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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 $v$?

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Functions on $\mathbb{R}$ or on the positive reals? In the latter case, $- \log x$ and $x$. –  Homology Jun 29 '10 at 23:26
    
Function on $\mathbb R$. Positivity is for $v$ only. –  Wadim Zudilin Jun 29 '10 at 23:28
    
Motivation for the question? –  Yemon Choi Jun 29 '10 at 23:35
    
@Yemon: might be a homework, might be curiosity. I don't see any deep context here but it's tricky. –  Wadim Zudilin Jun 29 '10 at 23:39
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Do we need $g$ to be non-vanishing? And what is this for? –  Homology Jun 29 '10 at 23:47

4 Answers 4

up vote 6 down vote accepted

Take $f(x)=(x+1)e^{-x}$ and $g(x)=e^x$, then $f(x)-v/g(x)=(x+1-v)e^{-x}$ and the derivative with respect to $x$ is $(v-x)e^{-x}$.

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Doesn't this work for any real $v$? (So, you do much more!) –  Wadim Zudilin Jun 29 '10 at 23:58
    
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.) –  Wadim Zudilin Jun 30 '10 at 0:03

By the arithmetic-geometric mean inequality, when $v$ is positive $$-|x|^{\frac{1}{2}}-\frac{v}{|x|^{\frac{1}{2}}}$$ is maximized at $x=v$ and $x=-v$.

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Nice! 123456789 (I needed some extra characters) –  Wadim Zudilin Jun 29 '10 at 23:57
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Nice! $ $ –  Steven Gubkin Jun 30 '10 at 1:47
    
It is not defined at $x=0$. I think the question was about functions on the whole real line. –  T.. Jun 30 '10 at 23:55

Let $f$ be arbitrary (but non-constant, real-valued, and differentiable), let $h$ be any antiderivative of $f'(x)/x$, and let $g=1/h$; then $f'(v)g'(v)=v$, so $f-v/g$ has a critical point at $x=v$. Now you can look for conditions under which that critical point is a maximum.

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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 $(v-x)M(x)$. 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:

  1. To have a maximum we need $M(x) \geq 0$, so $M$ can be interpreted as a density.

  2. 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 $\int M = 1$.

  3. $f$ is calculated as integral of $xM$, ie., an expected value of $x$.

  4. $1/g$ is calculated using the integral of $M$, ie., a probability.

So there might be a simple probability inequality lurking behind most of the solutions.

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