Suppose $f \in L^1 \cap C^1(\mathbf R)$, and assume $f'(x)$ goes to $0$ as $x$ goes to $\infty$. What is the best way to prove that $f(x)$ also goes to $0$ as $x$ goes to $\infty$.
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contradiction: $f(x)=3$ , $f(x)'=0$ – D.christian Jan 12 '16 at 07:21
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I forgot to mention that the domain is the real line. – hb12ah Jan 12 '16 at 07:23
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2Function defined by $f(x)=3$ is not in $L^1$ space. – hb12ah Jan 12 '16 at 07:26
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As $f^\prime\in C^1$ and $\lim_{x\to\infty}f^\prime(x)=0$, then $f^\prime$ is bounded in $[0,\infty]$. Say $|f^\prime(x)|\le M$ for $x\ge 0$. Now, it follows that $f$ is uniformly continuous in $[0,\infty)$
Now, a classic result implies that $\lim_{x\to\infty}|f(x)|=0$ (for example $f$ uniformly continuous and $\int_a^\infty f(x)\,dx$ converges imply $\lim_{x \to \infty} f(x) = 0$ )
