Do there exist nonconstant real valued functions $f$ and $g$ such that the expression: $$f(x) v/g(x)$$ is maximized at $x = v$ for all positive real $v$?

$\begingroup$ Functions on $\mathbb{R}$ or on the positive reals? In the latter case, $ \log x$ and $x$. $\endgroup$ – Homology Jun 29 '10 at 23:26

$\begingroup$ Function on $\mathbb R$. Positivity is for $v$ only. $\endgroup$ – Wadim Zudilin Jun 29 '10 at 23:28

$\begingroup$ Motivation for the question? $\endgroup$ – Yemon Choi Jun 29 '10 at 23:35

$\begingroup$ @Yemon: might be a homework, might be curiosity. I don't see any deep context here but it's tricky. $\endgroup$ – Wadim Zudilin Jun 29 '10 at 23:39

1$\begingroup$ Do we need $g$ to be nonvanishing? And what is this for? $\endgroup$ – Homology Jun 29 '10 at 23:47
Take $f(x)=(x+1)e^{x}$ and $g(x)=e^x$, then $f(x)v/g(x)=(x+1v)e^{x}$ and the derivative with respect to $x$ is $(vx)e^{x}$.

$\begingroup$ Doesn't this work for any real $v$? (So, you do much more!) $\endgroup$ – Wadim Zudilin Jun 29 '10 at 23:58

$\begingroup$ May I suggest you to take a more realistic name for MO? (I am very sorry for making a joke with your present one in a comment above but you don't give me an option to cite your answer correctly.) $\endgroup$ – Wadim Zudilin Jun 30 '10 at 0:03
By the arithmeticgeometric mean inequality, when $v$ is positive $$x^{\frac{1}{2}}\frac{v}{x^{\frac{1}{2}}}$$ is maximized at $x=v$ and $x=v$.

$\begingroup$ Nice! 123456789 (I needed some extra characters) $\endgroup$ – Wadim Zudilin Jun 29 '10 at 23:57

1

$\begingroup$ It is not defined at $x=0$. I think the question was about functions on the whole real line. $\endgroup$ – T.. Jun 30 '10 at 23:55
Let $f$ be arbitrary (but nonconstant, realvalued, and differentiable), let $h$ be any antiderivative of $f'(x)/x$, and let $g=1/h$; then $f'(v)g'(v)=v$, so $fv/g$ has a critical point at $x=v$. Now you can look for conditions under which that critical point is a maximum.
The following calculation suggests that a nice probability interpretation may exist for any solution one can construct.
$u(x) = f(x)  v/g(x)$ and all its $x$ derivatives are linear functions of $v$ with coefficients that are functions of $x$.
Thus, to have an extremum at $x=v$ the first derivative $u'$ must be of the form $(vx)M(x)$. Integrating the $v$degree 0 and 1 parts of this equation produces $f$ and (the reciprocal of) $g$. Algebraically this will be equivalent to Gerry's solution.
The interesting points are that:
To have a maximum we need $M(x) \geq 0$, so $M$ can be interpreted as a density.
The total mass $\int M$ has to be finite in order for $g(x)$ to exist on the whole real line. This is so that we can choose a constant of integration larger (in absolute value) than the total mass, when computing $1/g = C + \int M$. Thus, $M$ is a sort of probability measure, and literally is one when $\int M = 1$.
$f$ is calculated as integral of $xM$, ie., an expected value of $x$.
$1/g$ is calculated using the integral of $M$, ie., a probability.
So there might be a simple probability inequality lurking behind most of the solutions.