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 $\mathcal J(\epsilon):=J(y_0+ \epsilon h)$


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.


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 $0$ ?


1 Answer 1


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 $t=0$.

  • $\begingroup$ I see... :) But after that, why do we deduce that $\mathcal{J}'(0)=0$ although we are looking at the proposition of the proof that if $y_0$ is a local minimum then $\delta J(y_0,h)=0$ ? Because of the fact that $\mathcal{J}$ is a function of one variable? $\endgroup$
    – evinda
    May 24, 2015 at 11:18
  • $\begingroup$ Yes, if a differentiable function has a minimum at $0$, then the derivatives has to be zero. @evinda $\endgroup$
    – user99914
    May 24, 2015 at 11:20
  • $\begingroup$ A ok. And do we know that $\mathcal{J}$ is differentiable because of the fact that the first derivative of $J$ is defined? $\endgroup$
    – evinda
    May 24, 2015 at 11:22
  • $\begingroup$ Yes, you need to make assumptions on $J$ such that $\mathcal J$ is differentiable. $\endgroup$
    – user99914
    May 24, 2015 at 11:27
  • $\begingroup$ What other assumptions do we have to make on $J$ so that $\mathcal{J}$ is differentiable? $\endgroup$
    – evinda
    May 24, 2015 at 17:00

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