Multivariable Chain Rule for partially differentiable maps

Question

The following statement is a simple consequence of the multivariable chain rule:

Assume the subsets $X \subset \mathbb{R}^m$ and $Y \subset \mathbb{R}^n$ are open. Consider the maps $$g:X \to \mathbb{R}^n,\phantom{aaa} f:Y \to \mathbb{R}$$ where $g(X) \subset Y$. Let $g_1,...,g_n$ be the component maps of $g$, i.e., all the $g_i$'s map from $X$ to $\mathbb{R}$ and $$g(x)=(g_1(x),...,g_n(x))$$ for all $x \in X$. Assume that all the partial derivatives of the $g_i$'s exist at some point $\alpha \in X$. Assume further that $f$ is (Frechet-)differentiable at point $g(\alpha)$. Then, all the partial derivatives of $f \circ g$ at point $\alpha$ exist.

Short: The composition of a differentiable map with a partially differentiable map is partially differentiable.

My question: In the above setting, if $f$ is only partially differentiable at point $g(\alpha)$ (and not Frechet-differentiable), is the map $f\circ g$ still partially differentiable? If not, what is a counterexample? If yes, how to prove it?

Answer 1

Consider the function $f:\mathbb R^2\to\mathbb R$ defined by $$ f(x,y) = \begin{cases} 0, & (x,y) = (0,0) \\ \frac{xy}{x^2+y^2}, & (x,y)\ne (0,0). \end{cases}$$ It is partial differentiable at $(x,y) = (0,0)$, as $f$ is constant $0$ on the axes. But, $f$ is not even continuous at $(x,y)=(0,0)$.

In fact, the chain rule fails, as for $t\ne 0$ we have $$ f(t, t^2) = \frac{t^3}{t^2 + t^4}, $$ which derivative is $1$ and not $0$, as implied by the chain rule.

Answer 2

This is more of a reply to the user251257 answer than the original post, but I would like to see if I am thinking properly.

I do think that $$f(x,y)=\begin{cases} \dfrac{xy}{x^2+y^2},&(x,y)\ne(0,0)\\ 0,& (x,y)=(0,0) \end{cases} $$ is a good counterexample. If we take the limit along the line $y=kx$, then: \begin{align*} \lim_{(x,y)\to(0,0)}f(x,y) &=\lim_{x\to 0}\dfrac{x(kx)}{x^2+(kx)^2}\\ &=\lim_{x\to 0}\dfrac{k}{1+k^2}\\ &=\dfrac{k}{1+k^2} \end{align*} So we get a different limit along the line $y=kx$ for different values of $k$ as we approach the origin, so the function $f$ is not continuous at the origin. Therefore, the function is not differentiable at the origin (i.e., at $(0,0)$), so there is no guarantee that the chain rule will work.

Now, the answer decided to use: $$g(t)=\langle t,t^2\rangle$$ However, this is where I begin to see a problem, as: \begin{align*} f(g(t)) &=f(t,t^2)\\ &=\begin{cases} \dfrac{(t)(t)^2}{(t)^2+(t^2)^2},&t\ne0\\ 0,&t=0 \end{cases}\\ &=\begin{cases} \dfrac{t}{1+t^2},&t\ne0\\ 0,&t=0 \end{cases} \end{align*} Now, here is the graph of $f(g(t))$.

So, you can see that it is continuous and I don't see a point of singularity. Now I could be wrong, as I'm no expert.

Now, let's see what happens when we calculate $f'(g(0))$. \begin{align*} (f\circ g)'(0) &=\lim_{t\to 0}\dfrac{f(g(t))-f(g(0))}{t-0}\\ &=\lim_{t\to 0}\dfrac{\dfrac{t}{1+t^2}-0}{t}\\ &=\lim_{t\to 0}\dfrac{1}{1+t^2}\\ &=1 \end{align*} Note that an answer of 1 is quite reasonable when you look at the graph (the slope of the tangent line at the origin is 1). Now let's see if the chain rule gives the same answer. If $(x,y)\ne(0,0)$, then: \begin{align*} f_x(x,y) &=\dfrac{(x^2+y^2)y-xy(2x)}{(x^2+y^2)^2}\\ &=\dfrac{y(-x^2+y^2)}{(x^2+y^2)^2} \end{align*} Now, at $(x,y)=(0,0)$: \begin{align*} f_x(0,0) &=\lim_{h\to 0}\dfrac{f(0+h,0)-f(0,0)}{h}\\ &=\lim_{h->0}\dfrac{\dfrac{(h)(0)}{h^2+0^2}-0}{h}\\ &=\lim_{h->0}\frac{0}{h}\\ &=0 \end{align*} Hence: $$f_x(x,y)=\begin{cases} \dfrac{y(-x^2+y^2)}{(x^2+y^2)^2},&(x,y)\ne(0,0)\\ 0,&(x,y)=(0,0) \end{cases}$$ In similar fashion, one can show: $$f_y(x,y)=\begin{cases} \dfrac{x(x^2-y^2)}{(x^2+y^2)^2},&(x,y)\ne(0,0)\\ 0,&(x,y)=(0,0) \end{cases}$$ Now, if we were allowed to apply the chain rule, wit $x=t$ and $y=t^2$, we would start with: $$f_x(x,y)\dfrac{dx}{dt}+f_y(x,y)\dfrac{dy}{dt}=f_x(x,y)(1)+f_y(x,y)(2t),$$ then at $t=0$, both $x=0$ and $y=0$, and subbing these values gives us: $$f_x(0,0)(1)+f_y(0,0)(2(0))=(0)(1)+(0)(0)=0,$$ which differs from the correct answer of 1.

Finally, I would appreciate it if any one sees any errors in my argument.