In this [question](https://mathoverflow.net/questions/412729/can-we-invoke-almost-supermartingale-theorem-for-deterministic-sequences), there is a proof for deterministic version of "Almost Supermartingale"


**Question: Can we extend [[1](https://mathoverflow.net/questions/412729/can-we-invoke-almost-supermartingale-theorem-for-deterministic-sequences)] 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](https://mathoverflow.net/questions/412729/can-we-invoke-almost-supermartingale-theorem-for-deterministic-sequences)] besides stating non-random version of (almost) supermartingale?




---


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](https://mathoverflow.net/questions/412729/can-we-invoke-almost-supermartingale-theorem-for-deterministic-sequences)], 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?**



Proof of yet another extension of deterministic variant of "(Almost) Supermartingale" convergence theorem
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 $\beta\_k=0$.


