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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 knowledge 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: \begin{equation} \left(f(\xi)-f(\tilde{\xi})\right)^\top\left(\frac{\partial f}{\partial \xi}\left(\bar{\xi}\right)\right)\left(\xi-\tilde{\xi}\right) \label{1}\tag{1} \end{equation} Where $\xi, \bar{\xi}, \tilde{\xi} \in \mathbb{R}^n$ and are 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 \eqref{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?

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  • $\begingroup$ The inequality is certainly false in general. Something alike must be true, but not this one. $\endgroup$
    – Denis Serre
    Commented Mar 8, 2022 at 7:41

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

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

Such that taking the transpose of this inequality gives:

\begin{equation} \left(f(\xi)-f(\tilde{\xi})\right) \leq \left(\frac{\partial f}{\partial \xi} (\bar{\xi})\right) (\xi - \tilde{\xi}) \end{equation}

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

\begin{equation} 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}) \end{equation}

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

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  • $\begingroup$ 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! $\endgroup$
    – seaver
    Commented Sep 24, 2019 at 8:02

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