Do there exist nonconstant functions such that... - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T16:35:30Z http://mathoverflow.net/feeds/question/29978 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29978/do-there-exist-nonconstant-functions-such-that Do there exist nonconstant functions such that... pavel 2010-06-29T22:58:37Z 2010-06-30T20:33:40Z <p>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 $v$?</p> http://mathoverflow.net/questions/29978/do-there-exist-nonconstant-functions-such-that/29983#29983 Answer by Gjergji Zaimi for Do there exist nonconstant functions such that... Gjergji Zaimi 2010-06-29T23:37:32Z 2010-06-29T23:45:30Z <p>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 $x=-v$.</p> http://mathoverflow.net/questions/29978/do-there-exist-nonconstant-functions-such-that/29985#29985 Answer by Homology for Do there exist nonconstant functions such that... Homology 2010-06-29T23:46:23Z 2010-06-29T23:46:23Z <p>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 $(v-x)e^{-x}$.</p> http://mathoverflow.net/questions/29978/do-there-exist-nonconstant-functions-such-that/29997#29997 Answer by Gerry Myerson for Do there exist nonconstant functions such that... Gerry Myerson 2010-06-30T01:38:57Z 2010-06-30T01:38:57Z <p>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. </p> http://mathoverflow.net/questions/29978/do-there-exist-nonconstant-functions-such-that/30108#30108 Answer by T. for Do there exist nonconstant functions such that... T. 2010-06-30T20:33:40Z 2010-06-30T20:33:40Z <p>The following calculation suggests that a nice probability interpretation may exist for any solution one can construct.</p> <p>$u(x) = f(x) - v/g(x)$ and all its $x$ derivatives are linear functions of $v$ with coefficients that are functions of $x$. </p> <p>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.</p> <p>The interesting points are that:</p> <ol> <li><p>To have a maximum we need $M(x) \geq 0$, so $M$ can be interpreted as a density.</p></li> <li><p>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 $\int M = 1$.</p></li> <li><p>$f$ is calculated as integral of $xM$, ie., an expected value of $x$.</p></li> <li><p>$1/g$ is calculated using the integral of $M$, ie., a probability.</p></li> </ol> <p>So there might be a simple probability inequality lurking behind most of the solutions.</p>