In class, we were looking for extrema $\hat{y}$ of functional $\mathcal{L}$, while demanding that $\Phi(\hat{y}) = l$ for some other functional $\Phi$.

My professor proved the following statement: If the bounded extrema of functional $\mathcal{L}$ is obtained at point $\hat{y}(x) \in M$, then we have: $\ker (D\_{\hat{y}} \Phi) \subset \ker ( D\_{\hat{y}} \mathcal{L} ).$ The proof went as follows.

Let $v = v(x,t): [a,b] \times (- \varepsilon, \varepsilon) \to U$ be a path of functions $[a,b]\to U$, where $U$ is a Banach space, such that $v(x,0) = \hat{y}(x)$. Furthermore, let $v(x,t)$ lie entirely in $\Phi^{-1} =: M$. This means that $\Phi(v(x,t)) = l$. We can take the derivative with respect to time to obtain $ \frac{d}{dt} \big|\_{t=0} \Phi(v(x,t)) = (D\_{\hat{y}(x)} \Phi)(\dot{v}) = 0. $ This is true for $\dot{v}(x) = \frac{d}{dt} \big|\_{t=0} v(x,t)$ of any path $t \mapsto v(x,t)$ in $M$ that foes through $\hat{y}(x)$. Therefore: $\dot{v} \in \ker D\_{\hat{y}(x)} \Phi$ for any such path.

Now let's take a look at derivative of a composite function: $t \mapsto \mathcal{L}(v(x,t))$. We obtain $ \frac{d}{dt} \big|\_{t=0} \mathcal{L}(v(x,t)) = 0 $, where $v(x,0) = \hat{y}(x)$ is an extrema. Therefore $(D\_{\hat{y}(x)} \mathcal{L})(\dot{v}) = 0$. We saw that $\dot{v} \in \ker D\_{\hat{y}(x)} \Phi$ and by what we wrote above, it is also true that $(D\_{\hat{y}(x)} \mathcal{L})(\dot{v}) = 0 \Rightarrow \dot{v} \in \ker D\_{\hat{y}(x)} \mathcal{L}$

QED

I am absolutely confused as to how this proof works. Instead of choosing any element in $\ker (D\_{\hat{y}} \Phi) $ and showing that it is also in $\ker ( D\_{\hat{y}} \mathcal{L} )$, we just looked at paths that go through $\hat{y}(x)$ and showed that they are in both kernels. What am I missing? How is that enough?