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I need to show that $f:\mathbb{R}^2 \rightarrow \mathbb{R}, f(x,y)=x^2+y$ is continuous in $\mathbb{R}^2$ (everywhere) by $\epsilon - \delta$ definition of continuity.

First, I write the definition.

$f$ continuous $\Leftrightarrow \forall a=(a_{1},a_{2}) \in \mathbb{R}^2 \ \ \forall \epsilon >0 \ \ \exists \delta>0 \ \ \forall x=(x_{1},x_{2})\in \mathbb{R}^2 \ \ (\sqrt{(x_{1}-a_{1})^2+(x_{2}-a_{2})^2}<\delta \Rightarrow |f(x)-f(a)|<\epsilon)$

Than, I rewrite that for the problem,

$\sqrt{(x_{1}-a_{1})^2+(x_{2}-a_{2})^2}<\delta \Rightarrow |x_{1}^2+x_{2}-a_{1}^2-a_{2}|<\epsilon$

After, I tried many things to solve this problem. Especially, I think the triangle inequality but i couldn't find the solution.

What should I do? How can I solve this problem?

Thanks a lot for your helps.

matholympian
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  • You are not obligated to use the Euclidean metric. Frequently, there are other metrics that are easier to work with. – Doug M Feb 16 '21 at 22:21
  • It may help to grab a book or two on multivariable calculus where they do this kind of multivariable $\epsilon$-$\delta$ management by hand. That's maybe the only place you'd see it in a reference. In "real life," nobody proves that functions like this are continuous from the definitions; you'd use (fairly self-evident) continuity of the coordinate functions, alongside general theorems that products and sums of continuous functions are continuous, to immediately verify on sight, without epsilon wrangling. But many books do include examples from first principles. – leslie townes Feb 16 '21 at 22:22

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Choose $\delta = \min\{1, \frac{\varepsilon}{2(1+2|a_1|)}\}$. Now, $|x_1| \leq |x_1-a_1| + |a_1| \leq 1 + |a_1|$. So, $$|x_1^2-a_1^2+x_2-a_2| \leq |x_1-a_1| \cdot |x_1+a_1| + |x_2-a_2| \leq |x_1-a_1| \cdot (|x_1|+|a_1|) + |x_2-a_2| \leq \delta \cdot (1+2|a_1|) + \delta \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2(1+2|a_1|)} < \varepsilon$$

Tan
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