# Is it possible to find an upper bound and a lower bound for $(f(\xi)-f(\tilde{\xi}))^T(\frac{\partial f}{\partial \xi}(\bar{\xi}))(\xi-\tilde{\xi})$?

For a system engineering problem I have to solve the problem below, but since I am not a mathematician, I am not sure if I have enough knowlegde to solve it.

Problem definition: Let $$f(\xi) \in \mathcal{C}^1$$ be a smooth function with mapping $$f:\mathbb{R}^n \to \mathbb{R}^p$$ with $$p\le n$$. With this information, we have the term: $$$$\left(f(\xi)-f(\tilde{\xi})\right)^\top\left(\frac{\partial f}{\partial \xi}\left(\bar{\xi}\right)\right)\left(\xi-\tilde{\xi}\right) \qquad\qquad\text(1)$$$$ Where $$\xi, \bar{\xi}, \tilde{\xi} \in \mathbb{R}^n$$ and varying over time. Using the mean-value-theorem, we can prove that $$\left(f(\xi)-f(\tilde{\xi})\right)^\top\left(\frac{\partial f}{\partial \xi}\left(\bar{\xi}\right)\right)\left(\xi-\tilde{\xi}\right)\le \left(\xi-\tilde{\xi}\right)^\top \left(\frac{\partial f}{\partial \xi}\left(\bar{\xi}\right)\right)^\top\left(\frac{\partial f}{\partial \xi}\left(\bar{\xi}\right)\right) \left(\xi-\tilde{\xi}\right)$$ Thus, for $$(1)$$ we know that we can find an upperbound for all $$\xi, \bar{\xi}, \tilde{\xi} \in \mathbb{R}^n$$. However, for the problem, we require a lower bound for all $$\xi, \bar{\xi}, \tilde{\xi} \in \mathbb{R}^n$$ or show that the $$\le$$ sign can be changed to an equality sign for all $$\xi, \bar{\xi}, \tilde{\xi} \in \mathbb{R}^n$$. However, I do not know how to show it.

Could one of you help me out?

The same reasoning for the upper bound can be applied for the lower bound. Which was:

$$$$\left(f(\xi)-f(\tilde{\xi})\right)^{\top} \leq (\xi - \tilde{\xi})^{\top} \left(\frac{\partial f}{\partial \xi} (\bar{\xi})\right)^{\top}$$$$

Such that taking the transpose of this inequality gives:

$$$$\left(f(\xi)-f(\tilde{\xi})\right) \leq \left(\frac{\partial f}{\partial \xi} (\bar{\xi})\right) (\xi - \tilde{\xi})$$$$

Filling this into the left hand side of the original inequality gives:

$$$$0 \leq \left(f(\xi)-f(\tilde{\xi})\right)^{\top} \left(f(\xi)-f(\tilde{\xi})\right) \leq \left(f(\xi)-f(\tilde{\xi})\right)^{\top} \left(\frac{\partial f}{\partial \xi} (\bar{\xi})\right) (\xi - \tilde{\xi})$$$$

Now the left hand side has a lower bound of 0. Because $$x^{\top}x \geq 0$$.

• Since the mapping $f$ is multi dimensional, i.e. $\mathbb{R}^n \to \mathbb{R}^p$ the first statement cannot hold, since there is no ordening in vectors. 0 is obviously a trivial lowerbound. Thank you for the answer contribution! – seaver Sep 24 '19 at 8:02