1
$\begingroup$

I want to show that $f(x,y)=\frac{y}{(x^2+y^2)}$ is continuous on $[0,1]\times[1,2]$.

Let $\epsilon >0$ be given. I want to choose a $\delta>0$ such that for $||(x_1,y_1)-(x_2,y_2)||<\delta$ we have $|\frac{y_1}{(x_1^2+y_1^2)}-\frac{y_2}{(x_2^2+y_2^2)}|<\epsilon$.

The only thing I found is that \begin{equation}|\frac{y_1}{(x_1^2+y_1^2)}-\frac{y_2}{(x_2^2+y_2^2)}|\leqslant \frac{y_1(x_2^2+y_2^2)-y_2(x_1^2+y_1^2)}{(x_1^2+y_1^2)(x_2^2+y_2^2)}\leqslant \frac{2(1+4)-1(0+1)}{(0+1)(0+1)}=9.\end{equation}

I have no idea how to find such a $\delta$.

$\endgroup$
  • 2
    $\begingroup$ It's not continuous ! $\lim_{t\to 0^+}f(0,t)=+\infty$. $\endgroup$ – Surb Oct 5 '18 at 9:06
  • $\begingroup$ Correct with this domain, but that was a typo... I changed it now! $\endgroup$ – David W. Oct 5 '18 at 9:11
2
$\begingroup$

Your function is not defined at $(0,0)$. For the other points: fix $a=(x_0,y_0)$ and let $\epsilon >0$. Observe that if $\delta < \|a\|/2$ then for all $b=(x_1,y_1)$ such that $\|a-b\|< \delta$ we have $$\frac{1}{(\|a\|-\frac{\|a\|}{2})^2 }=\frac{4}{\|a\|^2} \geq \frac{1}{\|b\|^2}.$$ Observe that $|y_0 - y_1|\leq \| a-b\|$ and $$ |f(a)-f(b)| = \left| \frac{y_0}{\|a\|^2} - \frac{y_1}{\|b\|^2} \right| = \left| \frac{ y_0 \|b\|^2 - y_1 \| a\|^2}{\|a\|^2 \|b\|^2}\right| \leq \left| \frac{ |y_0|\cdot \left|\|b\|^2 - \| a\|^2 \right|+ |y_1 - y_0|\cdot \|a\|^2}{\|a\|^2 \|b\|^2}\right| \leq \frac{4}{\|a\|^4}\left( |y_0|\cdot \left|\|b\|^2 - \| a\|^2 \right|+ |y_1 - y_0|\cdot \|a\|^2 \right). $$ Choose $$\delta < \min \left\{ \frac{2 \epsilon}{5\|a\|}, \frac{\|a\|}{2}, \frac{\epsilon}{\|a\|^2} \right\}.$$

$\endgroup$
  • $\begingroup$ When I try to do this, I get $|f(a)-f(b)|\leqslant \frac{4|4y_0-y_1|}{||a||^2}$. Also, I don't get why you did say that $\delta <||a||/2$.. $\endgroup$ – David W. Oct 5 '18 at 10:51
  • 1
    $\begingroup$ You are right. I fixed the proof. I choose $\delta < \|a\|/2$ simply to avoid (0,0) and make estimates easier in some parts. $\endgroup$ – Nick Bottom Oct 5 '18 at 12:02

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.