Is there a "global" convexity locally around a minimum?

$$\newcommand{\til}{\tilde}$$

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 $F|_{[1-\epsilon,1+\epsilon]}$ is strictly convex, for some $\epsilon>0$.

Question: Does there exist a $\delta>0$ such that $F$ is convex at every point $y \in (1-\delta,1)$?

By convexity at $y\,\,$ I mean that for any $x_1,x_2>0, \alpha \in [0,1]$ satisfying $\alpha x_1 + (1- \alpha)x_2 =y$, $$ F(y)=F\big(\alpha x_1 + (1- \alpha)x_2 \big) \leq \alpha F(x_1) + (1-\alpha)F(x_2). \tag{1} $$ Equivalently, $F$ admits a supporting line at $y$, i.e. $\exists m \in \mathbb{R}$ such that $$ F(x) \ge F(y)+m (x-y) \, \, \, \text{ for every } \, \, x \in (0,\infty). \tag{2} $$

Edit: I have a proof for a positive answer if $F \in C^1$, and an incomplete argument for the case where we do not assume $F \in C^1$. I am interested to know the answer when $F$ is merely continuous.

(Feel free to skip over my proofs/arguments below. The question stands as it is.)

A (hopefully correct) proof assuming $F \in C^1$:

If there is no such $\delta$, there exists $y_n \in (0,1)$, $y_n \to 1$ such that $F$ is not convex at $y_n$. Thus, $\exists x_n \in (0,\infty)$ such that $$ F(x_n) < F(y_n)+F'(y_n) (x_n-y_n) \tag{3} $$ The tangent at $y_n$ is below the $x$-axis for $x>1$ (since $F$ is convex at $(y_n,1]$ it decreases more slowly than its tangent after the tangency point). Thus, $F>0$ is above its tangent which is negative for $x>1$, which implies $x_n \in (0,1)$.

We may assume that $x_n \to x_0$; taking limits on both sides of $(3)$, we obtain $F(x_0) \le F(1)=0$, since $|F'(y_n) (x_n-y_n)| \le |F'(y_n)| \to 0$.

This implies $x_0=1$, so both $x_n,y_n \to 1$. Looking again at inequality $(3)$, this contradicts the convexity of $F|_{[1-\epsilon,1+\epsilon]}$.

We have used $F'(y_n) \to F'(1)=0$. If we do not assume $F \in C^1$, we need to replace $F'(y_n)$ with slopes of supporting lines $m_n$, which do not necessarily converge to zero.

An incomplete proof without assuming $F \in C^1$:

Assume there is no such $\delta$. Then $\exists s_n \in [0,1]$, $s_n \to 1$ such that $F$ is not convex at $s_n$.

Thus $\exists x_n,y_n \in (0,\infty), \alpha_n \in [0,1]$, $x_n \le s_n \le y_n$ such that $$ s_n=\alpha_n x_n + (1- \alpha_n)y_n, \, \, \text{ and } \, \, F\left(s_n \right) > \alpha_n F(x_n) + (1-\alpha_n)F(y_n). \tag{4} $$

$x_n \le s_n \le 1$ so $x_n$ is bounded. W.L.O.G we may assume that $y_n \le 1$. Indeed, if $y_n >1$, we can replace it with $\til y_n=1$, and choose $\til x_n$ such that $\alpha_n \til x_n + (1- \alpha_n)\til y_n=s_n$. Then $x_n \le \til x_n \le s_n \le 1=\til y_n \le y_n$, hence $F(\til x_n) \le F(x_n), F(\til y_n) \le F(y_n)$. Then inequality $(4)$ holds with $x_n,y_n$ replaced by $\til x_n,\til y_n$.

We now have $x_n \le s_n \le y_n \le 1$, and $s_n \to 1$ so we may assume that $x_n \to x, y_n \to 1, \alpha_n \to\alpha$. Taking limits of inequality $(4)$ we get $$ 0=F(1) \ge \alpha F(x) + (1-\alpha)F(1) = \alpha F(x)\ge 0, $$ so $\alpha F(x)=0$. If $\alpha>0$, then $F(x)=0$ and $x=1$, so $x_n,y_n,s_n \to 1$, thus they eventually lie at $(1-\epsilon,1]$, which together with inequality $(4)$ contradicts the convexity of $F|_{[1-\epsilon,1+\epsilon]}$.

The problem is that if $\alpha=0$ we cannot deduce that $x=1$!

A counterexample: $$ F(x) = \begin{cases} \frac 74 - x & \text{ for } 0 \le x \le \frac 12 \, ,\\ x^2-4x+3 & \text{ for } \frac 12 \le x \le 1 \, ,\\ (x-1)^2 & \text{ for } 1 \le x \, . \end{cases} $$ $F$ is strictly decreasing on $[0, 1]$ and strictly convex on $[1/2, \infty)$.

Now assume that there are supporting lines $$ F(x) \ge F(y) + m_y(x-y) $$ for all $y \in (1-\delta, 1)$. Then $m_y \le F_-'(1) = -2$ so that for $0 < x < 1/2$ $$ F(x) \ge F(y) - 2(x-y) \implies F(x) \ge F(1) -2(x-1) = 2-2x $$ and that does not hold for $x$ close to zero.

Here is a plot with the function $F$ (blue) and the lower bound for all supporting lines (red):