Arnol'd's trivium problem #68

First, it's straightforward to show that we can restrict attention to functions that respect the radial symmetry of the problem. Since the gradient decomposes into a radial and a tangential component, given any function $u$ that isn't radially symmetric we can integrate the square of the radial component of the gradient along the radii, find the minimum of the resulting angular function (which exists since this a continuous function on a compact domain) and replicate the function values at that angle for all angles. Then the radial component of the integral hasn't increased and the tangential component is now zero, so the radially symmetric function thus constructed leads to a lower value of the objective functional.

So we can consider a function $u(r,\phi)=f(r)$, with $f(0)=0$, $f(1)=1$ and suitable behaviour at $r=0$ to ensure smoothness. The objective functional becomes

$$ \int_0^1f'(r)^2r\mathrm dr\;. $$

The Euler-Lagrange equation is $(2f'r)'=0$, so $f'r=C$ is a constant, and thus $f=C\log r+B$.

Here's where the ugly part begins. Since this is singular at the origin, we can't use it to get a full solution to the problem. (Presumably this is why it says $\inf$ and not $\min$.) We can cut the solution off when it hits zero, forgetting about the $C^\infty$ requirement for the moment, and use

$$ f(r)=\begin{cases}1-\log r/\log r_0&r\ge r_0\\0&\text{otherwise}\end{cases} $$

to obtain

$$ \int_{r_0}^11/(r\log r_0)^2r\mathrm dr=\left[-\frac{\log r}{\log^2 r_0}\right]_{r=r_0}^{r=1}=\frac1{\log r_0}\;, $$

which goes to $0$ for $r_0\to0$. Now we can convolve this function with a mollifier to make it $C^\infty$, and though the details of showing this rigorously seem tedious, it seems clear that we can change both the integral and the function value at $1$ by as little as we want, which shows that the infimum is $0$.