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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?

Asaf Shachar
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    I think $\phi:\Bbb R\to[0, \infty)$ given by $\phi(x) = \sqrt{|x|}$ will give you a counterexample by considering $c = 0$. To suit your requirement for $\phi:(0, \infty) \to [0, \infty)$, I think you can get it by shifting the graph. Something like $x\mapsto \sqrt{|x-1|}$ and $c = 1$. – Aryaman Maithani Jul 06 '20 at 08:22
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    Thanks, you are right of course! Convexity at a point $c$ always holds when $c$ is a global minimum of the function. So I guess I am really interested in the case where $\phi$ is strictly decreasing around $c$. (I have edited the question to reflect this). Thank you again for this observation. – Asaf Shachar Jul 06 '20 at 08:34
  • Maybe you could look at the Slater's inequality for convex function (1981).Maybe there is an interesting link see also the condition of the theorem.good day and good luck !! – Miss and Mister cassoulet char Jul 06 '20 at 12:17

2 Answers2

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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.)
The graph

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).

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

Aryaman Maithani
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  • Thank you very much. This is a very nice counter-example. I guess you defined the last part to be constant $-1$ (and not continued with $-x$) in order to make sure $\phi$ is non-negative, as I required? Also, I wonder how did you think about this example? Did you have any geometric/visual insight that you can describe? – Asaf Shachar Jul 06 '20 at 18:18
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    Yes, you are correct. (In fact, I only realised the non-negative constraint when I started typing the answer.) As for the second part, I was trying to stick to a counterexample like the first time where the limit exists as $\pm\infty$ and not something more pathologically weird. So I just fiddled with a semicircle on one side and something else on the other side such that the line segment joining points from one side to the other would always be above $0$. I also had in mind that the function would have to have an abrupt change and it can't be convex on the interval. – Aryaman Maithani Jul 06 '20 at 18:25
  • That being said, I am interested in seeing a function where the one sided limits truly do not exist. (That is, not even in the extended real line.) – Aryaman Maithani Jul 06 '20 at 18:26
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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 $-1< x \le 0$.

Asaf Shachar
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