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In my recent researches, I encountered functions $f$ satisfying the following functional inequality:

$$ (*)\; f(x)\geq f(y)(1+x-y) \; ; \; x,y\in \mathbb{R}. $$

Since $f$ is convex (because $\displaystyle f(x)=\sup_y [f(y)+f(y)(x-y)]$), it is left and right differentiable. Also, it is obvious that all functions of the form $f(t)=ce^t$ with $c\geq 0$ satisfy $(*)$. Now, my questions:

(1) Is $f$ everywhere differentiable?

(2) Are there any other solutions for $(*)$?

(3) Is this functional inequality well-known (any references (paper, book, website, etc.) for such functional inequalities)?

Thanks in advance

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    $\begingroup$ Regarding the two answers, I think it is more than noteworthy that this is a functional inequality instead of a functional equation, however it defines essentially (up to a constant factor) a unique function! I don't think there are many other functions that can be defined uniquely in such a way. $\endgroup$ – Wolfgang Jan 12 at 20:19
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    $\begingroup$ @Wolfgang Here is another example of a functional inequality that leads to a unique (up to choice of two constants) function: math.stackexchange.com/questions/777423/… $\endgroup$ – Reinstate Monica Jan 12 at 20:39
  • $\begingroup$ @ReinstateMonica thank you, nice example! $\endgroup$ – Wolfgang Jan 13 at 19:14
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    $\begingroup$ I don't recognize this particular problem, but there are several books out there that are geared toward mathematical competitions (Olympiad, Putnam) whose title contains "functional equations." For example, look for authors Andreescu or Parvardi or Efthimiou. Such books often have a section on functional inequalities so maybe you'll find your problem (or something similar) in one of those books. $\endgroup$ – Timothy Chow Jan 16 at 17:17
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Replace $x$ with $x+y$ to get $f(x+y)\ge f(y)(1+x)$ or $f(x+y)-f(y)\ge xf(y)$. Replace $y$ with $x+y$ and then interchange $x$ and $y$ to get $f(x+y)-f(y)\le xf(x+y)$. Together, $$ xf(y)\le f(x+y)-f(y)\le xf(x+y). $$ Dividing by $x$ and taking the limit as $x\to0$ implies that $f$ is differentiable with $f'=f$.

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    $\begingroup$ Thanks for your answer. Have you ever seen this functional inequality (or something like this)? $\endgroup$ – M.H.Hooshmand Jan 10 at 6:45
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    $\begingroup$ @M.H.Hooshmand I found the functional inequality $f(x+y)\ge f(y)(1+x)$ in one of the books mentioned in my earlier comment. Namely, it is Problem 17.11 in Topics in Functional Equations, 2nd edition, by Andreescu, Boreico, Mushkarov, and Nikolov, XYZ Press, 2015. $\endgroup$ – Timothy Chow Jan 26 at 21:01
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For any $x$ and for sufficiently large $n$ such that $1+x/n>0$, it holds that \begin{align} f(x) &\ge f\left (\frac{(n-1)x}n \right) (1+x/n)\\ &\ge f\left (\frac{(n-2)x}n \right)(1+x/n)^2 \\ &\ge \cdots \ge f(0) \left(1+ \frac x n\right)^n. \end{align} by substituting $(x,y)=(x,(n-1)x/n), ((n-1)x/n, (n-2)x/n),...$ in the given equation. In other words, $$ f(x) \ge \lim_{n\rightarrow \infty} f(0) \left(1+ \frac x n\right)^n = f(0)\cdot e^x. $$ On the other hand, for any $y$ and for sufficiently large $n$ such that $1-y/n>0$, we can similarly get the following inequality. \begin{align} f(y) &\le f\left( \frac{(n-1)y} n\right) / (1-y/n)\\ &\le \cdots \le f(0)/(1-y/n)^n. \end{align} It implies $f(y)\le f(0) \cdot e^y$. Combining these inequalities, we get that $f(x)=f(0) \cdot e^x$ is the only solution as you wanted.

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    $\begingroup$ Thanks for your answer. Have you ever seen this functional inequality (or something like this)? $\endgroup$ – M.H.Hooshmand Jan 10 at 6:46
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    $\begingroup$ @M.H.Hooshmand No, I didn't see this inequality before. $\endgroup$ – Hhan Jan 10 at 8:30

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