Proof of yet another extension of deterministic variant of "(Almost) Supermartingale" convergence theorem

Question

In this question, there is a proof for deterministic version of "Almost Supermartingale"

Question: Can we extend [1] as following? If yes, can we prove it?

Let the non-negative sequences be $\{V_1^k\}$, $\{V_2^k\}$, $\{S_{1,2}^k\}$, and $\{ U_{1,2}^k \}$ for $k=0,1,2,\ldots$. Let $\alpha_0, \alpha_1, \ldots$ and $\beta_0, \beta_1, \ldots$ be non-negative scalars satisfying $\sum_{k=0}^\infty \alpha_k < \infty$ and $\sum_{k=0}^\infty \beta_k < \infty$. Assume $$V_1^{k+1} + V_2^{k+1} \leq \left( 1 + \alpha_k \right) V_1^k + \left( 1 + \beta_k \right) V_2^k - S_{1,2}^k + U_{1,2}^k$$ and $$\sum_{i=1}^\infty U_{1,2}^i < \infty.$$ Then,

1. $V_1^k \rightarrow V_1^\infty$
2. $V_2^k \rightarrow V_2^\infty$
3. $\sum_{k=0}^\infty S_{1,2}^k < \infty$.

P.S.: Question: Is there any "name" of such convergence theorem in [1] besides stating non-random version of (almost) supermartingale?

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ATTEMPT:

We can split the above inequality \begin{align} V_1^{k+1} \leq \left( 1 + \alpha_k \right) V_1^k - {\color{red}{\frac{1}{2}}}S_{1,2}^k + {\color{red}{\frac{1}{2}}}U_{1,2}^k \tag{1} \end{align} and \begin{align} V_2^{k+1} \leq \left( 1 + \beta_k \right) V_2^k - {\color{red}{\frac{1}{2}}}S_{1,2}^k + {\color{red}{\frac{1}{2}}}U_{1,2}^k. \tag{2} \end{align}

From [1], we know that both $(1)$ and $(2)$ can have $V_1^k \to V_1^\infty$ and $V_2^k \to V_2^\infty$. Also, $\sum_{k=0}^\infty S_{1,2}^k < \infty$ follows directly from both $(1)$ and $(2)$.

Does this above approach make sense?

Answer 1

No. E.g., for $j=1,2$ and $k=0,1,\dots$, take $V_j^k=2+(-1)^{j+k}$, $S_{1,2}^k=0$, $U_{1,2}^k=0$, $\alpha_k=0$, and $\beta_k=0$.