MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $v$?

share|cite|improve this question
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
Do we need $g$ to be non-vanishing? And what is this for? – Homology Jun 29 '10 at 23:47
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 $(v-x)e^{-x}$.

share|cite|improve this answer
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 $x=-v$.

share|cite|improve this answer
Nice! 123456789 (I needed some extra characters) – Wadim Zudilin Jun 29 '10 at 23:57
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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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