Question: Given $f:\mathbb{R}^2 \to \mathbb{R}$ defined by $f(x, y) =
\begin{cases}
x, & \text{if $y=x^2$} \\
0, & \text{otherwise}
\end{cases}$, show $f$ is not differentiable at $(0, 0)$.
Attempt: I know a few things about $f$: it is continuous at $(0, 0)$ and has continuous directional derivatives (but am yet to prove these).
To prove non-differentiability, I need to show that there does not exist a linear mapping $A$ from $\mathbb{R}^2$ to $\mathbb{R}$ (which can be represented by the $2 \times 1$ matrix $\begin{bmatrix}
a \\
b \\
\end{bmatrix}$) so that $\lim_{h \to 0, h \in \mathbb{R}^2} \frac{\Vert f(x+h) – f(x) – Ah \Vert}{\Vert h \Vert} = 0$ where $x=(0, 0)$.
To do this, I considered the LHS of the equation and intend to show that it does not limit to $0$. Letting $h=(h_1, h_2)$ gives $$\lim_{(h_1, h_2) \to 0} \frac{\Vert f((0, 0)+(h_1, h_2)) – f(0,0) – A(h_1, h_2) \Vert}{\Vert (h_1, h_2) \Vert}=\lim_{(h_1, h_2) \to 0} \frac{\Vert f(h_1, h_2) – A(h_1, h_2) \Vert}{\sqrt{h_1^2+h_2^2}}$$ however I am unsure of how to further evaluate this since we do not know $f(h_1, h_2)$ and I am unsure of what $A(h_1, h_2)$ evaluates to.
Any help would be greatly appreciated.
Best Answer
Firstly, the matrix $A$ will be a $1 \times 2$ matrix, so $A=\begin{bmatrix}a&b\end{bmatrix}$. So $A\begin{bmatrix}h_1\\h_2\end{bmatrix}=ah_1+bh_2$.
Suppose $(h_1,h_2) \to (0,0)$ but $h_2 \neq h_1^2$. Then $f(h_1,h_2)=0$. This means $$\lim_{(h_1, h_2) \to 0} \frac{\Vert f(h_1, h_2) - A(h_1, h_2) \Vert}{\sqrt{h_1^2+h_2^2}}=\lim_{(h_1, h_2) \to 0} \frac{|ah_1+bh_2|}{\sqrt{h_1^2+h_2^2}}.$$ Further suppose we were approaching $(0,0)$ along the $x-$axis, i.e. $h_2=0$ and $h_1 \to 0$. Then the above limit $$\lim_{(h_1, h_2) \to 0} \frac{|ah_1+bh_2|}{\sqrt{h_1^2+h_2^2}}=|a|.$$ Likewise if we were approaching $(0,0)$ along the $y-$axis, i.e. $h_1=0$ and $h_2 \to 0$. Then the above limit $$\lim_{(h_1, h_2) \to 0} \frac{|ah_1+bh_2|}{\sqrt{h_1^2+h_2^2}}=|b|.$$ For differentiability, we want the limits to be $0$. So $|a|=|b|=0$
But we could approach $(0,0)$ along the path $y=x^2$ as well, i.e. $h_2=h_1^2$ and $(h_1,h_2) \to (0,0)$. In which case $f(h_1,h_2)=h_1$. Then,
$$\lim_{(h_1, h_2) \to 0} \frac{\Vert f(h_1, h_2) - A(h_1, h_2) \Vert}{\sqrt{h_1^2+h_2^2}}=\lim_{(h_1, h_2) \to 0} \frac{|(a-1)+bh_1|}{\sqrt{1+h_1^2}}=|a-1|.$$ For differentiability we want $|a-1|=0$ as well. But then $a=1$, a contradiction.