9

Update: The continuity condition is necessary, according to Ian Morris's comment in https://mathoverflow.net/questions/269064/for-a-continuous-function-f-mathbbr-to-mathbbr-does-fx-fy.

Miklós Schweitzer Competition 2015 Problem 8:

Problem 1: Prove that all continuous solutions of the functional equation $$[f(x) - f(y)]\left(f(\tfrac{x+y}{2}) - f(\sqrt{xy})\right) = 0, \forall x, y \in (0, +\infty)$$ are the constant functions.

http://www.math.u-szeged.hu/~mmaroti/schweitzer/schweitzer-2015-eng.pdf https://artofproblemsolving.com/community/c6h1224690p6149915 https://mathoverflow.net/questions/269064/for-a-continuous-function-f-mathbbr-to-mathbbr-does-fx-fy

To make the domain and range of $f$ clear (Thank @Calvin Lin for his valuable comments), I rephrased the problem above as follows.

Problem 1 (rephrased): Let $f : \ (0, \infty) \to \mathbb{R}$ be a continuous function satisfying $$[f(x) - f(y)]\left(f(\tfrac{x+y}{2}) - f(\sqrt{xy})\right) = 0, \forall x, y \in (0, +\infty).$$ Prove that $f$ is the constant function.

My question: If the continuity condition on $(0, \infty)$ is necessary?

Any comments are welcome and appreciated.

The solutions in the 2nd link above use the continuity condition. Terry Tao's proof in the 3rd link above also use the continuity condition. My solution (at the end) requires the continuity condition as well. On the other hand, the continuity condition is not necessary for the following problem (from a math exam):

Problem 2: Let $f : (0, \infty) \to (0, \infty)$ be a continuous function satisfying $f(\tfrac{x+y}{2}) - f(\sqrt{xy}) = 0,\ \forall x, y > 0$. Find all $f$.

Solution for Problem 2: $f$ is the constant function. For any $0 < b < a$, let $x = a + \sqrt{a^2 - b^2}$ and $y = a - \sqrt{a^2 - b^2}$. We have $f(a)=f(b)$. [Remark: We do not use the continuity condition in the proof.]

My solution for Problem 1:

Assume, for the sake of contradiction, that $f$ is not the constant function. WLOG, assume there exist two real numbers $0 < A < C$ with $f(A) < f(C)$. Since $f$ is continuous, there exists $A < B \le C$ such that $f(x) < f(B)$ for all $x$ in $[A, B)$.

Consider the sequence $$x_1 = \sqrt{AB}; \ x_{k+1} = \frac{x_k^2 + B^2}{2B}, \ k\ge 1.$$ By using Mathematical Induction, it is easy to prove that $x_k < x_{k+1} < B$ for all $k \ge 1$. Thus, $\lim_{k\to \infty} x_k$ exists. Let $L = \lim_{k\to \infty} x_k$. Then, $L = \frac{L^2 + B^2}{2B}$ and $L = B$. Thus, $\lim_{k\to \infty} x_k = B$.

Denote $a = x_{k+1}, b = x_k$ ($k\ge 1$). Let $X = a - \sqrt{a^2 - b^2}$ and $Y = a + \sqrt{a^2 - b^2}$. It is easy to verify that $X < Y$, $Y = B$, $\frac{X+Y}{2} = a$, and $\sqrt{XY} = b$. Thus, $X = \frac{b^2}{Y} \ge \frac{x_1^2}{B} = A$. Thus, $f(X) - f(Y) \ne 0$. From $[f(X) - f(Y)] [f(\frac{X+Y}{2}) - f(\sqrt{XY})] = 0$, we have $f(a) = f(b)$, i.e., $f(x_k) = f(x_{k+1})$ for all $k \ge 1$. Thus, $f(x_k) = f(x_1) = f(\sqrt{AB})$ for all $k \ge 1$.

Since $f$ is continuous, we have $f(B) = f(\lim_{k\to \infty} x_k) = \lim_{k\to \infty} f(x_k) = f(\sqrt{AB})$. Contradiction.

River Li
  • 37,323
  • 1
    @CalvinLin If $f(A) > f(C)$, how about letting $f \to -f$? – River Li May 04 '20 at 15:01
  • 1
    @CalvinLin Thanks. Actually, I have considered to revise the original problem to: Let $f : \ (0, \infty) \to \mathbb{R}$ be a continuous function satisfying $$[f(x) - f(y)]\left(f(\tfrac{x+y}{2}) - f(\sqrt{xy})\right) = 0, \forall x, y \in (0, +\infty).$$ Prove that $f$ is the constant function. But I think that the original problem may have the domain $(0, \infty)$ since the condition does not involve $(-\infty, 0]$ [we can let $f$ be constant on $(0, \infty)$ but not on $(-\infty, \infty)$]. Also, the original problem said "of functional equation", the domain should be $(0, \infty)$. – River Li May 04 '20 at 15:50
  • 1
    True, there is no information about the domain for negative numbers. And we can also extend the function to have range of all real numbers (then consider a restricted range if needed). – Calvin Lin May 04 '20 at 15:53
  • 1
    @CalvinLin Thanks. I will edit the question to make it clear. Also, for example, if $f : (0, \infty)\to (0, \infty)$, one can not let $f \to -f$, and we need to prove the case $f(A) > f(C)$. – River Li May 04 '20 at 16:05
  • 1
    Not necessarily. My comment meant "If the original problem is $ f: R^+ \rightarrow R^+$, then consider another problem of finding all functions $f : R^+ \rightarrow R$ with the same conditions, which are the constant functions. Hence the solutions to the original problem are the positive constant functions". – Calvin Lin May 04 '20 at 16:08
  • Just curious it is the Heron's algorithm ?(+1) for the subtle question ! – Miss and Mister cassoulet char Sep 03 '21 at 12:25
  • @ErikSatie Thanks. Any information for Heron's algorithm? – River Li Sep 03 '21 at 13:03
  • @RiverLi Oh well there is another name see https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method – Miss and Mister cassoulet char Sep 03 '21 at 13:15
  • @ErikSatie Yes. Thanks. – River Li Sep 03 '21 at 13:50

0 Answers0