Do the slopes of supporting lines tend to zero at a minimum point? Let  $F:(0,\infty) \to [0,\infty)$ be a continuous function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$. Suppose also that $F$ is convex on $[1-\epsilon,1]$, for some $\epsilon>0$.
Let $y_n \in [1-\epsilon,1)$, $y_n \to 1$, and suppose $m_n$ is a slope of a supporting line to $F|_{[1-\epsilon,1]}$ at $y_n$, i.e.
$$
F(x) \ge F(y_n)+m_n(x-y_n) \, \, \, \text{ for every } \, \, x \in [1-\epsilon,1].
$$
Is it true that $\lim_{n \to \infty}m_n=0$?
By plugging $x=1$, we get $0=F(1) \ge F(y_n)+m_n(1-y_n)$ so $m_n<0$.

If $F$ is differentiable at $y_n$, then this forces $m_n=F'(y_n)$, so if $F$ is $C^1$ near $x=1$, we have $m_n=F'(y_n) \to F'(1)=0$. However, I don't want to assume that $F$ is differentiable.
 A: Not under the hypotheses given.  Consider $F(x) = |x-1|$.  In this case, the unique supporting line for $y_n \in [1-\varepsilon, 1]$ is 1, so $m_n$ is a constant sequence.
More generally, I think you're after a sequential lower hemicontinuity condition for the subdifferential of $F\vert_{[1-\varepsilon, 1]}$.  As a correspondence the subdifferential will be (cyclically) monotone, hence every selection from it will be nondecreasing, but absent stronger conditions not every value of $\partial F\vert_{[1-\varepsilon, 1]}(0)$ will be obtainable as a limit (from the left) of a selection.
A: The slopes $m_n$ increase with increasing $y_n$ so that $\lim_{n \to \infty}m_n$ exists. But
$$
 F_-'(y_n) \le m_n \le F_+'(y_n) \le F_-'(1)
$$
so that limit is strictly negative if the left derivative $F_-'(1)$ is negative.
In fact one has $\lim_{n \to \infty}m_n = F_-'(1)$. If not, then $m_n \le a < F_-'(1)$ for all $n$. Then
$$
 F(x) \ge F(y_n)+m_n(x-y_n) \ge F(y_n) + a(x-y_n)
$$
for $1-\epsilon < x < y_n$. It follows that $F(x) \ge F(1) + a(x-1)$ for $1-\epsilon < x < 1$ and therefore $F_-'(1) \le a$, contrary to the assumption.
A: No.  Consider $F(x)=\lvert x-1\rvert+(x-1)^2$, which is strictly convex on $\mathbb{R}$ and differentiable except at $x=1$.  We have obviously $\lim_{n\to\infty} m_n=-1$ coming from the $\lvert x-1\rvert$ for $x<1$.
