Prove epsilon delta continuity for 4 real numbers.

Question

I am trying to prove continuity at $(a,b)$ for the function $f:\Bbb R^2→\Bbb R,$ where $d((a,b),(c,d))=\left|a-c\right|+2\left|b−d\right|$ is my metric and my function is $f(x,y)=x^2+y^2.$

Show that one can make a suitable choice of $\delta>0$ which is a function of $\epsilon$, $a$ and $b$ such that for a given $\epsilon>0$: $$\left|a-c\right|+2\left|b−d\right|<\delta$$ implies that $$\left|a^2+b^2-c^2-d^2\right|<\epsilon.$$

I attempted to use the triangle inequality on the above inequality to achieve:

$$\left|a^2+b^2-c^2-d^2\right| \le\left|a^2-c^2\right|+\left|b^2-d^2\right|$$ $$=\left|(a-c)(a+c)\right|+\left|(b-d)(b+d)\right|$$

but I don't know how to proceed from here.

Answer 1

Based on your comments, you'd like to demonstrate the pointwise continuity of $f(x,y) = x^2+y^2$ under the metric $d((x,y),(z,w)) = |x-z|+2|y-w|$ on $\mathbb{R}^2$.

To prove continuity at a given $(a,b)$, then, for each $\epsilon > 0$ we must identify a $\delta > 0$ such that

$$d((x,y),(a,b)) < \delta \implies |f(x,y)-f(a,b)| < \epsilon.$$

Suppose that $d((x,y),(a,b)) < \delta$. In particular, this separately implies that $|x-a| < \delta$ and $|y-b| < \delta$. You've noticed that the triangle inequality yields the first of the following inequalities, which we continue:

\begin{align} |f(x,y) - f(a,b)| & \leq |x-a||x+a| + |y-b||y+b| \\ & < (|x+a| + |y+b|) \delta \\ & \leq (|x|+|a| + |y|+|b|) \delta \end{align}

To complete the argument, we use that $|x-a| < \delta$ implies that $|x| \leq |a| + \delta$ (and similarly for $y$), meaning

$$ |f(x,y) - f(a,b)| < 2(|a| + |b| + \delta) \delta$$

From this, we see that choosing $\delta = \min \left( |a|+|b|, \frac{\epsilon}{4(|a|+|b|)} \right)$ yields the desired inequality (check this) so long as $(a,b) \neq (0,0)$. To establish continuity at the origin, we may simply take $\delta = \sqrt{\epsilon/2}$.

Answer 2

Letting $h:=b-a$ and $k:=d-c,$ your problem becomes: given $a,b\in\Bbb R$ and $\epsilon>0,$ find $\delta>0$ such that for all $h,k\in\Bbb R^2:$

$$|h|+2|k|<\delta\implies|h||2a+h|+|k||2b+k|<\epsilon.$$ Now, let$$M:=\max(|2a|,|b|).$$Then, $$|h|+2|k|<\delta\implies|2a||h|+|2b||k|+h^2+k^2<M\delta+\delta^2.$$ It is therefore sufficient to choose $\delta>0$ such that $$\delta^2+M\delta-\epsilon<0,$$ i.e. $$\delta\le\frac{\sqrt{M^2+4\epsilon}-M}2.$$