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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.

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2 Answers 2

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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.

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  • $\begingroup$ :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. $\endgroup$
    – Adam
    Oct 10, 2019 at 10:53
  • $\begingroup$ 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 $t$. $\endgroup$ Oct 10, 2019 at 11:42
  • $\begingroup$ I'm a bit confused, because when $h\rightarrow o$ , $f_{h}^{'}$ is differentiable a.e. $\endgroup$
    – Adam
    Oct 10, 2019 at 11:48
  • $\begingroup$ 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. $\endgroup$ Oct 10, 2019 at 11:58
  • $\begingroup$ Thank you. I think there is no problem. $\endgroup$
    – Adam
    Oct 10, 2019 at 12:19
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Denoting $f'$ either the right or the left derivative, for $0<x<y$ 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 $f(y)-yf'(y)\le f(x)-xf'(x)$.

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  • $\begingroup$ Did you show that it is decreasing? Could you please explain it a bit more? $\endgroup$
    – Adam
    Oct 10, 2019 at 21:20
  • $\begingroup$ Yes, what exactly is not clear? $\endgroup$ Oct 11, 2019 at 7:15
  • $\begingroup$ Why for $0<x<y$ we have $f^{'}(x)\leq \frac{f(y)-f(x)}{y-x}\leq f^{'}(y)?$ $\endgroup$
    – Adam
    Oct 11, 2019 at 10:12
  • $\begingroup$ 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 $(a,b)$ $$f_-'(x)\le f_+'(x) \le {f(y)-f(x)\over y-x} \le f_-'(y)\le f_+'(y).$$ $\endgroup$ Oct 11, 2019 at 11:07

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