Let $f:(0, \infty) \rightarrow \mathbb{R}$ be a convex and continuous function. We know that $\partial^{e} f(. )$(either right or left derivative) is non-decreasing and upper semi-continuous function. So, $f^{''}$ is differentiable a.e. and it is non negative.


Let $g(t)=f(t)-t\partial^{e} f(t)$ that is a upper semi-continuous map. I got this point that $g^{'}(t)=-tf^{''}(t)$ for a.e. $t>0$. I want to show that $g$ is decreasing.


My attempt:
I wanted to use Goldowsky-Tonelli theorem, but the map is not continuous. Does one help me to get the result?if it's not true, under which assumption is true.


Goldowsky-Tonelli theorem: Let $f$ be a continuous function that has a derivative at each point of $\mathbb{R}$ except on countable set, and $f^{'} \geq 0$ a.e., then $f$ is a nondecreasing function.



Goldowsky-Tonelli theorem for upper semi continuous function
Let's assume that $\partial^{e} f(. )$ denotes the right derivative (the left derivative can be handled similarly).


Claim: The function $g(t)=f(t)-t\partial^{e} f(t)$ is indeed (weakly) decreasing for any convex $f:(0, \infty) \rightarrow \mathbb{R}$ (Note that continuity of $f$ on an open interval follows from convexity, and that $f$ is differentiable outside a countable set [1]).


The claim can be proved in several ways. One of them involves using the Goldowsky-Tonelli theorem as you suggested (this theorem is proved, e.g., in Section 5.1, page 102 in [1]).


Proof of claim: Fix $h>0$. For $t>0$, let $f\_h(t):=\frac{1}{h} \int\_t^{t+h} f(s) ds$, so that $f\_h'(t)=
\frac{1}{h} \Bigl (f(t+h)-f(t)\Bigr)$ holds for all $t>0$, and define 
$$g\_h(t)=f\_h(t)-t f\_h'(t) =f\_h(t)-(t/h) \Bigl (f(t+h)-f(t)\Bigr) \, .$$
 Then $g\_h:(0, \infty) \rightarrow \mathbb{R}$ is continuous, $g\_h'(t)$ exists 
for all $t$ outside a countable set, and $g\_h'(t)=-(t/h)(f'(t+h)-f'(t)) \le 0$ for a.e. $t>0$. By Goldowsky-Tonelli applied to $-g\_h$, the function $g\_h$ is weakly decreasing
on $(0,\infty)$, i.e., $g\_h(t) \le g\_h(u)$ for $t>u$. Taking $h \downarrow 0$ in the last inequality proves the claim.


Remark: To avoid using the Goldowsky-Tonelli theorem, one could define $f\_h$ as a convolution of $f$ with a smooth positive function supported on $(0,h)$, instead of convolution with a step function.


[1] <https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable>


[2] Kannan, R. and Krueger, C.K., 2012. Advanced analysis: on the real line. Springer Science & Business Media. 


