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On account of a SE question , I raised the following question.

Let $f:D \to \mathbb R^2$ be a function in two variables. How would we go about setting up an epsilon-delta proof?

Let $f$ for example be given by $f(x,y) = (x+y)^2$ on the domain $D = \mathbb R^2$.

Edit. I am aiming at the kind of continuity that goes like the following for a one-variable function $f$. For every $\varepsilon > 0$, we have an $\delta >0$ such that $|x-a| < \delta$ implies $|f(x)-f(a)| < \varepsilon$.

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    $\begingroup$ (Of course, once you know that composition of cont. functions is cont. and sum of cont. functions is continuous...) $\endgroup$
    – Grigory M
    May 31, 2014 at 8:25

3 Answers 3

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We show that the given function is continuous at any point $(a,b)$. So given any $\epsilon\gt 0$, we want to produce a number $\delta$ such that $$|(x+y)^2-(a+b)^2|\lt \epsilon$$ whenever $d((x,y),(a,b))\lt \delta$. Here $d((x,y),(a,b))$ is the Euclidean distance between $(x,y)$ and $(a,b)$, that is, $\sqrt{(x-a)^2+(y-b)^2}$. Note that if $\sqrt{(x-a)^2+(y-b)^2}\lt \delta$ then $|x-a|\lt \delta$ and $|y-b|\lt \delta$.

We have $$(x+y)^2-(a+b)^2=((x+y)-(a+b))(x+y+a+b).\tag{1}$$ By choosing $\delta$ to be small, we have very direct control over the size of $|(x+y)-(a+b)|$, and can force it to be as small as we wish. But we must also get control over the size of $|x+y+a+b|$, to make sure it does not get too big. That is what the rest of the calculation is devoted to.

Suppose that $|x-a|\lt \delta$ and $|y-b|\lt \delta$ and $\delta\lt 1$. Then $|x|\lt |a|+1$, and $|y|\lt |b|+1$. It follows that $|x+y+a+b|\lt 2+2|a|+2|b|$. Thus from (1) we have $$|(x+y)^2-(a+b)^2|\lt (2\delta)(2+2|a|+2|b|).\tag{2}$$ It follows from (2) that if $0\lt \delta\lt \frac{\epsilon}{2(2+2|a|+2|b|)}$ then $|(x+y)^2-(a+b)^2|\lt \epsilon$.

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  • $\begingroup$ What I always have found cryptic in epsilon-delta proofs was step $(1)$. We somehow factor an expression to make the argument work. Did you write this out first, where you expecting it, or did you just noticed it right away? $\endgroup$ May 31, 2014 at 4:24
  • $\begingroup$ With polynomials, there will always be some kind of factorization, because of the theorem that $a$ is a root of $P(t)$ if and only if $t-a$ divides $P(t)$. In several variables, things get more complicated, but still factorization will occur. In our case, it was a simple factorization, because $(x+y)^2$ is a polynomial in the one variable $x+y$. $\endgroup$ May 31, 2014 at 5:14
  • $\begingroup$ We needn't make the assumption $\delta \lt 1$. Assuming $|x-a|\lt \delta$ and $|y-b|\lt \delta$, we can arrive at \begin{align*}|(x+y)^2-(a+b)^2| &\leq (|x-a|+|y-b|)|x+y+a+b| \\ &= (|x-a|+|y-b|)|(x-a)+2a+(y-b)+2b| \\ &\leq (|x-a|+|y-b|)(|x-a|+|2a|+|y-a|+|2b|) \\ &\lt 4\delta(\delta + |a|+|b|)\ . \end{align*} Solving the equation $$\varepsilon = 4\delta(\delta+|a|+|b|) $$ for $\delta$, we can conclude with $$\delta=\bigg| \frac{-(|a|+|b|)\pm \sqrt{(|a|+|b|)^2+\varepsilon}}{2}\bigg|\ .$$ $\endgroup$ Feb 26, 2015 at 3:52
  • $\begingroup$ @MusséRedi: An argument like yours will work, but we really don't want to take the $-$ in the $\pm$ above. $\endgroup$ Feb 26, 2015 at 6:38
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Let's show it is continuous at $(a,b)$. Then you calculate:

$|f(x,y) - f(a,b)| = |((x-a) + (y-b) + (a+b))^2 - (a+b)^2| \leq (x-a)^2 + (y-b)^2 + 2|x-a||y-b| + 2|x-a||a+b| + 2|y-b||a+b)| < (x-a)^2 + (y-b)^2 + (x-a)^2 + (y-b)^2 + 4(|a|+|b|)\cdot \sqrt{(x-a)^2 + (y-b)^2} = 2\left((x-a)^2 + (y-b)^2\right) + 4(|a|+|b|)\cdot \sqrt{(x-a)^2 + (y-b)^2} < \left(2 + 4|a| + 4|b|\right)\cdot \sqrt{(x-a)^2 + (y-b)^2}$, if we preset $\sqrt{(x-a)^2 + (y-b)^2} < 1$. From here you can see the answer as you take: $\delta = \text{min} \{1, \dfrac{\epsilon}{2 + 4|a|+4|b|}\}$ for a given $\epsilon > 0$. Done.

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  • $\begingroup$ Could you organize the post more carefully? $\endgroup$ May 31, 2014 at 4:34
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I see that the answers claim they are continuous, but it appears $F(x,y) = (x+y)^2$ is not uniformly continuous.

By definition of uniformly continuous. A function f is uniformly continuous on a set $E$ if and only if, corresponding to each $e > 0$, a number $d > 0$ can be found such that $|F(x,y) - F(a,b)| < e$ whenever $(x,y)$ and $(a,b)$ are in $E$ and $|(x,y) - (a,b)| < d$.

Let $a = x + d/2$ and $b = y + d/2$, and suppose $e=1$

It follows that \begin{align*} |F(x,y) - F(a,b)| =& |(x + y)^2 - (a + b)^2| \\ =& |(x + y)^2 - (x + y + d)^2| \\ =& | x^2 + 2xy + y^2 - (x^2 + 2xy + 2dx + y^2 + 2dy + d^2)| \\ =& | 2dx + 2dy + d^2| \\ =& 2d|x + y| + d^2 \\ <& 1 \end{align*}

but you can see that picking $x$ and $y$ arbitrarily large, this is a contradiction.

This implies that $F(x,y)=(x+y)^2$ is not uniformly continuous.

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