I have difficulty proving the following claim from a paper (a free version is here, see Lemma 2.4 on page 9):
Let in a Banach space $X$ a sequence $\{x_n\}_{n=1}^\infty$ be given. Assume that for any $k,l$ we have $$ ||x_k − x_l||^2 = y_k − y_l + \theta_{k,l}, $$ where $\{y_n\}_{n=1}^\infty$ is a convergent sequence of real numbers and the $\theta_{k,l}$ satisfy the property $$ \liminf\limits_{l\to \infty} \max\limits_{k<l}\theta_{k,l} = 0. \tag{1} $$ Then $\{x_n\}_{n=1}^\infty$ converges.
I understand that the aim is to show that the $x_n$ form a Cauchy sequence which implies convergence in a Banach space.
I was able to handle the special case where all $y_j$ are zero but I don't know how to proceed when $\{y_n\}$ is a general convergent sequence.
Here is my work: If all $y_j$ are zero, $(1)$ becomes $$ 0 = \liminf\limits_{l\to \infty} \max\limits_{k<l}||x_k − x_l||^2. \tag{1'} $$ Claim: The sequence $\{x_n\}$ is constant and therefore convergent. Assume by contradiction that there exist $n<m$ such that $x_n \ne x_m.$ Let $$ \epsilon=||x_n - x_m|| > 0 $$ Then for all $l > m$ it holds that \begin{align*} \max\limits_{k<l}||x_k − x_l|| &\ge \max\{||x_n − x_l||, ||x_m − x_l||\} \\ &\ge \frac{1}{2}(||x_n − x_l|| + ||x_m − x_l||) \\ &\ge \frac{1}{2}||x_n − x_m|| = \epsilon/2. \end{align*} In particular, $$ \liminf\limits_{l\to \infty} \max\limits_{k<l}||x_k − x_l||^2 \ge (\epsilon/2)^2, $$ in contradiction to $(1')$.
Here is the citation for the paper: Temlyakov, V.N., Weak greedy algorithms, Adv. Comput. Math. 12, No.2-3, 213-227 (2000). ZBL0964.65009.
If the sequence is not a Cauchy sequence, then you can pick $\varepsilon >0$ and sequences $(n_l)_{l\geq 1}, (m_l)_{l\geq 1}$ such that $n_l<m_l$ and $$ \Vert x_{n_l} - x_{m_l} \Vert^2 \geq \varepsilon.$$ Thus, we get $$ \varepsilon \leq \vert y_{n_l} - y_{m_l} \vert + \max_{k<m_l} \Theta_{k, m_l}.$$ If you take the limes inferior for $l\rightarrow \infty$ you obtain the contradiction $\varepsilon \leq 0$.