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Suppose $f$ is twice-differentiable on $(0, \infty), f^{\prime \prime}$ is bounded on $(0, \infty)$, and $f(x) \rightarrow 0$ as $x \rightarrow \infty$. Prove that $f^{\prime}(x) \rightarrow 0$ as $x \rightarrow \infty$.
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We shall prove an even stronger statement. If $f(x) \rightarrow L$ as $x \rightarrow \infty$ and $f^{\prime}(x)$ is uniformly continuous on $(0, \infty)$, then $f^{\prime}(x) \rightarrow 0$ as $x \rightarrow \infty$.

For, if not, let $x_{n} \rightarrow \infty$ be a sequence such that $f\left(x_{n}\right) \geq \varepsilon>0$ for all $n$. (We can assume $f\left(x_{n}\right)$ is positive by replacing $f$ with $-f$ if necessary.) Let $\delta$ be such that $\left|f^{\prime}(x)-f^{\prime}(y)\right|<\frac{\varepsilon}{2}$ if $|x-y|<\delta$. We then have $f^{\prime}(y)>\frac{\varepsilon}{2}$ if $\left|y-x_{n}\right|<\delta$, and so
$\left|f\left(x_{n}+\delta\right)-f\left(x_{n}-\delta\right)\right| \geq 2 \delta \cdot \frac{\varepsilon}{2}=\delta \varepsilon$
But, since $\delta \varepsilon>0$, there exists $X$ such that
$|f(x)-L|<\frac{1}{2} \delta \varepsilon$
for all $x>X$. Hence for all large $n$ we have
$\left|f\left(x_{n}+\delta\right)-f\left(x_{n}-\delta\right)\right| \leq\left|f\left(x_{n}+\delta\right)-L\right|+\left|L-f\left(x_{n}-\delta\right)\right|<\delta \varepsilon$
and we have reached a contradiction.
The problem follows from this result, since if $f^{\prime \prime}$ is bounded, say $\left|f^{\prime \prime}(x)\right| \leq M$, then $\left|f^{\prime}(x)-f^{\prime}(y)\right| \leq M|x-y|$, and $f^{\prime}$ is certainly uniformly continuous.
by Diamond (66,769 points)

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