If solution of $f'(x) /g'(x)= \lambda $ is an extremum point of $y = f(x) - \lambda \,g(x) $ then can it be shown that $f(x)$ and $g(x)$ are monotone functions?
Extremum of a monotonic function combination
0
$\begingroup$
maxima-minima
-
0I think this question is unclear. What do you mean by monotonously changing function? Just monotone? Locally? What are the hypotheses on $f$, $g$? – 2017-01-29
-
0$f(x)-λg(x)=0$ implies $\frac{f(x)}{g(x)}=λ$ and so $\frac{g'(x)}{g(x)}=\frac{f'(x)}{f(x)}$ Hint: domain of $\ln{x}$: $x>0$ – 2017-01-29
-
0What even is an extremum point of $f(x)-\lambda g(x)=0$? Is that the constant, zero function? – 2017-01-29
-
0Sorry, edited it. – 2017-01-29
1 Answers
0
Not even locally, no. A 'simple' counter-example can be obtained by taking $\lambda=0$ and $f$ given by
$$f(x)=x^2\cdot \exp\left({\sin\left(\frac1x\right)}\right),$$
where of course $f(0)=0$. Then $f(x)>0$ when $x\neq 0$, so that $x=0$ is a (global) minimum, but $f$ is never monotone near $x=0$.
You can check that $f$ is everywhere differentiable, although its derivative is not continuous. That said, if you take $f(x)=x^4\cdot \exp\left({\sin\left(\frac1x\right)}\right)$, then all of these hold and additionally $f'$ is everywhere continuous (differentiable in fact).
-
0I was implying non-zero $ \lambda$. For $ \lambda = 1,f = x, g = -1/x, f- \lambda g $ has minimum at $ x = 1$ – 2017-01-29
-
0You have probably not yet asked this clearly yet. Let $h$ be any function, say $h(x)=x^2\cdot \exp\left({\sin\left(\frac1x\right)}\right)$. Let $\alpha,\beta\neq0$ and let $f=\alpha\cdot h$ and $g=\beta\cdot h$. Then with $\lambda =\alpha/\beta$, every $x$ is a solution to $f'/g'=\lambda$, so the answer still applies. – 2017-01-29
-
0What is true only in the immediate neighborhood,need not be so in entire domain, right? – 2017-01-29
-
0That is also true, but even if you were only concerned about them being locally monotone (as opposed to globally monotone, which is stronger), that is still not true (with those hypotheses). – 2017-01-29