Question: Is it possible to have a continuously differentiable function $f:\mathbb{R}\rightarrow\mathbb{R}$ with the property that (1) $f$ is bounded, (2) $f^{\prime}(x)\leq-\nu$ for some $\nu>0$.

My thoughts: No, since $f^{\prime}(x)\leq-\nu$ implies $f(x)-f(x_0)\leq -\nu(x-x_0)$

This question came about from reading the text Linear System Theory Example 7.10 where such a function is implied to exist.

  • $\begingroup$ Your thoughts are correct (for example, by the mean value theorem). However, the title of the question suggests the weaker condition $f'(x)<0$ for all $x \in \mathbb{R}$ (rather than $f'(x)\leq -v$ for all $x \in \mathbb{R}$). You can find examples to fit the weaker condition of being continuously differentiable, bounded, and having strictly negative derivative everywhere (just draw a picture). $\endgroup$ – Michael Sep 21 '17 at 1:49

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