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Given a sequence $(a_n)$ in $\mathbb{R}^+$ such that for all $n, m$: $$ (+) \quad a_{n+m} \leq K (a_n + L) $$ where $K, L$ are positive constants, prove that $(a_n)$ is Cauchy.

I had an idea, but it doesn't really work. Maybe the solution has nothing to do with this, but here it goes anyway.

Chose some $\epsilon > 0$ and assume that: $(*)$ there exists an $N$ such that $2 K(a_N + L) < \epsilon$.

Take any $n, m > N$. Then: $$ |a_n - a_m| \leq 2 K(a_N + L) $$ which is justified by $(+)$ together with the triangle inequality. By $(*)$ we immediately get that $|a_n - a_m| < \epsilon$, as desired.

However assumption $(*)$ seems completely unrealistic. In particular, it would mean that $a_N < \epsilon / (2 K) - L$ and if $L$ were bigger than $\epsilon / (2 K)$, this would be impossible as $a_N$ has to be positive.

Any help would be greatly appreciated.

EDIT: As Doug M pointed out, the sequence might not be Cauchy after all. The assumption has to be stronger, namely: $$ (**) \quad a_{n+1} \leq (1 + b_n)a_n + c_n $$ where $(b_n), (c_n)$ are non-negative, summable real sequences.

Ana Borges
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2 Answers2

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Are you sure you have this written down correctly?

Suppose $a_n = \sin n+1$... not a Cauchy sequence. $K, L = 3$ then $a_{n+m} < K(a_n + L) \implies a_{n+m} < 3$ which is true.

My hunch is that when we have $K, L$ properly defined then.

The sketch will be something along the lines of

$\{a_n\}$ is bounded.

There exists a convergent sub-sequence.

$a_{n_k}$ converges to $a$

$K(a + L) < \epsilon$

Doug M
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I now have an idea of why this works, taken from a paper by Qihou called "Iterative Sequences for Asymptotically Quasi-nonexpansive Mappings with Error Member" (Lemma 2).

Start by showing that for all $n, m$: $$ (++) \quad a_{n + m} \leq e^{\sum_{i = n}^{n + m - 1} b_i} a_n + e^{\sum_{i = n}^{n + m - 1} b_i} \sum_{i = n}^{n + m - 1} c_i $$ which, is done by repeatedly applying $(**)$ and noticing that $1 + x \leq e^x$, by the Taylor series expansion.

Then we can chose $n$ big enough such that $e^{\sum_{i = n}^{n + m - 1} b_i}$ is as close to $1$ as we want (because $(b_n)$ is summable, and hence for $n$ big enough, the sum of the $b_i$'s starting from $n$ will be close to $0$) and also such that $\sum_{i = n}^{n + m - 1} c_i$ is as close to $0$ as we want (same argument). This means that for $n$ big enough, $a_n, a_{n+1}, a_{n+2}, ...$ will be a monotone sequence. And the fact that $(a_n)$ is bounded gives us that it is Cauchy.

Ana Borges
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