TITLE: Minimizing $\int_0^1|f'(t)|^2dt$ subject to $f(0)=0$ and $f(1)=1$.
QUESTION [6 upvotes]: Say you want to minimize 
$$
\int_0^1 \!|f'(t)|^2 \,dt,
$$
subject to $f(0)=0$ and $f(1)=1$, amongst infinitely differentiable functions. Is this even possible? My guess is the minimizing function $f(x)=x$.
My line of thinking...
A function $f$ minimizes $\int_0^1|f'(t)|^2\,dt$ iff it minimizes $\int_0^1 (1+|f'(t)|^2)\, dt$, iff it minimizes $\int_0^1\sqrt{1+|f'(t)|^2}\,dt$. Since this last expression is the arc length formula, the question is equivalent to finding the shortest continuous function from $(0,0)$ to $(1,1)$, which is just $f(x)=x$, and this is the unique function which minimizes the integral.
Is this line of reasoning sound? Maybe I overlooked some subtlety.

REPLY [3 votes]: Alternatively, you can apply integral Cauchy-Schwarz
which is $\int_0^1 h^2(x)dx\cdot\int_0^1 g^2(x)dx\ge \left(\int_0^1 h(x)g(x)dx\right)^2$ for $h(x)=f'(x)$ and $g(x)=1$ to get
$$\int_0^1 f'^2(x)dx\cdot\int_0^1 1 dx\ge \left(\int_0^1 f'(x)dx\right)^2=(f(1)-f(0))^2=1.$$
Equality holds when $f'(x)=const$ almost everywhere which together with boundary conditions implies that $f(x)=x.$