# Proving continuity by epsilon-delta proof for a function of two variables.

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 $|f(x)-f(a)| < \varepsilon$.

• (Of course, once you know that composition of cont. functions is cont. and sum of cont. functions is continuous...) May 31, 2014 at 8:25

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 $|(x+y)^2-(a+b)^2|\lt \epsilon$.

• 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? May 31, 2014 at 4:24
• 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 $x+y$. May 31, 2014 at 5:14
• 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 $$\delta=\bigg| \frac{-(|a|+|b|)\pm \sqrt{(|a|+|b|)^2+\varepsilon}}{2}\bigg|\ .$$ Feb 26, 2015 at 3:52
• @MusséRedi: An argument like yours will work, but we really don't want to take the $-$ in the $\pm$ above. Feb 26, 2015 at 6:38

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.

• Could you organize the post more carefully? May 31, 2014 at 4:34

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.