Why does the generalized Weierstrass theorem not apply to $\Psi(u) = \int_{-1}^1 (xu'(x))^2dx$?

Question

Let $$\Psi(u) = \int_{-1}^1 (xu'(x))^2dx$$

for $u \in \{u \in C^1[-1,1], u(-1) = 0, u(1)=1 \}$.

We have that $\Psi$ has an $\inf$ (0) but $0$ is not a $\min$. Why does Weierstrass theorem not apply?

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Weierstrass: Let $\Phi$ be coercive and a weakly lower semicontinuous functional on a reflexive space, then $u$ has a min.

Answer 1

The reason for the failure depends on the topology you are using.

• The space $X = C^1([-1,1])$ is not reflexive.
• In $X = W^{1,2}(-1,1)$, the functional is not coercive.

You can, however, work in a weighted Sobolev space: $$X := \{v \in L^2(-1,1) \mid v|_{(-1,0)} \text{ and } v|_{(-1,0)} \text{ are weakly differentiable and } \|v\|_{X} < \infty\},$$ where $$\|v\|_X^2 := \|v\|_{L^2(-1,1)}^2 + \int_{-1}^0 x^2 \, v'(x)^2 \, \mathrm{d} x + \int_0^1 x^2 \, v'(x)^2 \, \mathrm{d} x.$$ In this space, your functional achieve its minimum at $u = \chi_{(0,1)}$.