Bound of derivative of Lipschitz continuous function

Question

I have a function $f(x) = g(x)^Tx$, where $f: \mathbb R^n\to \mathbb R$, $g(x):\mathbb R^n \to \mathbb R^n$, $x\in \mathbb R^n$. Furthermore assume $g(0) = 0$, $\|g(x)\|$ is bounded and also the partial derivative $\|\partial g(x)/\partial x_i\|$ is bounded for all $x \in \mathbb R^n$.

Since $g(x)$ is bounded, it holds that $\|f(x)\| \leq \|g(x)\|_{\max}\,\|x\|$ and if $\|g(x)\|_{\max}=:L$, then $L$ can be considered as a Lipschitz constant.

Now, with this information, I am looking for an upper bound for $\partial f(x)/\partial x_i$. It is possible to find a bound $\|\partial f(x)/\partial x_i\| \leq \tilde L$ independent of $x$?

Since $\partial f(x)/\partial x_i = (\partial g(x)/\partial x_i) x + g(x)^T e_i$ it is clear that $\|\partial f(x)/\partial x_i \| \leq \|(\partial g(x)/\partial x_i)\|_{\max} \|x\| + L$.

I am wondering if I could obtain a better result. I considered using the mean value theorem, but at this point I got stuck.

Can anybody give me a hint how to say from this something about the bound of $\|\partial f(x)/\partial x_i \|$? Also, if I have an error in the derivation please let me know.

Answer 1

We have

\begin{align} \frac{\partial f}{\partial x_i}(x) &= \lim_{t\to 0}\, \frac{f(x+t\cdot e_i)-f(x)}t \\&= \lim_{t\to 0}\, \frac{{g(x+t\cdot e_i)}^T(x+t\cdot e_i)-{g(x)}^Tx}t \\&= \lim_{t\to 0}\, \frac{\left({g(x+t\cdot e_i)}^T-{g(x)}^T\right)x}t +{g(x+t\cdot e_i)}^Te_i \\&= {\left(\lim_{t\to 0}\, \frac{g(x+t\cdot e_i) - g(x)}t\right)}^T x \,+\,g(x)^Te_i \\&= {\left(\frac{\partial g}{\partial x_i}(x)\right)}^T x \,+\,g(x)^Te_i \end{align}

Since $g$ is bounded, $g(x)^Te_i$ also is. However, it not enough to know that $\frac{\partial g}{\partial x_i}$ is bounded to conclude that the term ${\left(\frac{\partial g}{\partial x_i}(x)\right)}^T x$ also is. It would seem one needs some decay condition on $\frac{\partial g}{\partial x_i}(x)$ as $\lVert x \rVert$ increases.

Answer 2

Let $k$ be any constant vector and $g(x)=k\sin(||x||)$

Then $g(x)$ is bounded with $g(0)=0$. The partial derivative is given by: $$ \frac{\partial g}{ \partial x_i} = \frac{k_i x_i}{||x||}\cos(||x||) $$

So also bounded. Thus satisfying your conditions.

However even for the one-dimensional case with $k=1$: $$ f(x) = x \sin(|x|) $$

$$ f'(x) = \sin(|x|)+\text{sign}(x) x\cos(|x|) $$

We have that $f'(x)$ is unbounded. Granted that this function is not continuously differentiable on 0. I'm thus suspecting that at least more conditions should be given so that there could exist the requested bound.