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Given a function $f : U \to \mathbb R^m$ defined on an open $U \subset \mathbb R^n$, its directional derivative in direction $v \in \mathbb R^n$ is defined as $$\frac{\partial f}{\partial v}(x) = \lim_{h \to 0}\frac{f(x+ hv) -f(x)}{h} .$$

Here we approach $x$ on the line through $x$ having direction $v$, but we can also approach $x$ on more general curves. Let $\varphi : (-\epsilon, \epsilon) \to \mathbb R^n$ be a function which is continuous at $0$ and has value $\varphi(0) = 0$. Then define $$\frac{\partial f}{\partial \varphi}(x) = \lim_{h \to 0}\frac{f(x+ \varphi(h)) -f(x)}{h} .$$

If $\varphi$ is differentiable at $0$ and $f$ is differentiable at $x$, we get $$\frac{\partial f}{\partial \varphi}(x) = \frac{\partial f}{\partial \varphi'(0)}(x) \tag{1}.$$ This is a consequence of the chain rule. Consider the function $F = f \circ \varphi_x$, where $\varphi_x(h) = x + \varphi(h)$. We get $$\frac{\partial f}{\partial \varphi}(x) = F'(0) = Df(\varphi_x(0))(\varphi_x'(0)) = Df(x)(\varphi'(0)) = \frac{\partial f}{\partial \varphi'(0)}(x) .$$

Question 1. If we do not assume that $f$ is differentiable at $x$, but only that $\dfrac{\partial f}{\partial \varphi'(0)}(x)$ exists, can we prove that $\dfrac{\partial f}{\partial \varphi}(x)$ exists and satisfies $(1)$?

Question 2. If $\varphi$ is not differentiable at $0$, is it possible that $\dfrac{\partial f}{\partial \varphi}(x)$ exists?

Kritiker der Elche
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2 Answers2

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  1. No.

Consider

$f(x,y)= \begin{cases}x \text{ if } |y| \leq x^2 \\ 0 \text{ otherwise} \end{cases}$

and consider $\phi(x)= (x, x^4)$. Then in $x_0 =(0,0)$ you have:

$\partial_\phi f (0,0)=\lim_{h \to 0} \frac{f (h,h^4)}{h} =1$

but its derivative in $(0,0)$ in the direction $(1,0)= \phi' (0)$ is 0.

  1. Yes

    Ok I am cheating, but take $f \equiv 0$ (I have to think a little about more interesting cases)

tommy1996q
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  1. No. Let $x=(0,0)$, $f=a$ on the line $(a,b)$ and $0$ everywhere else, and $\phi(t)=(t,t^2)$. You can use a similar idea where $f$ is continuous.

  2. Of course it's possible that $\frac{\partial f}{\partial\phi}$ exists, as long as you're allowed to choose the $f$: just make it constant. However, in general it will not. Just take two unit vectors from $x$ on which $f$ has linearly independent directional derivatives and make $\phi$ pivot between the two.

Thomas Anton
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