Consider Banach Space $L^{p}(U)$ where $U$ is bounded open subset in $\mathbb{R}^{n}$. Take a bounded sequence $\{u_{m}\}_{m}^{\infty}$ in $L^{p}(U)$. Consider a subsequence $\{u_{m_{j}}\}_{j}^{\infty} \subset \{u_{m}\}_{m}^{\infty}$ such that $\text{limsup}_{j,k \rightarrow \infty}||u_{m_{j}}-u_{m_{k}}||_{L^{p}(U)} \leq \delta$.
What is the standard diagonal argument to extract a subsequence $\{u_{m_{l}}\}_{l}^{\infty} \subset \{u_{m}\}_{m}^{\infty}$ satisfying:
$\text{limsup}_{j,k \rightarrow \infty}||u_{m_{j}}-u_{m_{k}}||_{L^{p}(U)} = 0$
with $\delta = 1,\frac{1}{2},\frac{1}{3},...$
Thanks!
You have a sequence that is not only bounded in $L^p(U)$, you have a sequence that is bounded in $W^{1,p}(U)$. That is an important difference, since that also gives bounds for the (weak) derivatives, and the long first part of the proof establishes that under these conditions, you can, for every $\delta > 0$, extract a subsequence such that
$$\limsup_{j,k\to\infty} \lVert u_{m_j} - u_{m_k}\rVert_{L^q(U)} \leqslant \delta.$$
Now we come to the standard diagonal sequence construction. Note that a subsequence of a sequence $x_m$ corresponds to a strictly increasing function $\sigma \colon \mathbb{N}\to\mathbb{N}$, per $x_{m_k} = x_{\sigma(k)}$.
So extracting the first subsequence for $\delta = 1$ produces a sequence
$$u_{m_k} = u_{\sigma_1(k)} = u_{1,k}$$
of the original sequence with $\limsup\limits_{j,k\to\infty} \lVert u_{1,j} - u_{1,k}\rVert \leqslant 1$. Now the sequence $(u_{1,k})_{k\in\mathbb{N}}$ is of course bounded in $W^{1,p}(U)$, so satisfies the same assumptions, and hence by the long argument before, we can extract a subsequence
$$u_{2,k} = u_{1,\sigma_2(k)} = u_{\sigma_1(\sigma_2(k))}$$
of $u_{1,k}$ with $\limsup\limits_{j,k\to\infty} \lVert u_{2,j}-u_{2,k}\rVert \leqslant \frac12$. That is still a subsequence of the original (of course), hence staisfies the assumptions, and we can extract a further subsequence... Recursively, for every $m\in \mathbb{N}$, we can extract a subsequence $(u_{m+1,k})_{k\in\mathbb{N}}$ of $(u_{m,k})_{k\in\mathbb{N}}$ with $\limsup\limits_{j,k\to\infty} \lVert u_{m+1,j} - u_{m+1,k}\rVert \leqslant \frac{1}{m+1}$.
Since in general you must expect that $\lim\limits_{m\to\infty} \sigma_1(\dotsc(\sigma_m(k))\dotsc) = \infty$ for some $k$, you can't take the "limit" of these subsequences, therefore you take the diagonal sequence
$$v_k = u_{k,k} = u_{\sigma_1(\dotsc(\sigma_k(k))\dotsc)},$$
which is a subsequence of the original, and since $(v_k)_{k\in\mathbb{N}}$ is - except possibly for the first $m$ terms - a subsequence of $(u_{m,k})$, you have $\limsup\limits_{j,k\to\infty} \lVert v_j - v_k\rVert \leqslant \frac1m$ for every $m$, and that means that $(v_k)$ is a Cauchy sequence. By completeness of $L^q(U)$, it converges, so the original sequence contains a subsequence that converges in $L^q$.