Question about bounded/unbounded function Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function, and let $\sup_{x\in \mathbb{R}}{f'(x)<0}$.
I have some ideas but I'm not sure about the following questions:
Can I affirm that $f$ is bounded or unbounded? Why?
Furthermore, from the $\sup$ information can I tell that $f$ is monotonically decreasing and Lipschitz continuous?
 A: $f$ may be bounded and may not be bounded. For example, $f(x)=-x$.
A: Ok finally converting my comment into an answer. 

Let $k=\sup\, \{f'(x) \mid x\in\mathbb{R} \} $ and then we have $f'(x) \leq k<0$. Since the derivative is negative the function $f$ is strictly decreasing. Even more us true. Let $x>0$ then $f(x) - f(0)=xf'(\xi)$ via mean value theorem and this means that $$f(x) \leq f(0)+kx$$ The RHS of the above inequality tends to $-\infty$ as $x\to\infty $ and hence by the above inequality $f(x)\to -\infty$ as $x\to\infty $. Thus $f$ is unbounded. 
Further note that $$|f(x)-f(y) |=|x-y||f'(\xi) |\geq |k||x-y|$$ and hence it is quite clear that $f$ is not necessarily Lipschitz continuous. A simple example is $f(x) = - x^{3}-2x$ and clearly here $k=-2$ and one can see that it is not Lipschitz continuous because the derivative $f'$ itself is unbounded. 
A: It doesn't matter that $f'$ is negative. Just assume $|f'(x)| \ge a$ for all $x,$ for some positive constant $a.$ Then by the mean value theorem
$$|f(n)-f(0)| = |f'(c_n)\cdot n| \ge |a|\cdot n$$
for $n=1,2,\dots$ As $n\to \infty,$ the right side $\to \infty.$ This shows $f$ is not bounded.
