0
$\begingroup$

What's the condition to exist a real global maxima or minima of a function $f:\mathbb{R}\mapsto \mathbb{R}$?

At first I thought that to test if this is true I could take the image of any of its derivatives, and if any of them is a proper subset of $\mathbb{Z}$ then the claim is true. This conclusion is false, but I get the feel that it has something to do with derivatives.

I'm not asking where they are, I just want to know the conditions for $\exists y\in\mathbb{R}\;\forall x\in\mathbb{R}:f(x)<y\lor y<f(x)$ to be true.

$\endgroup$
  • $\begingroup$ Are you assuming $f$ is at least twice differentiable? If you want to look at ALL functions, it's a pretty tough question. $\endgroup$ – John Hughes Mar 2 '17 at 14:03
  • $\begingroup$ BTW, what you've written is surely not what you want according to the title, for all you need for that to be true is that $f$ is not surjective, for then you pick $y$ to be any point not in $f$'s image. $\endgroup$ – John Hughes Mar 2 '17 at 14:04
  • $\begingroup$ Your condition "$\exists y\in\mathbb{R}\;\forall x\in\mathbb{R}:f(x)<y\lor y<f(x)$" is weaker than "exist a local extremum". Take $f(x) = \arctan x$. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 2 '17 at 15:04
0
$\begingroup$

I've found out that what I'm looking for is what proposition implies $\exists y\;\forall x:(f(x)<y\lor y<f(x))$ and by definition $f(x)<y\lor y<f(x)\longleftrightarrow y\neq f(x)$ so there must exist an element in the codomain not in the image of $f$, so $f$ must not be surjective.

Supposing we don't know the definition of $a\neq b$:

$$X\implies\exists y\;\forall x:(f(x)<y\lor y<f(x))$$

taking the contrapositive

$$\lnot\exists y\;\forall x:(f(x)<y\lor y<f(x))\implies \lnot X$$

We found that the inverse of everything that $\lnot\exists y\;\forall x:(f(x)<y\lor y<f(x))$ implies is an answer. Simplifying the formula

\begin{align} \lnot\exists y\;\forall x&:(f(x)<y\lor y<f(x))\\ \forall y\;\lnot\forall x&:(f(x)<y\lor y<f(x))\\ \forall y\;\exists x&:\lnot(f(x)<y\lor y<f(x))\\ \forall y\;\exists x&:(\lnot(f(x)<y)\land\lnot( y<f(x)))\\ \forall y\;\exists x&:(f(x)\leq y\land y\leq f(x))\\ \forall y\;\exists x&:y=f(x)\\ \end{align}

Inverting this we get

\begin{align} \lnot\forall y\;\exists x&:y=f(x)\\ \exists y\;\lnot\exists x&:y=f(x)\\ \exists y\;\forall x&:y\neq f(x)\\ \end{align}

So this there must exists a number in the codomain that is not in the image of $f$, hence $f$ must not be surjective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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