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Consider a function $F(x,y)$, where $x \in {\mathbb R}^n$, $y \in {\mathbb R}^m$, and $F(x,y) \in {\mathbb R}^n$. Additionally, $F(x,y)$ is continuously differentiable w.r.t. $x$ but locally Lipschitz continuous w.r.t. $y$ (not necessarily differentiable).

The question is: is the partially derivative $\frac {\partial F(x, y)}{\partial x}$ continuous w.r.t. $y$?

(Without locally Lipschitz condition, $y\sin(x/y)$ is a counterexample.)

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  • $\begingroup$ @NormalHuman Thanks for your reminder, and I have changed the title and tags accordingly. $\endgroup$
    – Ryan
    Oct 22, 2015 at 0:58
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    $\begingroup$ @NormalHuman Thank you for the nonexample. $\endgroup$
    – Ryan
    Oct 22, 2015 at 1:16
  • $\begingroup$ @NormalHuman How about F is locally lipschitz w.r.t. $y$? $\endgroup$
    – Ryan
    Oct 22, 2015 at 5:02
  • $\begingroup$ @NormalHuman No problems :-) $\endgroup$
    – Ryan
    Oct 22, 2015 at 21:31

1 Answer 1

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Counterexample: $$F(x,y)=\begin{cases} y\,\phi (x/y)\quad &\text{ if }y\ne 0 \\ 0 &\text{ if }y = 0 \end{cases}$$ where $\phi : \mathbb{R}\to\mathbb{R}$ is a $C^1$-smooth function which is zero outside of $[-1,1]$, and such that $\phi'(0)\ne 0$. Something like $\phi(x) = x\max(1-x^2, 0)^2$, for example.

  1. Smoothness with respect to $x$ is obvious.
  2. The Lipschitz property with respect to $y$ follows from the boundedness of the partial derivative $$\left|\frac{\partial F}{\partial y}\right| = \left|-\frac{x}{y}\phi'(x/y)\right| \le \sup|\phi'|$$ where the inequality $|x/y|\le 1$ can be used because $\phi'(x/y)=0$ when $|x/y|>1$.
  3. Yet, the $x$-derivative is not continuous. It is equal to zero when $y=0$, and $$\left|\frac{\partial F}{\partial x}\right| = \phi'(x/y)$$ when $y\ne 0$. Continuity fails as $x=0$, $y\to0$.
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  • $\begingroup$ Awesome! Great Answers! Thank you very much! $\endgroup$
    – Ryan
    Nov 2, 2015 at 10:00

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