# If $f(x_0) - f(x) \geq C(x-x_0), ~C>0$ then $f$ is decreasing?

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?

• 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? – Isaac Browne Jul 16 at 23:18
• please, who is downvoting comment to improve the question, I am not new in this site – L.F. Cavenaghi Jul 16 at 23:18
• @IsaacBrowne, you are correct, I shall edit – L.F. Cavenaghi Jul 16 at 23:18

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 $$f(x)=-x+\cos(1000x)-1$$.
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.