1
$\begingroup$

every convex function $f$ preserves arithmetic mean: $f(\frac{x+y}{2})\le\frac{f(x)+f(y)}{2}$, this is a particular case of the convexity condition.

Does convex function $f$ preserve geometic mean: $f(\sqrt{xy})\le\sqrt{f(x)f(y)}$ ? (of course here $f\colon(0,\infty)\to(0,\infty)$)

the answer is no, but there is no increasing function among counterexampes I know.

so the question is: does every increasing convex function $f\colon(0,\infty)\to(0,\infty)$ preserve geometic mean?

$\endgroup$
2
$\begingroup$

In general, you have exact equality if the function is of the form $x^\alpha$ for some $\alpha >1$. So, every function that increases slower than an exponential type function is a good candidate: for example, choose $f(t)=t \ln(t+1)$. This works because $\sqrt{f(x)f(y)} = \sqrt{xy}\sqrt{\ln(x+1)\ln(y+1)}$ and $f(\sqrt{xy}) = \sqrt{xy} \ln(\sqrt{xy}+1)>\sqrt{xy}\ln(\sqrt{xy}) = \sqrt{xy}{\ln x + \ln y\over 2}$. So the result follows from the arithmetico-geometric inequality for $x$ and $y$ sufficiently large.

$\endgroup$
3
$\begingroup$

Let $f$ be an increasing convex function. $$\sqrt{xy}\le \frac{x+y}{2}\implies f(\sqrt{xy})\le f\left(\frac{x+y}{2}\right)\le \frac{f(x)+f(y)}{2}$$ Now, if we define $g(x)=e^{f(x)}$, then $g$ has the property that $$g(\sqrt{xy})\le \sqrt{g(x)g(y)}$$ Now, $g$ is increasing and $g'(x)=f'(x)g(x)\implies g''(x)=f''(x)g(x)+f'(x)g'(x)>0\implies g$ is convex. Thus $g$ is an increasing convex function that has the desired property.

So, for example, if we take $f(x)=x^2$, then $g(x)=e^{x^2}\implies g(\sqrt{xy})=e^{xy}\le e^{(x^2+y^2)/2}=\sqrt{g(x)g(y)}$ as desired.

So every convex function of the form $e^{f(x)}$ where $f$ is a convex increasing function preserves geometric mean.

$\endgroup$
  • $\begingroup$ Nice observation. $\endgroup$ – Michael Feb 11 '15 at 17:25
2
$\begingroup$

The condition is equivalent to $g(t) = \ln (f(e^t))$ being convex. Assuming twice differentiability, and with $x = e^t$, we get $$ g''(t) = \dfrac{f''(x) x^2}{f(x)} + \dfrac{f'(x) x}{f(x)} - \dfrac{f'(x)^2 x^2}{f(x)^2}$$ For a counterexample it suffices to have (at some point) $f''(x) = 0$ with $f'(x) x > f(x) > 0$. For example, $f(x) = x - 1$ would do for $x > 1$.
To make $f$ convex, increasing and positive on $(0,\infty)$, take $f(x) = \max(\epsilon x, x - 1)$ for some $\epsilon \in (0,1)$.

EDIT: We can then show directly: if $y - 1 > \epsilon y$ and $y < x$, $f(\sqrt{xy}) = \sqrt{xy} - 1 > \sqrt{f(x) f(y)} = \sqrt{(x-1)(y-1)}$ since $\sqrt{xy} > 1$ and $$(\sqrt{xy}-1)^2 = (x-1)(y-1) + (\sqrt{x}-\sqrt{y})^2 > (x-1)(y-1)$$

$\endgroup$
1
$\begingroup$

Counterexample: Take any convex increasing function $f:(0,1)\rightarrow (0,1)$ that satisfies:

\begin{align} &f(1) = 0.1\\ &f(2) = 1 \\ &f(4) = 3 \end{align}

Then $f(\sqrt{1}\sqrt{4}) > \sqrt{f(1)f(4)}$ because $1 > \sqrt{0.3}$. You can get such a function by assuming $f(0)=0$ and filling in peicewise linearly. Note that the slopes over the intervals $(0,1]$, $[1,2]$, $[2,\infty)$ are $0.1, 0.9, 1$, respectively, and so the function $f(x)$ is indeed convex increasing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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