Analysing functionals having no local extrema In the calculus of variations, how do we analyse functionals for which there are no local extrema? In basic calculus, functions not having local extrema can often be seen to diverge to an infinity (e.g. the function given by $f(x) = x$) or approach an asymptote (e.g. arctan). Figuring out the behaviour is usually not so hard because the ambient space $\mathbb{R}^n$ is somehow quite comprehensible. But when we're dealing with function spaces on which functionals are defined, how can we get a grip on the behaviour without local extrema? 
A nice example for illustration is $J(y) = \int_{-1}^1 x \sqrt{1 + (y')^2} dx$.
Thanks in advance.
 A: The saddle-like critical points of a functional can also be of interest, especially for functionals that involve Lagrange multipliers coming  from optimization constraints. The functional in your example has a critical point at $y=0$, which is not a local extremum.   
Also, for a functional that does not attain its (finite) supremum or infimum, it is interesting to look at the sequences $f_n$ such that  $J(f_n)$ tends to the extremal value. Sometimes one sees that although the sequence $f_n$ does not have a limit under the original topology of the function space, it has a limit in some weaker topology, and possibly in a larger space. For example, consider the functional $M(f)=\int_0^1 x f(x)\,dx$ defined on the closed unit ball of $L^1([0,1])$. The supremum of $M$ is $1$, but it is not attained. Considering the sequence $f_n=n\chi_{[1-1/n,1]}$, for which $M(f_n)\to 1$, we observe that $f_n$ converge to point mass at $1$ in the sense of the weak* convergence of measures.  

Philosophical remarks. We  study functionals not for their own sake, but to understand the nature of functions on which they act. If the functional is somehow "physically relevant", the functions which realize or approach its extrema may be "physically relevant" too. For this reason, I don't  find the functional you gave "nice". It looks as if someone forgot the absolute value sign around $x$. By making $y$ wiggly to the left of zero or to the right of zero, I can make $J$ attain any value I want. The functional does not identify any particular function or class of functions to me. 
In contrast, $\int_{-1}^1 \sqrt{1+(y')^2}$ is bounded from below. Its  extrema form an important class of functions: affine functions. The functional $\int_{0}^1 x\sqrt{1+(y')^2}$ is even more interesting  to  minimize: it is bounded from below, but it is not obvious what the minimizer will be. But having a functional that's not bounded from either side immediately takes the fun out of the game.
