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I cant seem to prove this thing. By working on the definition: a function f is continuous at $p_0 \in D$ if for every $\epsilon >0, \exists \delta >0$ such that $\mid f(p) - f(p_0)\mid < \epsilon$ whenever $\mid p-p_0 \mid < \delta$.

Well I tried working backwards. Let $p = (x,y)$ and $p_0 = (x_0 , y_0)$

  • $\mid M(p) - M(p_0) \mid = \mid xy - x_0y_0 \mid = \mid (x-x_0)y + x_0(y-y_0) \mid < \epsilon$

If $\mid p - p_0 \mid < \delta$

$\Rightarrow \mid \delta y + \delta x_0\mid = \delta \mid y + x_0 \mid < \epsilon$

Now is it alright for me to choose $\delta = \frac{\epsilon}{y + x_0}$?

I'm confused, what if $y+x_0 = 0$ at some point in the disk? or is this possible.

llawliet_78
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1 Answers1

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We have \begin{eqnarray} |x_1 y_1 - x_2 y_2| &\le& |x_1 y_1 - x_1 y_2| + | x_1 y_2- x_2 y_2| \\ &=& |x_1||y_1-y_2| + |y_2||x_1-x_2| \\ &\le& r (|x_1-x_2|+|y_1-y_2|) \end{eqnarray}

copper.hat
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  • you sir, are a genius!!! thanks so much for this!! – llawliet_78 Oct 23 '15 at 15:13
  • @llawliet_78: I wish :-). It is one of the standard tricks, worth remembering. – copper.hat Oct 23 '15 at 15:14
  • sorry I don't see why $|x_1| * + |y_2|** \leq r (* + **) $, can you explain, thanks. – Lilian Hernández Aug 27 '18 at 04:11
  • I'm not sure what particular step is causing you concern, we have $|x_1| \le r$ and $|y_2| \le r$. – copper.hat Aug 27 '18 at 04:20
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    @LilianHernández: Since $|x_1|\leq r$, we can multiply both sides of the inequality by $|y_1-y_2|$, getting that $|x_1||y_1-y_2| \leq r|y_1-y_2|$.

    Repeating the same process for $|y_2| \leq r$, we multiply both sides of the inequality by $|x_1-x_2|$ to get

    $|y_2||x_1-x_2| \leq r|x_1-x_2|$. We add both of the inequalities together to get

    $|x_1||y_1-y_2| + |y_2||x_1-x_2| \leq r|y_1-y_2| + r|x_1 -x_2|=r(|y_1-y_2|+|x_1-x_2|)$

     – Skm Sep 17 '18 at 02:04
  • Thank you so much for taking your time, now It's way clearer. – Lilian Hernández Dec 16 '18 at 06:15