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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 \(x \rightarrow \infty\).
in Mathematics by Diamond (66,769 points) | 65 views

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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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