in (0,\infty)$ be fixed.

Suppose that "$\phi$ is convex at $c$". i.e. for any $x_1,x_2>0, \alpha \in [0,1]$ satisfying $\alpha x_1 + (1- \alpha)x_2 =c$, we have $$ \phi(c)=\phi\left(\alpha x_1 + (1- \alpha)x_2 \right) \leq \alpha \phi(x_1) + (1-\alpha)\phi(x_2) . $$

Assume also that $\phi$ is strictly decreasing in a neighbourhood of $c$.

Do the one-sided derivatives $\phi'_{-}(c),\phi'_{+}(c)$ necessarily exist?

Edit:

As pointed by Aryaman Maithani if $c$ is a global minimum of $\phi$, then clearly $\phi$ is convex at $c$, but there should be no reason to expect for existence of one-sided derivatives. (e.g. $\phi(x)=\sqrt{|x|}, c=0$).

Edit 2:

In the example described here, the left derivative does not exist. Can we create an example where the right derivative does not exist?

Answer 1

Define $\phi:(-1, \infty) \to [-1, \infty)$ as $$\phi(x) = \begin{cases} \sqrt{1 - (1+x)^2} & x \le 0\\ -x & 0 \le x \le 1 \\ -1 & 1 \le x\end{cases}$$

A graph is shown below. (Courtesy of Desmos.)

Clearly, $\phi$ is continuous and strictly decreasing in $(-1, 1)$. Thus, choosing $c = 0$ satisfies the conditions. (It has to be shown that $\phi$ is convex at this point but that is simple.)
However, the limit $\displaystyle\lim_{x\to0^-}\phi'(x)$ does not exist (as a real number).

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To meet the conditions of your domain and codomain, consider $\tilde \phi := [x \mapsto \phi(x-1)+1].$

Answer 2

This answer is merely an attempt to fill in the details in the example described here. Convexity of $\phi$ at $0$ means that

$$ 0=\phi(0) \leq \alpha \phi(x) + (1-\alpha)\phi(y), \tag{1} $$ for every $-1< x \le 0 \le y \le 1$ satisfying $$ \alpha x + (1- \alpha)y =0. \tag{2} $$ In particular, for every $-1<x \le 0 \le y \le 1$, we should have $$ 0 \le \alpha \sqrt{1 - (1+x)^2} + (1-\alpha)(-y)=\alpha\big( \sqrt{1 - (1+x)^2} +x\big). $$ This is equivalent to $$ x^2+x=x(x+1) \le 0, $$ which holds since $-1<x\le 0$.

Now, suppose that $-1< x \le 0 \le 1 \le y $. The inequality $(1)$ holds if and only if $$ 0\leq \alpha \sqrt{1 - (1+x)^2} + (\alpha-1). $$

we also have $0 \ge -\alpha x=(1-\alpha)y\ge (1-\alpha) \Rightarrow (\alpha-1) \ge \alpha x$, so $$ \alpha \sqrt{1 - (1+x)^2} + (\alpha-1) \ge \alpha \big(\sqrt{1 - (1+x)^2} + x\big) \ge 0 $$ holds as before for $-1< x \le 0$.