# Proving non-differentiability of $f:\mathbb{R}^2 \to \mathbb{R}$

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.

• continuous directional derivatives imply differentiability.
– Zest
Jun 1, 2020 at 2:31
• @Zest Do you think it would be easier to prove directional derivative continuity than the approach I am taking now? Jun 1, 2020 at 2:33
• 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? Jun 1, 2020 at 2:48

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.

• But you know what $a$ and $b$ must be if $f$ is to be differentiable at $(0,0)$. Jun 1, 2020 at 2:49
• No, what you’re writing seems incorrect. This limit must be $0$ in every case, by definition of differentiability. Jun 1, 2020 at 3:01
• I agree with Ted. I am suspicious of this claim. Jun 1, 2020 at 3:05
• @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. 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 $$\Delta x \rightarrow 0$$.