There are the given conditions:
$$f: Df \rightarrow R, Df \subset R,\; x_0 \in \operatorname{Int} Df, $$ $f$ is differentiable at a point $x_0$ and $x_o$ is a loc. extremum. Then $f'(x_0)=0$
Proof:
If $x_0 \in \operatorname{Int} Df$, then $\exists \epsilon_1 >0: U _{\epsilon_1}(x_0) \subset Df$
$\exists \epsilon_2>0: $ that in set $U_{\epsilon_2}(x_0) \, \cap Df $ the conditions for a local extremum are fulfilled.
- What are these conditions?
$$\epsilon=\min\{\epsilon_1,\epsilon_2\}$$
- Why can't we have just one $\epsilon$ from the beginning? I assume that the condition for local extrema can be fulfilled outside of the Definition range, but I still don't quite get the logic behind it.
A case when $x_0$ - local max.
$$\forall x \in U_{\epsilon}(x_0 ) \subset Df \, \, f(x) \le f(x_0) $$
- Is $f(x)$ less than OR equal to $f(x_0)$, because there is a change that $f(x)$ is at the end point of interval where the $f$ is defined, so it can by given conditions be equal to loc. extrema?
If $f(x)$ is differentiable at $x_0 $, then $f'(x_0)=\lim_{\Delta x \to 0}\frac{f(x_0+\Delta x) - f(x_0)}{\Delta x}$
We denote $\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}= \lambda(x_0,\Delta x)$
So $$\lim_{\Delta x \to 0^-}\lambda(x_0,\Delta x)=lim_{\Delta x \to 0^+}\lambda (x_0, \Delta x )$$
$\Delta x : x_0 + \Delta x \in U_\epsilon(x_0)$
- I assume that $\Delta x \in U_\epsilon(x_0)$ because $\Delta x $ tends to 0, but maybe there is some other reason as to why this is?
1) $0<\Delta x < \epsilon$
$f(x_0 + \Delta x) \le f(x_0)$
$\frac{f(x_0+\Delta x) - f(x_0}{\Delta x} \le 0$
2) $0> \Delta x > -\epsilon$
$f(x_0 + \Delta x ) \le f(x_0)$
$\frac{f(x_0 + \Delta x) -f(x_0)}{\Delta x }\ge 0$
$$\frac{f(x_0+\Delta x) - f(x_0)}{\Delta x}=\frac{f(x_0+\Delta x ) - f(x_0)}{\Delta x}$$
That's why : $$\lim_{\Delta x \to 0}\frac{f(x_0+\Delta x) - f(x_0) }{\Delta x }=f'(x_0)=0$$
The end of the proof is clear to me, but I would like some clarification about the middle section of the proof.
Your statement is mixed up. We are given in the hypothesis of the theorem that $x_0$ is a local extremum. There are no "conditions" needed for this. What it should say is
$\exists \epsilon_2 > 0$ such that $x_0$ is the global maximum or minimum of $f$ on the interval $U_{\epsilon_2}(x_0) = (x_0 - \epsilon_2, x_0 + \epsilon_2)$.
Which is just the definition of "local extremum".
The condition for local extrema cannot be fulfilled outside the domain of $f$, since the definition requires $f$ to be defined at the extremum point. However, it does not require that $f$ be defined everywhere on a neighborhood of that point. For example, you could have the function $g : [0,1] \to \Bbb R : x \mapsto x$. The maximum of $g$ is at $x = 1$, but $g$ is not defined on a full neighborhood of $1$. Thus simply saying that $x_0$ is a local extremum of $f$ does not guarantee that $f$ is defined on a neighborhood of $x_0$.
Since this proof requires that $f$ be defined on a neighborhood of $x_0$ to work, the theorem hypothesis includes the condition that $x_0$ is in the interior of the domain. $\epsilon_1$ comes from the condition that $x_0\in\operatorname{Int} Df$. $\epsilon_2$ comes from the condition that $x_0$ is a local extremum. Since neither condition implies the other, the author has to call them out separately, then take the smaller value to be assured of an interval where both conditions are true.
These are open intervals. They do not include endpoints. There are two reasons for the $\le$: First, $x_0 \in U_{\epsilon}(x_0)$, so $x$ could be $x_0$. Second, the condition for local maximum does not exclude other nearby points with the same value. It just excludes the possibility of nearby points with a higher value. For example, every point is a local maximum and a local minimum for a constant function, even though nearby points have the same value.
What that notation is supposed to mean (it is really, really poor in my opinion), is that the domain of the expression in the limits is restricted to all values of $\Delta x$ such that $x_0 + \Delta x \in U_\epsilon(x_0)$ (note that it is not necessary for $\Delta x$ itself to be in this interval). That is, it is a restriction we are placing on the values of $\Delta x$ that we are taking the limit over, not an additional statement that we are claiming is true.
However, this same requirement is much more easily and understandably stated as just $|\Delta x| < \epsilon$, since $U_\epsilon(x_0)$ is the interval $(x_0 - \epsilon, x_0 + \epsilon)$.