Let $F:(0,1) \to [0,\infty)$ be a strictly decreasing smooth function.
Suppose that $F$ is convex on $(0,a)$ and strictly concave on $(a,1)$.
I want to prove that we cannot make $F$ convex on $(0,1)$ by decreasing its value on $(a,1)$.
Formally, let $G:(0,1) \to \mathbb R$ be continuous, $G=F$ on $(0,a]$, and $G \le F$ on $(a,1)$.
How to prove that $G$ cannot be convex on $(0,1)$?
$F$ is strictly concave on $(a, 1)$, so that $$ F(x) < F(a) + F'(a)(x-a) $$ for $a < x < 1$.
If $G$ coincides with $F$ on $(0, a]$ and is $\le F$ on $[a, 1)$ then $$ G(x) \le F(x) < F(a) + F'(a)(x-a) = G(a) + G_-'(a)(x-a) $$ for $a < x < 1$, so that $G$ is not convex. Here $G_-'(a)$ denotes the left-sided derivative of $G$ at $a$, which is equal to $F'(a)$.
The idea is that a concave function lies below the tangent at the graph at $x=a$, and this does not change if the function values are decreased.
Remark: The statement becomes wrong if $f$ is not required to be differentiable at $x=a$. An example is $$ F(x) = \begin{cases} 1-x & \text{if } 0 < x \le 1/2 \\ 2x(1-x) &\text{if } 1/2 \le x < 1 \end{cases} $$ and $G(x) = 1-x$.