I am following a proof that some sequence is Cauchy's, and end of proof is omitted so I skipped it thinking it's easy but when I took a closer look I realized that I don't understand an expression. Here how it goes: $\Delta_{n,m}$ and $x_m$ are some partial sums.
Take $m>n$, it is shown that $$\Delta_{n,m}\le x_{m+1}$$ where $$\liminf_{n \to \infty}x_{n}=0$$
from this two follows $$\liminf_{m \to \infty} \max_{m>n}\Delta_{n,m}=0$$
and in the end this is used to show that since $$ \|f_k-f_l \|^2=y_k-y_l-2\Delta_{k,l} $$
and $y_n$ converges $\{f_n\}_{n=1}^{\infty}$ is Cauchy's
My guess is that because of limit inferior we find some $r$ such that $\Delta_{r,k}$ and $\Delta_{l,r}$ are small enough and use $\|f_k-f_r +f_r - f_l \|$ with triangle inequality and that $y_i$ and $y_j$ are close, but I don't understand how they got $\liminf_{m \to \infty} \max_{m>n}\Delta_{n,m}=0$ and how it is used later to finish proof
Note that $\Delta_{n,m}\le x_{m+1}$ depends only on greater of them 2 ( $m$ in this case ) since we have that $\liminf$ of $x_m$ is 0 there are infinitely many $x_m$ close to 0, so we can find some $r$ such that $\Delta_{n,r}$ and $\Delta_{m,r}$ are smaller then $\varepsilon/16$ ( $r$ is such that $x_{r+1}<\varepsilon/16$ ).
use that $\|a+b\|^2\le\|a\|^2+\|b\|^2+2\|a\|\|b\|$ where $a=f_n-f_r$ and $b=-(f_m-f_r)$
finally use that $y_n$ is Cauchy's and that (for $r>m>n$ big enough) we have $|y_n-y_r|<\varepsilon/8$ and $|y_m-y_r|<\varepsilon/8$
add that up and get $\|f_n-f_m\|^2\le \varepsilon$