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The following information is given:

For a function $f$ satisfying $f(x)>0$, we define the geometric mean, $F$, by: $$F(y)=\exp\left(\frac1y\int_0^y\ln f(x)dx\right)\,\,(y>0)$$ i) Prove that for $a>0,a\ne1$: $$F(y)=a^{\frac1y\int_0^y\log_a > f(x)dx}$$ ii) The functions $f$ and $g$ satisfy $f(x)>0$ and $g(x)>0$, and the function $h$ is defined by $h(x)=f(x)g(x)$. Their geometric means are $F,G,H$ respectively. Show that: $$H(y)=F(y)G(y)$$ iii) Prove that, for any positive number $b$, the geometric mean of $b^x$ is $\sqrt{b^y}$

iv) Prove that, if $f(x)>0$ and the geometric mean of $f(x)$ is $\sqrt{f(y)}$ then $f(x)=b^x$ for some positive number $b$

I have done parts i),ii),iii) but I do not see how part iv) differs from iii) in terms of it being the same question done in reverse.

This is how I did iii):

$$f(x)=b^x$$ $$F(y)=\exp\left(\frac1y\int_0^y\ln(b^x)dx\right)=\exp\left(\frac{\ln(b)}{y}\int_0^yxdx\right)=e^{\frac{y\ln(b)}2}=\sqrt{b^y}$$

Thanks in advance.

Jean Marie
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Henry Lee
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1 Answers1

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(iv) is not the same question as (iii). You have to assume that the function $f:[0, \infty) \to (0, \infty)$ satisfies $$ \tag {*} \exp\left(\frac 1y\int_0^y\ln f(x) \, dx\right) = \sqrt{f(y)} $$ for all $y > 0$, and prove the existence of a positive constant $b$ such that $f(x) = b^x$ for all $x \ge 0$.

Let us also assume that $f$ is continuous, so that the integral is well-defined (as Riemann or Lebesgue integral). $(*)$ is equivalent to $$ \int_0^y\ln f(x) \, dx = \frac 12 y \ln f(y) $$ and differentiating this relation with respect to $y$ gives $$ \ln f(y) = \frac 12 \left(\ln f(y) + y \frac{f'(y)}{f(y)} \right) \\ \iff \ln f(y) = y \frac{f'(y)}{f(y)} \\ \iff \left( \frac{\ln f(y)}{y} \right)' = 0 $$ for all $y > 0$. It follows that $\frac{\ln f(y)}{y}$ is constant, and finally that $f(y) = b^y$ for some positive constant $b$.

Martin R
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