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.

  • $\begingroup$ continuous directional derivatives imply differentiability. $\endgroup$
    – Zest
    Jun 1, 2020 at 2:31
  • $\begingroup$ @Zest Do you think it would be easier to prove directional derivative continuity than the approach I am taking now? $\endgroup$
    – Viv4660
    Jun 1, 2020 at 2:33
  • 2
    $\begingroup$ Remember that if $f$ is differentiable at $(0,0)$, then $A$ is the matrix of partial derivatives at $(0,0)$. What are those partial derivatives? $\endgroup$ Jun 1, 2020 at 2:48

2 Answers 2


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.

  • $\begingroup$ But you know what $a$ and $b$ must be if $f$ is to be differentiable at $(0,0)$. $\endgroup$ Jun 1, 2020 at 2:49
  • $\begingroup$ No, what you’re writing seems incorrect. This limit must be $0$ in every case, by definition of differentiability. $\endgroup$ Jun 1, 2020 at 3:01
  • $\begingroup$ I agree with Ted. I am suspicious of this claim. $\endgroup$ Jun 1, 2020 at 3:05
  • $\begingroup$ @TedShifrin and Carlo: I made a horrible mistake. Thanks for pointing out. I don't know what I was thinking. I have edited it. Please let me know if there are errors remaining. Thanks. $\endgroup$
    – Anurag A
    Jun 1, 2020 at 3:09

$ f^{'}_{x}(0, 0) = f^{'}_{y}(0, 0) = 0$.

For differentiation we need $f(\Delta x, \Delta y) = f(0,0) +f^{'}_{x}(0, 0) \cdot \Delta x + f^{'}_{y}(0, 0)\cdot \Delta y + o(\sqrt{\Delta x ^{2} + \Delta y ^{2}}) $ for $(\Delta x, \Delta y) \rightarrow (0, 0)$

So we need zero limit for fraction $\dfrac{f(\Delta x, \Delta y)}{\sqrt{\Delta x ^{2} + \Delta y ^{2}}}$.

But for set $y=x^2$ we have $\dfrac{\Delta x}{\sqrt{\Delta x ^{2} + \Delta x ^{4}}}$, which have not zero limit when $\Delta x \rightarrow 0$.


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