Lemma 36.12.4.label For each $m, n \in \natp$ with $m \le n$,
- (1)
$\sup_{t \in [0, \infty)}\paren{1/n + t}^{-2}t \le n/4$.
- (2)
For every $t \in [0, \infty)$, $m^{-1}(m^{-1}+ t)^{-1}\ge n^{-1}(n^{-1}+ t)^{-1}$.
- (3)
For every $t \in [0, \infty)$, $(1/n + t)^{-1}t = 1 - (1/n + t)^{-1}/n$.
Proof. (1): For each $t \in [0, \infty)$,
\begin{align*}0 \le \paren{t - \frac{1}{n}}^{2}&= t^{2} - \frac{2t}{n}+ \frac{1}{n^{2}}= t^{2} + \frac{2t}{n}+ \frac{1}{n^{2}}- \frac{4t}{n}\\ 0&\le \paren{t + \frac{1}{n}}^{2} - \frac{4t}{n}\end{align*}
so $\frac{4t}{n}\le \paren{t + \frac{1}{n}}^{2}$ and $\frac{n}{4}\ge \paren{\frac{1}{n} + t}^{-2}t$.
(2): For each $t \in [0, \infty)$,
\[\frac{1}{m}\paren{\frac{1}{m} + t}^{-1}= \frac{1}{1 + mt}\ge \frac{1}{1 + nt}= \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}\]
(3): For every $t \in [0, \infty)$,
\begin{align*}\paren{\frac{1}{n} + t}^{-1}t&= \paren{\frac{1}{n} + t}^{-1}\braks{\paren{\frac{1}{n} + t} - \frac{1}{n}}\\&= \paren{\frac{1}{n} + t}^{-1}\paren{\frac{1}{n} + t}- \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}\\&= 1 - \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}\end{align*}
$\square$
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