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$\{x_n \}$ is a positive sequence defined for $n=0,1,2, \cdots $ and satisfies $x_n\geqq \dfrac{1}{2}(x_{n-1}+x_{n+1})$ for $n\geqq 1 \cdots ☆.$

Define $\{y_n\}_{n=1}^{\infty}\ $ by $y_n=x_n-x_{n-1}$.

Then, prove that $\{y_n\}$ is a bounded sequence.

Due to the definition of $\{y_n\},$ $y_{n+1}\leqq y_n$ holds for all $n\geqq 1.$ Thus $\{y_n\}$ is bounded above.

But I cannot prove that $\{y_n\}$ is bounded below.

This seem to be proved by contradiction.

Suppose $\{y_n\}$ is unbounded below.

Then, for small enough $K<0$, there is $N\in \mathbb N$ s.t. $n\geqq N\Rightarrow y_n<K \ \cdots (\ast)$.

Letting $n=N, N+1, N+2$ in $(\ast)$, we get $y_N=x_N-x_{N-1}<K, y_{N+1}=x_{N+1}-x_{N}<K, y_{N+2}=x_{N+2}-x_{N+1}<K$.

Using these and ☆, it seems possible that to lead contradiction, but I couldn't.

How can I lead contradiction ?

daㅤ
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1 Answers1

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A non-negative “concave” sequence is necessarily increasing, i.e. $y_n$ is bounded below by zero.

This can be proved by contradiction, and you were already on the right track. Assume that $K = y_N < 0$. Then $$ x_{n+1} - x_{n} = y_n \le K $$ for all $n \ge N$, and therefore $$ 0 \le x_n \le x_N + K(n - N) \, . $$ This is impossible because the right-hand side is negative for sufficiently large $n$.

Alternative: We can rewrite the above to make it a direct proof: Fix $N$ and define $K = x_{N+1}-x_N$. Then, as before, $$ 0 \le x_n \le x_N + K(n - N) $$ for $n \ge N$, so that $$ K \ge -\frac{x_N}{n-N} $$ and for $n \to \infty$ it follows that $K \ge 0$.

Remark: The same is true for concave functions: If $f:[0, \infty) \to \Bbb R$ is non-negative and concave then it is increasing (see for example Non-negative concave functions are non-decreasing).

Martin R
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