# Goldowsky-Tonelli theorem for upper semi continuous function

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.

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.

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

• :Thank you for your anwser, but your proof shows that $g$ is weakly decreasing a.e. .Because when $h\rightarrow 0$ then $g(t)$ decrease a.e.
Oct 10, 2019 at 10:53
• The proof shows $g$ is weakly decreasing everywhere. There is no a.e. in the conclusion. As $h$ tends to zero, $g_h(t)$ tends to $g(t)$ for all $t$. Oct 10, 2019 at 11:42
• I'm a bit confused, because when $h\rightarrow o$ , $f_{h}^{'}$ is differentiable a.e.
Oct 10, 2019 at 11:48
• Since $f$ is continuous, $f_h$ tends to $f$ everywhere by fundamental theorem of calculus. That is all you need. $f_h$ is differentiable everywhere. If there is still a confusing step please indicate exactly where. Oct 10, 2019 at 11:58
• Thank you. I think there is no problem.
Denoting $$f'$$ either the right or the left derivative, for $$0 we have $$\displaystyle f'(x)\le\frac{f(y)-f(x)}{y-x} \le f'(y)$$, so $$\displaystyle {f(y)-f(x)} \le y f'(y)-xf'(y)\le y f'(y)-xf'(x)$$, and $$f(y)-yf'(y)\le f(x)-xf'(x)$$.
• Why for $0<x<y$ we have $f^{'}(x)\leq \frac{f(y)-f(x)}{y-x}\leq f^{'}(y)?$
• This is elementary. Recall $f:(a,b)\to\mathbb{R}$ is convex iff the incremental ratio ${f(y)-f(x)\over y-x}$ is increasing in both variables. As a consequence, at any point $x$, $f$ has both right derivative, $f_+'(x)=\inf_{y>x}{f(y)-f(x)\over y-x}$ and left derivative $f_-'(x)=\sup_{z<x}{f(z)-f(x)\over z-x}$, and for all $x<y$ in $(a,b)$ $$f_-'(x)\le f_+'(x) \le {f(y)-f(x)\over y-x} \le f_-'(y)\le f_+'(y).$$ Oct 11, 2019 at 11:07