# Show from every positive sequence decreasing to zero, we can build a sequence satisfying a concavity condition

Let $\{x_n\}_{n=-\infty}^{\infty}$ be a positive sequence decreasing to zero as $|n| \to \infty$.
Show there is a sequence $\{y_n\}$ satisfying \begin{align} y_n >& x_n \tag{1}\\ y_{n-1}+y_{n+1}-2y_n \ge& 0 \tag{2} \end{align}

I have attempted this question with this approach where $y_n=c_n$.
The method gives me (1).
But not being able to get $\frac{1}{n} \ge c_n-c_{n+1} \ge \frac{1}{n+1}$, the method fails for (2).

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There's something wrong with (1). If there are $N$ $x_n$ values $> x_0$, but all $y_n > x_n$, then there are at least $N+1$ $y_n$ values $> x_0$ (i.e. all $y_n$ where $x_n > x_0$, and also $y_0$). But then $\{y_n\}$ can't be a subsequence of $\{x_n\}$. – Robert Israel Oct 16 '12 at 2:52
@RobertIsrael I see your point, I relax this requirement in the question. However, as $x_n$ is a decreasing sequence, I think your example doesn't apply, but your point is still valid. – Nicolas Essis-Breton Oct 16 '12 at 10:29

$$y_n=x_0+2^n\qquad (n\in\mathbb Z)$$