3
$\begingroup$

Let $h : \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function defined by $$h(x,y) =\begin{cases} x^2 + y^2 & : (x,y) \in \mathbb{Q} \times \mathbb{Q} \\[1ex] 0 & : \mbox{otherwise}\end{cases}$$

Show that $h$ is continuous only at $(0,0)$, and differentiable there.

I can show the continuity of $h$ at $(0,0)$. Also, I can show the discontinuity of $h$ at rational pair which is not $(0,0)$. However, I cannot show the discontinuity of other points. Also, the differentiation. Could anyone give a hint ?

$\endgroup$
1
$\begingroup$

HINT:

If $(x_0,y_0)\neq(0,0)$ then in a small disk with centre in $(x_0,y_0)$ there are numbers greater than some $\varepsilon$.

Partial derivatives are equal to 0 in origin.

$\endgroup$
0
$\begingroup$

The discontinuity at $(a,b)\in{\Bbb R}\times{\Bbb R}\setminus{\Bbb Q}\times{\Bbb Q}$ is easy: take a sequence $(x_n,y_n)\in{\Bbb Q}\times{\Bbb Q}$ s.t. $(x_n,y_n)\to(a,b)$. About the differentiability: only is possible at $(0,0)$ (why?). Start studying the partial derivatives at $(0,0)$.

$\endgroup$
0
$\begingroup$

It's a quite simple argument.   Every rational coordinate pair $(x,y)$ will be adjacent to points with an irrational ordinate.   Thus the equation with be discontinuous at every rational ordinate pair where $h(x,y)\neq 0$.

If the function is not continuous at these points, then it follows that it is not differentiable there.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.