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I am reading a paper in analysis that appears to conclude some argument from the following, is it true? I could not prove or disprove it.

Let $f : [x_0,b) \to \mathbb{R}, b < \infty$ be a smooth function. If there is $C > 0$ such that $$f(x_0) - f(x) \geq C(x-x_0),~\forall x \in [x_0,b)$$ can I guarantee that $f$ is decreasing?

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  • $\begingroup$ I think you need some clarification on your variables. Am I correct in thinking that $x_0$ and $b$ are fixed constants in this problem, while $x$ varies between them? $\endgroup$ – Isaac Browne Jul 16 at 23:18
  • $\begingroup$ please, who is downvoting comment to improve the question, I am not new in this site $\endgroup$ – L.F. Cavenaghi Jul 16 at 23:18
  • $\begingroup$ @IsaacBrowne, you are correct, I shall edit $\endgroup$ – L.F. Cavenaghi Jul 16 at 23:18
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For example, with $x_0=0$, $f(x_0)=0$, and $C=1$, this merely states $f(x)\le -x$. We cannot conclude that $f$ is decreasing. Try $f(x)=-x+\cos(1000x)-1$.

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You can rewrite your condition as $$f(x) \leq f(x_0) - C(x - x_0)$$ This just says that the graph of $f(x)$ lies below the line of slope $-C$ containing $(x_0, f(x_0))$. For any $C$ it's easy to draw a graph of such a function that is not decreasing... just add a few "wiggles" to a decreasing function.

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