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Can someone help me please with this problem?

If the function $f:\mathbb{R}^+\rightarrow\mathbb{R}$ satisfies the equation $f\Big(\frac{x+y}{2}\Big)+f\Big(\frac{2xy}{x+y}\Big)= f(x)+f(y)$, then it satisfies also $2f(\sqrt{xy})=f(x)+f(y)$.

I tried to collect more formulas from substitutions like $\sqrt{xy}\rightarrow x$, $x\rightarrow\frac{x+y}{2}$ but didn't succeed.

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  • $\begingroup$ If you are looking for continuous functions I can provide a solution. $\endgroup$ – clark May 3 '13 at 16:39
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The statement is equivalent to

The value of the expression $f(x) + f(y)$ remains the same if we replace $x$ and $y$ by their arithmetic and harmonic means:

$$\text{AM}(x,y) = \frac{x+y}{2}\quad\text{ and }\quad\text{HM}(x,y) = \frac{1}{\frac12 ( \frac{1}{x} + \frac{1}{y}) }$$

It is known that for any $x,y \in \mathbb{R}_{+}$, we have:

$$\text{AM}(x,y) \ge \text{GM}(x,y) = \sqrt{xy} \ge \text{HM}(x,y)$$

It is also known that if we repeatly apply $\text{AM}$ and $\text{HM}$ to a pair of numbers, they will converge to GM. More precisely, if we construct two sequences $(x_n), (y_n), n\in \mathbb{N}$ by:

$$(x_n, y_n) = \begin{cases} (\max(x,y),\;\min(x,y)), & n = 0\\ (\text{AM(x,y)},\;\;\text{HM(x,y)}), & n > 0 \end{cases}$$

we will find

$$x_0 \ge x_1 \ge x_2 \ge \cdots \ge x_n \cdots \ge \text{GM}(x,y) \ge \cdots \ge y_n \ge \cdots \ge y_2 \ge y_1 \ge y_0$$

and $\displaystyle \lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n = \text{GM}(x,y) = \sqrt{xy}$. If $f$ is continuous, this will imply

$$f(x) + f(y) = f(x_1) + f(y_1 ) = \cdots = \lim_{n->\infty} f(x_n) + f(y_n) = 2f(\sqrt{xy})$$

The part that $(x_n,y_n)$ is sandwiching $\text{GM}(x,y)$ are the standard $\text{AM}, \text{GM}, \text{HM}$ stuff, I will skip their justification. Let me demonstrate why $x_n, y_n$ converges to same limit. Notice

$$x_{n} - y_{n} = \frac{x_{n-1}+y_{n-1}}{2} - \frac{2 x_{n-1} y_{n-1}}{x_{n-1}+y_{n-1}} = \frac{(x_{n-1}-y_{n-1})^2}{2(x_{n-1}+y_{n-1})}\\ = \frac12 \left|\frac{x_{n-1}-y_{n-1}}{x_{n-1}+y_{n-1}}\right|(x_{n-1} - y_{n-1}) \le \frac12 (x_{n-1} - y_{n-1}) $$ We can conclude for general $n$, $|x_n - y_n | \le 2^{-n}|x-y|$ and hence $x_n, y_n$ converges to same limit as $n \to \infty$. Since $x_n$ and $y_n$ are sandwiching $\text{GM}(x,y)$, the common limit is $\text{GM}(x,y) = \sqrt{xy}$.

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Hints:

$\dfrac{f(x)+f(y)}{2} \ge\sqrt{f(x)f(y)}$

Here we have $\dfrac{f(x)+f(y)}{2} =2f(\sqrt{xy})$

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  • 2
    $\begingroup$ How do you relate the inequality $\frac{f(x)+f(y)}{2}\geq \sqrt{f(x)f(y)}$ of the function values to the argument $\sqrt{xy}$? $\endgroup$ – Albert May 3 '13 at 15:24
  • $\begingroup$ Have you heard of AM-GM inequality? The inequality crossed my mind and it possibly does help? $\endgroup$ – Inceptio May 3 '13 at 15:25
  • $\begingroup$ You shouldn't mix up an inequality for $f(x)$ and $f(y)$ with a with a search for a relation between the arguments $x$, $y$, $\frac{x+y}{2}$, $\frac{2xy}{x+y}$ and $\sqrt{xy}$. $\endgroup$ – Albert May 3 '13 at 15:34
  • $\begingroup$ You surely can have the inequality which give bounds and thus help solve the problems.And the hint is incomplete. $\endgroup$ – Inceptio May 3 '13 at 15:38
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    $\begingroup$ am-gm only holds for non-negative reals, and $f$ can take values in all of $\mathbb{R}$ $\endgroup$ – cats May 4 '13 at 3:24

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