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Problem of the Week # 280 - Sep 12, 2017

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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,198
Here is this week's POTW:

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Let $f : [a,\infty)\to \Bbb R$ be a continuous function that satisfies the inequality $\displaystyle f(x) \le A + B\int_a^x f(t)\, dt$, where $A$ and $B$ are constants with $B < 0$. If $\displaystyle \int_a^\infty f(x)\, dx$ exists, show that $\displaystyle \int_a^\infty f(x)\, dx \le -A/B$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,198
Congratulations to Opalg for his correct solution.

Let \(\displaystyle F(x) = \int_a^x\!\!\! f(t)\,dt\) and let \(\displaystyle L = \int_a^\infty\!\!\! f(t)\,dt\). Then $F(x) \to L$ as $x\to\infty.$

Since $F$ is differentiable, with derivative $f$, it follows from the mean value theorem that (for each integer $n \geqslant a$) $F(n+1) - F(n) = f(x_n)$ for some $x_n \in (n,n+1).$ But $F(n+1) - F(n) \to L-L = 0$ as $n\to\infty$. Therefore $f(x_n) \to0.$

Now let $n\to\infty$ in the inequality \(\displaystyle f(x_n) \leqslant A + B\int_a^{x_n}\!\!\!f(t)\,dt\) to get $0\leqslant A + BL$. This inequality gets flipped when we divide though by the negative number $B$, giving $0 \geqslant \frac AB + L$, or $L\leqslant -\frac AB.$

Ackbach Comment: Opalg avoids the problem of $f(x)\not\to 0$ as $x\to\infty$ by choosing a subsequence $\{x_n\}\to \infty$ and $\{f(x_n)\}\to 0$ based on the Mean Value Theorem. Very clever!
 
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