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Let $(a_n)$ be a bounded sequence and this inequality holds:

$2a_n\leq a_{n-1}+a_{n+1}$

Prove, that $b_n=a_{n+1}-a_n$ converges to zero.

My Proof:

$2a_n\leq a_{n-1}+a_{n+1}$

$\iff2(a_{n+1}-b_n)\leq a_n-b_{n-1}+a_{n+1}$

$\iff a_{n+1}-b_n\leq\frac{1}{2}a_n-\frac{1}{2}b_{n-1}+\frac{1}{2}a_{n+1}$

$\iff -b_n\leq\frac{1}{2}a_n-\frac{1}{2}b_{n-1}-\frac{1}{2}a_{n+1}$

$\iff 2b_n\geq b_{n-1}+(a_{n+1}-a_n)$

$\iff b_n\geq b_{n-1}\Rightarrow b_n$ is monotonically increasing

Since $a_n$ is bounded, $a_{n+1}-a_n$ is also bounded, so $b_n$ is bounded.

Since $b_n$ is monotonically increasing and bounded, $\lim\limits_{n \to \infty}b_n$ exists.

Furthermore, $$ \lim\limits_{n \to \infty}b_n = \lim\limits_{n \to \infty}(a_{n+1}-a_n)= \lim\limits_{n \to \infty}a_{n-1}-\lim\limits_{n \to \infty}a_n $$ so $\lim\limits_{n\to\infty}a_n$ exists.

Therefore, $\lim\limits_{n\to\infty} a_{n-1} - \lim\limits_{n\to\infty}a_n = 0$, which makes $b_n$ a null sequence.

Is this proof correct? Particularly, I'm not sure about the second part.

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If a sequence is (midpoint-)convex and bounded, it is weakly decreasing.

If we assume $a_{n+1}-a_{n}=D>0$, from $a_{n+2}-a_{n+1} \geq a_{n+1}-a_{n}$ and so on we get $a_{n+m}\geq a_{n}+Dm$. By taking $m$ big enough, we contradict the fact that the sequence is bounded.

So we have a bounded and weakly decreasing sequence, i.e. a converging sequence by the monotone convergence theorem. Any convergent sequence of real numbers is a Cauchy sequence, hence $a_{n+1}-a_{n}$ converges to zero.

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This equality: $$\lim_{n \to \infty}(a_{n+1}-a_n)=\lim_{n \to \infty}a_{n-1}-\lim_{n \to \infty}a_n$$ requires the limit $$\lim_{n \to \infty}a_n$$ exists, so you cannot deduce this limit exists this way.

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