3
$\begingroup$

Let $g: \mathbb{R^2} \to \mathbb{R}$.

Function g

How can I prove that $g$ is continuous in its origin, but not totally differentiable?

If I take

$$g(\frac{1}{n},\frac{1}{n}) = \frac{1}{n\sqrt{2}} \to 0 \text{ for } n \to \infty $$

Or rather:

$$|x,y| \leq \frac{1}{2} (x^2 + y^2) $$

from which we can follow

$$|g(x,y)| \leq \frac{1}{2} \sqrt{x^2 + y^2}$$

which proves continuity of $g$ in the origin $(0,0)$.

But how can I show that this function is not total differentiable?

Can I do the following estimation?

$$|g(x,y) - 0| = |y| \cdot \frac{x \cdot y}{\sqrt{x^2 + y^2}} \leq |y| \cdot 1 = |y| $$

From which it follows, that

$$|y| \leq |\sqrt{x^2 + y^2}| = ||(x,y)|| $$

$\endgroup$
-1
$\begingroup$

By polar coordinates as $r \to 0$

$$\frac{xy}{\sqrt{x^2 + y^2}}=r \cos \theta \sin \theta$$

we see that $f(x,y)$ is continuous but since $f_x(0,0)=f_y(0,0)=0$ by definition of differential we have

$$\lim_{(h,k)\to (0,0)} \frac{\frac{hk}{\sqrt{h^2 + k^2}}}{\sqrt{h^2 + k^2}}=\frac{hk}{h^2 + k^2}$$

which doesn't exist.

Refer also to the related

$\endgroup$
2
$\begingroup$

Note that $x^2+y^2\ge 2|xy|$. Hence, we have

$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \le \frac{\sqrt{|xy|}}{\sqrt 2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy