Superlinearity and <continuity> make $\Phi$ bounded below by a finite constant $M$. Consequently \b[$I(u)\geq M(b-a)$ for every admissible <function>]. The affine endpoint interpolant has finite energy by <continuity> on its compact range, so the infimum is finite. Some other admissible energies can be infinite; this does not affect the bounded-below or minimizing-<sequence> argument.

Solution

Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2007 / B / 8 / a / i / Solution

 Solution (source code)