Weierstrass's counterexample to the Dirichlet principle

Counterexample

Consider the integral functional $$J(u) = \int_{-1}^{1} x^2 \, u'(x)^2 \, dx$$ on the set $\mathcal{A}$ of functions $u \in C^1([-1, 1])$ with $u(-1) = -1$ and $u(1) = 1$. Then $\inf_{\mathcal{A}} J = 0$, but no $u \in \mathcal{A}$ has $J(u) = 0$, so $J$ has no minimiser on $\mathcal{A}$.

Check that the infimum is $0$. For $\varepsilon > 0$, put $u_\varepsilon(x) = \arctan(x / \varepsilon) / \arctan(1 / \varepsilon)$, which lies in $\mathcal{A}$. With $c_\varepsilon = \arctan(1 / \varepsilon)$, one has $u_\varepsilon'(x) = \varepsilon / \big(c_\varepsilon (\varepsilon^2 + x^2)\big)$. Since $x^2 \le \varepsilon^2 + x^2$, $$J(u_\varepsilon) \le \frac{1}{c_\varepsilon^2} \int_{-1}^{1} \frac{\varepsilon^2}{\varepsilon^2 + x^2} \, dx = \frac{2 \varepsilon}{c_\varepsilon},$$ which tends to $0$ as $\varepsilon \to 0$, because $c_\varepsilon \to \pi / 2$.

Check that $0$ is not attained. If $J(u) = 0$, then $x u'(x) = 0$ for all $x$, by continuity of the integrand. So $u' = 0$ for $x \ne 0$, hence also at $0$, and $u$ is constant. This contradicts $u(-1) = -1 \ne 1 = u(1)$.