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.


2 Answers 2


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.

  • $\begingroup$ :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
    Commented 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 $t$. $\endgroup$ Commented 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
    Commented 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. $\endgroup$ Commented Oct 10, 2019 at 11:58
  • $\begingroup$ Thank you. I think there is no problem. $\endgroup$
    – Adam
    Commented Oct 10, 2019 at 12:19

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 $f(y)-yf'(y)\le f(x)-xf'(x)$.

  • $\begingroup$ Did you show that it is decreasing? Could you please explain it a bit more? $\endgroup$
    – Adam
    Commented Oct 10, 2019 at 21:20
  • $\begingroup$ Yes, what exactly is not clear? $\endgroup$ Commented 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
    Commented 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 $(a,b)$ $$f_-'(x)\le f_+'(x) \le {f(y)-f(x)\over y-x} \le f_-'(y)\le f_+'(y).$$ $\endgroup$ Commented Oct 11, 2019 at 11:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.