In Feynman's lecture on least action, he mentions that a perturbation at an extremum of the action functional would cause a zero first order change in the action. Intuitively this makes sense because I find it analogous to the Taylor expansion of a function - at extrema, $f^\prime = 0$ this the first order term disappears. My difficulty is with the fact that the action is a functional, and thus I don’t really understand this rigorously.
http://www.feynmanlectures.caltech.edu/II_19.html
Furthermore, the Wikipedia derivation of the Euler Lagrange equation seemed to make sense to me. However, I am unable to connect that derivation to statement made by Feynman.
The proof is very easy if you know the multivariable chain rule (for the Calculus of Variations, you need to know the infinite-dimensional version... which is still the same) and the single variable version of this theorem. The statement
made by Feynman would be phrased in the following manner by a Math person:
Note that the differential $df_{x_0}$ is itself by definition a continuous linear transformation from $V$ into $\Bbb{R}$. So, what we have to show is that for every "perturbation" $\xi \in V$, we have $df_{x_0}(\xi) = 0$. So, to prove this, pick any $\xi \in V$. Now, define the curve $\gamma : \Bbb{R} \to V$ by $\gamma(t) = x_0 + t\xi$. Now, consider the composite mapping $f \circ \gamma: \Bbb{R} \to \Bbb{R}$. It is clear that $\gamma$ is differentiable, and by assumption, $f$ is also differentiable. Hence, by the chain rule, $f \circ \gamma$ is differentiable at the origin.
Next, note that since $x_0$ is a local extremum of $f$, it follows that $0$ is a local extremum for $f \circ \gamma$. Since $f\circ \gamma$ is a simple mapping from $\Bbb{R} \to \Bbb{R}$, we can use the simple single-variable calculus version of the theorem to conclude that $(f \circ \gamma)'(0) = 0$. Hence, \begin{align} 0 &= (f \circ \gamma)'(0) \\ &= df_{\gamma(0)}\big( \gamma'(0)\big) \tag{chain rule} \\ &= df_{x_0}(\xi) \tag{by definition of $\gamma$} \end{align}
This is precisely what we wanted to prove! Since $\xi$ was arbitrary, this completes the proof that $df_{x_0} = 0$.
The theorem proven above is very general. In the particular context of the principle of "least" action, here's how we interpret the above theorem. The Banach space $V$ is usually a set of curves in the configuration space of the system. We interpret $f$ to be the action (I realize the typical notation is $S$ or $J$), and what I called $x_0$ is going to be a " local extremal curve" for the action $f$.
By the way, the Euler Lagrange equations have absolutely nothing to do with the actual statement Feynman made. The statement made by Feynman translated word-for-word into more pedantic and technical mathematical jargon is exactly what I have written in the Theorem above.
What the Euler Lagrange equations tell you is that if the action $S: V \to \Bbb{R}$ is of the specific form \begin{align} S(\varphi) := \int_a^b L(t, \varphi(t), \varphi'(t)) \, dt \end{align} where $L: \Bbb{R} \times \Bbb{R} \times \Bbb{R} \to \Bbb{R}$ is a smooth function, then the condition $dS_{\varphi} = 0$ is equivalent to the Euler Lagrange equations: \begin{align} \dfrac{d}{dt} \bigg( \big(\partial_3L\big) \big(t, \varphi(t), \varphi'(t) \big) \bigg) - \big(\partial_2 L \big)(t, \varphi(t), \varphi'(t)) = 0 \end{align} Where $\partial_iL$ means the $i^{th}$ partial derivative of $L$.