replace supremum by a sequence

Question

I am reading the note https://www.ceremade.dauphine.fr/~lewin/data/conc-comp.pdf. I have a question showing in the following picture. My question is how to choose a sequence $z=\{z_n\}\subset \mathbb R^d$ such that the equality holds.

Answer 1

It seems to me that the limit on the left hand side also needs to be a lim sup, unless there's some relevant context not given here in in the question. Because, for instance, infinitely many of the $u_n$ could be $0$.

But with that small change, you can do this by a diagonalization argument. For each integer $k$ there exists arbitrarily large $n_k$ such that $$\left|\limsup_{n \to \infty} \sup_{z \in \mathbb{Z}^d} \int_{C_z} |u_{n}|^2 - \sup_{z \in \mathbb{Z}^d} \int_{C_z} |u_{n_k}|^2 \right| < \frac{1}{k},$$ by definition of lim sup. Then for each fixed $k$, there exists a sequence $z_{k,j}$ such that $$\left|\sup_{z \in \mathbb{Z}^d} \int_{C_z} |u_{n_k}|^2 - \int_{C_{z_{j,k}}} |u_{n_k}|^2\right| < \frac{1}{j}.$$ Then the diagonal sequence $z_{k,k}$ will satisfy $$\lim_{k \to \infty} \int_{C_{z_{k,k}}}|u_{n_k}|^2 = \limsup_{n \to \infty} \sup_{z \in \mathbb{Z}^d} \int_{C_z} |u_{n}|^2$$