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$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\al}{\alpha}$

Let $\Phi:\R^2 \to \R$. Let $\al:I \to \R^2$ be a path which is differentiable at $t=0$, and $\dot \al(0) \neq (0,0)$.

Assume that $\Phi \circ \alpha$ is differentiable at $t=0$.

Is it true that $\Phi$ has a directional derivative at the point $\al(0)$ at the direction $\dot \al(0)$?

Remarks:

1) If $\Phi$ is Lipschitz, the answer is yes (and the directional derivative coincides with $\Phi \circ \alpha'(0)$). Thus, in order to find a counter-example, we need to search for non-Lipschitz functions.

2) In this question, I gave what I thought to be a counter-example, which turned out to be wrong (example 2).

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Asaf Shachar
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2 Answers2

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Consider $\alpha(t)=(t,t^2)$. Notice that $\alpha'(t)=(1,2t)$, hence $\alpha'(0)=(1,0)$.

Now, several things can go wrong:

  1. $\Phi(x,0)$ may not be continuous at $0$, while $\Phi\circ\alpha$ can be constant. For instance: $$\Phi(x,y)=\begin{cases}1&\text{if }y=x^2\vee (x,y)\in\Bbb Q^2\\0&\text{otherwise}\end{cases}$$

  2. $\Phi(x,y)$ can be continous, $\Phi\circ\alpha$ can be a polynomial but $\Phi(x,0)$ may not be differentiable at $x=0$ as a function $\Bbb R\to\Bbb R$. For instance, $$\Phi(x,y)=\begin{cases}x^2&\text{if }y\ge x^2\\ \lvert x\rvert+\dfrac{y}{x^2}(x^2-\lvert x\rvert)&\text{if }0<y<x^2\\\lvert x\rvert&\text{if }y\le 0\end{cases}$$

  3. $\Phi$ can be continous, $\Phi\circ\alpha$ can be a polynomial, $\dfrac{\partial \Phi}{\partial x}(0,0)$ can exist, but $(\Phi\circ\alpha)'(0)\ne \dfrac{\partial \Phi}{\partial x}(0,0)$.

    For instance, consider $$\Phi(x,y)=\begin{cases}0&\text{if }y\le 0\\(\operatorname{sgn}x)\sqrt{y}&\text{if }0<y<x^2\\x&\text{if }y\ge x^2\end{cases}$$ $(\Phi\circ\alpha)'(t)\equiv 1$, but $\dfrac{\partial\Phi}{\partial x}(x,0)\equiv 0$.

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Let $\Phi(x,0) = \sin(1/x), x\ne 0,$ $\Phi (x,y) =0$ elsewhere. Define $\alpha (t) = (t,t^2).$ Note $\alpha(0) = (0,0),\alpha'(0)=(1,0).$ We have $\Phi\circ \alpha \equiv 0,$ so $(\Phi\circ \alpha)'(0) = 0.$ Does $D_{\alpha'(0)}\Phi (0,0)$ exist? No, because if it did, it would just be the partial derivitave of $\Phi$ with respect to $x$ at $(0,0).$ But $\Phi(x,0)$ is not even continuous at $x=0,$ so there's no chance for that partial derivative.

zhw.
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