Definition:
Let $J: A \to \mathbb{R}$ be a functional , where $A \subset V$ and $(V, ||\cdot||)$ a linear space with norm.
Let $y_0 \in A$ and $h \in V$ such that $y_0+ \epsilon h \in A $ for $\epsilon$ small enough.
The first derivative of the functional $J$ at $y_0$ in the direction $h$ is defined as follows:
$$\delta J(y_0,h)= \mathcal{J}'(0)=\frac{d}{d \epsilon} J(y_0+ \epsilon h)|_{\epsilon=0}$$
where $\mathcal J(\epsilon):=J(y_0+ \epsilon h)$
Proposition:
Let $J, y_0, A, V$ as above.
If $y_0$ is a local minimum for the functional $J$ at $A$ (in respect to $||\cdot||$) then
$$\delta J(y_0,h)=0$$
for each direction $h$ for which the first derivative is defined.
Proof:
Let $y_0$ be a local minimum for the functional $J$ at $A$ and let a direction $h \in V$ for which the first derivative of $J$ at $y_0$ is defined.
Then the function $\mathcal{J}: (- \epsilon, \epsilon) \to \mathbb{R}$ (for a suitable small $\epsilon_0$) $$\epsilon \to \mathcal{J}(\epsilon):=J(y_0+ \epsilon h)$$ that is a real function of one real variable has local minimum at $0$. Thus $\mathcal{J}'(0)=0$.
Could you explain me how we deduce that if $y_0$ is a local minimum for $J$ then $\mathcal{J}$ has a local minimum at $0$ ?
If $y_0$ is a local minimum, then there is $\delta>0$ so that $J(y) \ge J(y_0)$ whenever $||y-y_0||<\delta$.
Now fix $h\neq 0$ and consider the function $\mathcal{J}: (- \epsilon, \epsilon) \to \mathbb{R}$, $\mathcal{J}(t) = J(y_0 + th)$. So if $\epsilon <\frac{\delta}{||h||}$, we have (for all $|t|<\epsilon$)
$$||y_0 + th - y_0|| = |t|\cdot ||h|| <\delta \Rightarrow J(y_0 +th) \ge J(y_0)$$
which is the same as $\mathcal J(t) \ge \mathcal J(0)$. Thus $\mathcal J$ has a local minimum at $t=0$.