$g$ - Convex function

Question

For $x>0$, let $y=f(x)$ be a function satisfying $f'''(x) \geq 0$ (third derivative). Prove that $g(x)=\dfrac{f(x)-f(2m-x)}{x-m}$ is a convex function on $(0;m)$, where $m>0$. I take the second derivative as follows: $$ g^{\prime \prime}(x)=\frac{(x-m)^2 f^{\prime \prime}(x)-2(x-m) f^{\prime}(x)+f(x)-f(2 m-x)-2 f^{\prime}(x)+2 f^{\prime}(2 m-x)}{(x-m)^4} $$ But I don't know why the derivative above has a non-negative value.

Answer 1

$f'''(x) \geq 0$ means that $f'$ is convex, therefore it is advantageous to represent $g$ in terms of of $f'$: $$ f(2m-x) - f(x) = \int_x^{2m-x} f'(t) \, dt = 2(m-x)\int_0^1 f'\bigl(2sm + (1-2s) x\bigr) \, ds $$ where we have substituted $t= x + 2(m-x) s = 2ms + (1-2s)x$. It follows that $$ \tag{$*$} g(x) = 2 \int_0^1 f'\bigl(2sm + (1-2s) x \bigr) \, ds \, . $$ If we require $f'''$ to be continuous then we can compute $g''$ by differentiating under the integral: $$ g''(x) = 2 \int_0^1 (1-2s)^2 f'''\bigl(2sm + (1-2s) x \bigr) \, ds \ge 0 \, . $$ But even without that requirement one can argue that for each fixed $s \in [0, 1]$ the function $$ x \mapsto f'\bigl(2sm + (1-2s) x \bigr) $$ is convex, and that implies the convexity of the integral in $(*)$.