How does this proof on constrained extrema even work? 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?
 A: The proof aims to show that if y^(x) is a bounded extrema of functional L, then the kernel of Dy^Φ is contained in the kernel of Dy^L. The starting point of the proof is to consider a path v(x,t) that lies in M and passes through y^(x) at t=0. By definition, the derivative of Φ with respect to time at t=0 is zero along this path, which means that v˙(x) lies in the kernel of Dy^Φ. Next, the derivative of the composite function t↦L(v(x,t)) is calculated at t=0 and shown to be zero, implying that v˙(x) also lies in the kernel of Dy^L. Hence, it is concluded that the kernel of Dy^Φ is contained in the kernel of Dy^L. The reason this is enough to conclude that the kernel of Dy^Φ is contained in the kernel of Dy^L is because the kernel of a linear operator is a linear subspace, and it is closed under addition and scalar multiplication. Hence, if a point v˙(x) lies in the kernel of Dy^Φ and is also in the kernel of Dy^L, then it follows that any linear combination of such points will also lie in the kernel of both Dy^Φ and Dy^L. In conclusion, the proof you mentioned shows that if y^(x) is a bounded extrema of functional L, then any element in the kernel of Dy^Φ will also be in the kernel of Dy^L.
A: The fact that $v(x,t)$ lies entirely in $\Phi^{-1}(l)$ implies that its tangent vector is in $ker(D_{\hat{y}}\Phi)$. This comes from a standard characterization of the tangent space of the inverse image. Your professor is using a definition of tangent space that implies that all tangent vectors come as the tangent vectors to paths. So this part of the argument adresses your question.
