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.

  • $\begingroup$ 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

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.


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