For sufficiently small arguments, why does change of function has the same sign as differential? From I. M. Gelfand, S. V. Fomin - Calculus of Variations (2000) page 13: 
"A necessary condition for the differentiable functional
$J[y]$ to have an extremum for y = y_0 is that its variation vanishes for $y = y_0$,
i.e., that $\delta J[h] = 0$ for $y = y_0$ and all admissible $h$."
He proceeds proving the following theorem by the following argument.
"To be explicit, suppose $J[y]$ has a minimum for $y = y_0$.
According to the definition of the variation $\delta J[h]$, we have
$\Delta J [h] = \delta J[h] + \epsilon ||h||$, where $\epsilon \to 0$ as $||h|| \to 0$. Thus, for sufficiently small $||h||$ the sign of $\delta J[h]$ will be the same as the sign of $\Delta J[h]$. "
I have trouble understanding the last statement. 
I understand that for $J[y]$ to have a minimum we have by definition that there is some $\delta>0$ such that for all $||h|| < \delta$ we have $\Delta J[h] \geq 0$. So I would like to find $\delta_2 > 0$ that would guarantee me that $\delta J[h] \geq 0$ for all $||h|| <$ min$(\delta_1, \delta_2)$. Unfortunately, I don't see how. I feel like we have to assume continuity of $J[y]$ but I am not positive.
Any help or suggestions are appreciated!
 A: The reason is that $\delta J$ is linear in $h$ whereas the second term in $\Delta J$ is super-linear and, as such, negligible compared to the linear term. Think of a function of $h\in\mathbb{R}$ of the form
$$
f(h)=Ah+\epsilon h
$$
with $A\neq 0$ and $\epsilon\to 0$ as $h\to 0$. For $|h|$ small enough you have
$$
|\epsilon h|\le|Ah|/2,
$$
because $\lim\limits_{h\to 0}\epsilon=0$ and you can choose $|h|$ small such that $|\epsilon|<|A|/2$. For such $h$, $f(h)\in [Ah/2,3Ah/2]$ so it has the same sign as $Ah$. 
The same argument works for any similar situation (multivariable calculus, calculus of variations, etc.)
A: The statement that $J$ is differentiable at $y_0$ is more explicitly expressed as: for every $\varepsilon > 0$, there exists an $\eta>0$ such that for all admissible $h$, if $\lVert h-a \rVert < \eta$ then
\begin{align}
|\Delta J_{y_0}(h) - \delta J_{y_0}(h)| \leq \varepsilon \lVert h \rVert
\end{align}
Equivalently, 
\begin{align}
\delta J_{y_0}(h) - \varepsilon \lVert h \rVert  \leq \Delta J_{y_0}(h)  \leq \delta J_{y_0}(h) +  \varepsilon \lVert h \rVert
\end{align}
With this you can see that if for example $\delta J_{y_0}(h)>0$, then by choosing $\varepsilon> 0$ small enough The LHS of the inequality is positive, and hence $\Delta J_{y_0}(h)$ will also be positive.

I read the proof they offered for this, and I feel that it is a little more complicated than it needs to be, and it is also worded in a slightly confusing manner. All you really need to know is the chain rule, and the single variable version of this theorem:

Single Variable Version: Let $U$ be an open subset of $\Bbb{R}$ containing the point $y_0$, and let $f: U \to \Bbb{R}$ be a given function, which is differentiable at $y_0$. If $f$ has a local extremum at $y_0$, then $f'(y_0) = 0$.

The proof of this is pretty easy, and I assume you might have seen it somewhere, but I'll write it anyway. Let's just consider the case of minimum (the maximum case can be deduced by considering the function $-f$). If $t > 0$, then
\begin{align}
\dfrac{f(y_0+t) - f(y_0)}{t} \geq 0 \tag{$*$}
\end{align}
(the numerator is $\geq 0$ by hypothesis, and since the denominator is $>0$, the quotient is $\geq 0$). Hence, taking the limit $\lim_{t \to 0^{+}}$, and using the fact that the double sided limit $\lim_{t \to 0}$ is already known to exist, we can conclude using $(*)$ that
\begin{equation}
f'(y_0) = \lim_{t \to 0} \dfrac{f(y_0+t) - f(y_0)}{t} = \lim_{t \to 0^+} \dfrac{f(y_0+t) - f(y_0)}{t} \geq 0
\end{equation}
However, if we repeat this for $t<0$, then we find that
\begin{align}
\dfrac{f(y_0+t) - f(y_0)}{t} \leq 0
\end{align}
($\leq 0$ since the denominator is negative, while numerator is $\geq 0$) Hence,
\begin{equation}
f'(y_0) = \lim_{t \to 0} \dfrac{f(y_0+t) - f(y_0)}{t} = \lim_{t \to 0^-} \dfrac{f(y_0+t) - f(y_0)}{t} \leq 0
\end{equation}
It follows that $f'(y_0) = 0$. This completes the single variable case.

For the case you are interested in, suppose $J$ has a local extremum at $y_0$. We want to show that for every admissible $h$, $\delta J_{y_0}(h) = 0$. To do this, pick any $h$. Now, define the function $\lambda:I \subset \Bbb{R} \to V$, where $I$ is a small open interval containing $0$ and $V$ is the vector space of curves by
\begin{equation}
\lambda(t) = y_0 + th
\end{equation}
By assumption, $J$ has a local extremum at $y_0$. This implies the composite mapping $J \circ \lambda : I \to \Bbb{R}$ has a local extremum at $0$. Since we assumed $J$ is differentiable at $y_0$, and $\lambda$ is clearly everywhere differentiable, $J \circ \lambda$ being a composite function is differentiable at $0$ (by the chain rule). Hence, by the single variable case, we know that $(J \circ \lambda)'(0) = 0$. Using the chain rule, we get:
\begin{equation}
0 = (J \circ \lambda)'(0) = \delta J_{y_0}(\lambda'(0)) = \delta J_{y_0}(h)
\end{equation}
Since $h$ was arbitrary, it follows that $\delta J_{y_0} = 0$, which is what we wanted to prove.

