Counter example to these false claims

Question

False claims:

$(a)$ Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ be a function.If the restriction of $f$ to each of its coordinates are continuous at $0$, then $f$ is continuous at $0$.

$(b)$ If $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is a differentiable function, then the partial derivatives of $f$ exists and are continuous.

---

My attempt

$(a)$ not sure what "restriction of $f$ to each of its coordinates" means...

$(b)$ Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ be defined as $f(x,y)= (y\ln|x|)$ The partial derivatives

$d_x = y/x$

$d_y = \ln|x|$

exist but they are not continuous at $0$.

Answer 1

I post this answer since I've typed it.

For part (a), take $f(x,y) = \frac{xy}{x^2+y^2+1}$ and consider the limit as $(x,y)$ approaches zero. Clearly, $f$ verifies the hypothesis of the question as $f(0,y) = f(x,0) = 0$.

• $\lim\limits_{y\to0} f(0,y) = 0 = \lim\limits_{x\to0} f(x,0)$
• $\lim\limits_{h\to0} f(h,h) = \frac{h^2}{h^2+h^2+1} = \frac12 \ne 0$

So $f$ is not continuous at $(0,0)$.

As @Peter Franek has says, take $$f(x,y) = \begin{cases} x^2 \sin\frac1x &\text{if } x \ne 0 \\ 0 &\text{if } x = 0. \end{cases} $$

It's differentiable because

• at $(x,y)$ with $x \ne 0$, it's a product of $x^2$ (differentiable everywhere) and $\sin\frac1x$, which is a composition of an everywhere differentiable function and $\frac1x$. As $x \ne 0$, $\frac1x$ is differentiable.
• at $(0,y)$, we set $h = (h_1,h_2)$, and we have $$\lim_{||h||\to0}\frac{f(h_1,y+h_2)-f(0,y)}{||h||} = \lim_{||h||\to0}\frac{f(h_1,y+h_2)}{\sqrt{h_1^2+h_2^2}}.$$
  • if $h_1 = 0$, we have $$\lim_{||h||\to0}\frac{f(h_1,y+h_2)-f(0,y)}{||h||} = \lim_{||h||\to0}\frac{f(h_1,y+h_2)}{\sqrt{h_1^2+h_2^2}}=0.$$
  • if $h_1 \ne 0$, we have \begin{align} \lim_{||h||\to0}\frac{f(h_1,y+h_2)-f(0,y)}{||h||} &= \lim_{||h||\to0}\frac{f(h_1,y+h_2)}{\sqrt{h_1^2+h_2^2}} \\ &= \lim_{||h||\to0}\frac{h_1^2\sin\frac1{h_1}}{\sqrt{h_1^2+h_2^2}} \\ &= \lim_{||h||\to0} h_1 \cdot \frac{h_1}{\sqrt{h_1^2+h_2^2}} \cdot \sin\frac1{h_1} \end{align} The first term tends to zero, the second and third terms are bounded by one, so the whole limit goes to zero whenever $x = 0$, no matter $h_1 = 0$ or not.

This proves the differentiability of $f$ on $\Bbb R^2$, and thus the existence of partial derivatives. But when $x \ne 0$, $$\frac{\partial f}{\partial x} = 2x\sin\frac1x - \cos\frac1x,$$ which is not continuous at $x = 0$. (If we take $x_n = \frac{1}{(n+\frac12)\pi}$ and $x'_n = \frac{1}{n\pi}$, $\cos\frac1{x_n}$ and $\cos\frac1{x'_n}$ don't agree with each other.) So $\frac{\partial f}{\partial x}$ is not continuous at $x = 0$.

Answer 2

a) Let $f(x,y) = {xy \over x^2 + y^2}$ if $(x,y) \neq (0,0)$ and $f(0,0) = 0$ Then $f$ is not continuous at $x=0$.

b) Let $f(x,y) = x^2 \sin(1/x)$ if $x \neq 0$ and $f(0,y) = 0$. Then $f$ is differentiable, but $\partial f / \partial x$ is not continuous at $x=0$