[proofplan] The proof is a chain-rule computation for the Legendre transform. First we differentiate the Hamiltonian $H(x,y,p)=p\cdot v(x,y,p)-L(x,y,v(x,y,p))$ and use the defining relation $p=\partial_vL(x,y,v)$ to cancel all terms involving derivatives of the inverse velocity map. The two resulting identities, $\partial_pH=v$ and $\partial_yH=-\partial_yL$, convert the Euler-Lagrange equation into Hamilton's equations and convert Hamilton's equations back into the Euler-Lagrange equation. [/proofplan]

[step:Derive the differential identities for the Hamiltonian]Let \begin{align*} V:W\to U \end{align*} denote the map \begin{align*} V(x,y,p):=(x,y,v(x,y,p)). \end{align*} Since $W$ is relatively open by hypothesis, and since $L\in C^2(U;\mathbb R)$ and $v\in C^1(W;\mathbb R^n)$, the Hamiltonian $H\in C^1(W;\mathbb R)$. Thus the following partial derivatives are taken on the relatively open domain $W$. Fix $(x,y,p)\in W$, and put \begin{align*} \tilde v:=v(x,y,p). \end{align*} By the defining property of the inverse Legendre map, \begin{align*} p=\partial_vL(x,y,\tilde v). \end{align*} For each $j\in\{1,\dots,n\}$, the chain rule gives \begin{align*} \frac{\partial H}{\partial p_j}(x,y,p)=\tilde v_j+\sum_{i=1}^n p_i\frac{\partial v_i}{\partial p_j}(x,y,p)-\sum_{i=1}^n \frac{\partial L}{\partial v_i}(x,y,\tilde v)\frac{\partial v_i}{\partial p_j}(x,y,p). \end{align*} Using $p_i=\frac{\partial L}{\partial v_i}(x,y,\tilde v)$ for every $i$, the two sums cancel, so \begin{align*} \frac{\partial H}{\partial p_j}(x,y,p)=\tilde v_j. \end{align*} Since this holds for each component $j$, \begin{align*} \partial_pH(x,y,p)=v(x,y,p). \end{align*} For each $j\in\{1,\dots,n\}$, the chain rule with respect to the coordinate $y_j$ gives \begin{align*} \frac{\partial H}{\partial y_j}(x,y,p)=\sum_{i=1}^n p_i\frac{\partial v_i}{\partial y_j}(x,y,p)-\frac{\partial L}{\partial y_j}(x,y,\tilde v)-\sum_{i=1}^n \frac{\partial L}{\partial v_i}(x,y,\tilde v)\frac{\partial v_i}{\partial y_j}(x,y,p). \end{align*} Again using $p_i=\frac{\partial L}{\partial v_i}(x,y,\tilde v)$, the two sums cancel, and therefore \begin{align*} \frac{\partial H}{\partial y_j}(x,y,p)=-\frac{\partial L}{\partial y_j}(x,y,\tilde v). \end{align*} Thus \begin{align*} \partial_yH(x,y,p)=-\partial_yL(x,y,v(x,y,p)). \end{align*}[/step]

[guided]The Hamiltonian depends on $p$ both directly through the dot product $p\cdot v(x,y,p)$ and indirectly through the velocity $v(x,y,p)$. The point of the Legendre transform is that the indirect terms cancel because $p$ is exactly the velocity gradient of $L$. Define the map \begin{align*} V:W\to U \end{align*} by \begin{align*} V(x,y,p):=(x,y,v(x,y,p)). \end{align*} Because $W$ is relatively open by hypothesis, the partial derivatives of $H$ are taken on a genuine relative domain. The assumptions $L\in C^2(U;\mathbb R)$ and $v\in C^1(W;\mathbb R^n)$ imply, by the chain rule for $C^1$ maps, that $H\in C^1(W;\mathbb R)$. Fix $(x,y,p)\in W$ and define \begin{align*} \tilde v:=v(x,y,p). \end{align*} The inverse Legendre relation says precisely that \begin{align*} p=\partial_vL(x,y,\tilde v). \end{align*} We first differentiate $H$ with respect to $p_j$, where $j\in\{1,\dots,n\}$. Since \begin{align*} H(x,y,p)=p\cdot v(x,y,p)-L(x,y,v(x,y,p)), \end{align*} the product rule applied to the dot product gives one direct term $\tilde v_j$ and one indirect term involving $\partial v_i/\partial p_j$. The chain rule applied to $L(x,y,v(x,y,p))$ gives another indirect term involving the same derivatives of $v$. Therefore \begin{align*} \frac{\partial H}{\partial p_j}(x,y,p)=\tilde v_j+\sum_{i=1}^n p_i\frac{\partial v_i}{\partial p_j}(x,y,p)-\sum_{i=1}^n \frac{\partial L}{\partial v_i}(x,y,\tilde v)\frac{\partial v_i}{\partial p_j}(x,y,p). \end{align*} Now substitute the defining identity $p_i=\frac{\partial L}{\partial v_i}(x,y,\tilde v)$. The two sums are identical with opposite signs, hence \begin{align*} \frac{\partial H}{\partial p_j}(x,y,p)=\tilde v_j. \end{align*} Since this holds for every component $j$, we obtain the vector identity \begin{align*} \partial_pH(x,y,p)=v(x,y,p). \end{align*} The derivative with respect to $y$ is similar, except there is no direct derivative of $p$ because $p$ is held fixed when taking the partial derivative $\partial_yH$. For $j\in\{1,\dots,n\}$, the chain rule gives \begin{align*} \frac{\partial H}{\partial y_j}(x,y,p)=\sum_{i=1}^n p_i\frac{\partial v_i}{\partial y_j}(x,y,p)-\frac{\partial L}{\partial y_j}(x,y,\tilde v)-\sum_{i=1}^n \frac{\partial L}{\partial v_i}(x,y,\tilde v)\frac{\partial v_i}{\partial y_j}(x,y,p). \end{align*} Again the Legendre relation $p_i=\frac{\partial L}{\partial v_i}(x,y,\tilde v)$ cancels the two sums. What remains is \begin{align*} \frac{\partial H}{\partial y_j}(x,y,p)=-\frac{\partial L}{\partial y_j}(x,y,\tilde v). \end{align*} Since this holds for every component $j$, \begin{align*} \partial_yH(x,y,p)=-\partial_yL(x,y,v(x,y,p)). \end{align*} These two identities are the only special facts about the Legendre transform needed in the rest of the proof.[/guided]

[step:Convert the Euler-Lagrange equation into Hamilton's equations] Assume $y\in C^2([a,b];\mathbb R^n)$, assume $(x,y(x),y'(x))\in U$ for every $x\in[a,b]$, and define \begin{align*} p:[a,b]\to\mathbb R^n \end{align*} by \begin{align*} p(x):=\partial_vL(x,y(x),y'(x)). \end{align*} Since $L\in C^2(U;\mathbb R)$ and $y\in C^2([a,b];\mathbb R^n)$, the curve $p$ is of class $C^1$. By the definition of $W$, the point $(x,y(x),p(x))$ belongs to $W$ for every $x\in[a,b]$. The inverse Legendre relation gives \begin{align*} v(x,y(x),p(x))=y'(x). \end{align*} Using $\partial_pH=v$ from the previous step, we obtain \begin{align*} y'(x)=\partial_pH(x,y(x),p(x)). \end{align*} This is the first Hamilton equation. The Euler-Lagrange equation states \begin{align*} \partial_yL(x,y(x),y'(x))-\frac{d}{dx}\partial_vL(x,y(x),y'(x))=0. \end{align*} Since $p(x)=\partial_vL(x,y(x),y'(x))$, this is equivalent to \begin{align*} p'(x)=\partial_yL(x,y(x),y'(x)). \end{align*} Using $y'(x)=v(x,y(x),p(x))$ and $\partial_yH=-\partial_yL(\cdot,\cdot,v(\cdot,\cdot,\cdot))$ from the previous step, we get \begin{align*} p'(x)=-\partial_yH(x,y(x),p(x)). \end{align*} This is the second Hamilton equation. [/step]

[step:Convert Hamilton's equations back into the Euler-Lagrange equation] Conversely, assume $y,p\in C^1([a,b];\mathbb R^n)$, assume $(x,y(x),p(x))\in W$ for all $x\in[a,b]$, and assume Hamilton's equations hold. The first Hamilton equation and the identity $\partial_pH=v$ give \begin{align*} y'(x)=v(x,y(x),p(x)). \end{align*} Since $v\in C^1(W;\mathbb R^n)$, $y,p\in C^1([a,b];\mathbb R^n)$, and $(x,y(x),p(x))\in W$ for all $x\in[a,b]$, the composition $x\mapsto v(x,y(x),p(x))$ is of class $C^1$. Hence $y'\in C^1([a,b];\mathbb R^n)$, so $y\in C^2([a,b];\mathbb R^n)$. Therefore $(x,y(x),y'(x))\in U$ and, by the defining property of the inverse Legendre map, \begin{align*} p(x)=\partial_vL(x,y(x),y'(x)). \end{align*} Because $p\in C^1([a,b];\mathbb R^n)$, we may differentiate this identity along the curve and obtain \begin{align*} p'(x)=\frac{d}{dx}\partial_vL(x,y(x),y'(x)). \end{align*} The second Hamilton equation and the identity $\partial_yH=-\partial_yL(\cdot,\cdot,v(\cdot,\cdot,\cdot))$ give \begin{align*} p'(x)=-\partial_yH(x,y(x),p(x))=\partial_yL(x,y(x),v(x,y(x),p(x))). \end{align*} Using $v(x,y(x),p(x))=y'(x)$, this becomes \begin{align*} p'(x)=\partial_yL(x,y(x),y'(x)). \end{align*} Combining the two displayed identities for $p'(x)$ yields \begin{align*} \frac{d}{dx}\partial_vL(x,y(x),y'(x))=\partial_yL(x,y(x),y'(x)). \end{align*} Equivalently, \begin{align*} \partial_yL(x,y(x),y'(x))-\frac{d}{dx}\partial_vL(x,y(x),y'(x))=0. \end{align*} Thus $y$ satisfies the Euler-Lagrange equation, and the equivalence is proved. [/step]