Proof that $F$ has a global minimun in $y=u$ where $u(x)=\frac{x-x_0}{x_1-x_0}$

Question

Let $x_0,x_1\in \mathbb{R}$ such that $x_0<x_1$ and $F$ the set of functions $y\in C^{1}([x_0,x_1];\mathbb{R})$ such that $y(x_0)=0$ and $y(x_1)=1$ Let $y\in F$ $$F[y]=\int_{x_0}^{x_1} (y')^2dx$$ Proof that $F$ has a global minimum in $y=u$ where $u(x)=\frac{x-x_0}{x_1-x_0}$

Using the Euler-Langrange equations it is easy to prove that it is a local minimum but I need it to be a global minimum and the Euler-Lagrange equations do not help in this case. How can I prove this?

I need to prove that $F[u]<F[y]$ for all $y \in F$.

Answer 1

To avoid confusion I will use $f$ to denote the functional in question.

Note that $f$ is convex and so we have $f(y)-f(u) \ge d f(u, y-u)$, where $df$ is the Gateaux derivative of $f$ at $u$ in the direction $y-u$.

Hence $u$ is a global $\min$ iff $d f(u, y-u) \ge 0$ for all $y \in F$.

It is not difficult to show that $d f(u, y-u) = 2 \int_{x_0}^{x_1} u'(x) (y'(x)-u'(x)) dx$, and substituting the actual value for the first instance of $u'$ we get $df(u,y-u) = { 1\over x_1-x_0} ((y(1)-u(1))-(y(0)-u(0))$ and if $y \in F$ we get $df(u,y-u) = 0$. Hence $u$ is a global minimiser.

Answer 2

You can argue that Legendre's condition $f_{y'y'}=2>0$ holds strictly, and there are no conjugate points (w.r.t. the Jacobi equation) to $0$ in $[0,1]$. Therefore the second variation is strongly positive definite at every point, your functional is convex and the only extremal is the global minimum