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 $\delta$.

  • 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

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 $$\delta < \min \left\{ \frac{2 \epsilon}{5\|a\|}, \frac{\|a\|}{2}, \frac{\epsilon}{\|a\|^2} \right\}.$$

  • $\begingroup$ 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

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