Proving Continuity in Higher Dimensions

Question

Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $f(x,y) = (x^2-y^2, 2xy)$. Show that $f$ is continuous.

In one dimension ($\mathbb{R}$), using epsilon-delta proofs is a very straightforward process. You start with $|f(x) - f(x_0)| < \varepsilon$ and you manipulate the inequality algebraically until you can obtain $|x-x_0| <$ [expression]. Then you let $\delta = $ [expression]. And doing the algebra in reverse gives the conclusion.

Now, in more than one dimension (i.e. when the codomain is multi-dimensional), it becomes less obvious how to use this same process to obtain a $\delta$ in terms of $\varepsilon$. I show my work below.

Proof Attempt.

In order to show that $f$ is continuous, we show that

$$ \forall \, \varepsilon > 0, \exists \, \delta > 0 \, \forall \, j_0 \in \mathbb{R}^2, \|j - j_0 \| < \delta \implies \|f(j) - f(j_0) \| < \varepsilon$$

where $j_0 \in \mathbb{R}^2$ and for fixed $j \in \mathbb{R}^2$.

So we start with $\|f(j) - f(j_0)\| < \varepsilon$.

$$ \| (x^2-y^2, 2xy) - (x_0^2 - y_0^2, \, 2x_0y_0) \| < \varepsilon$$

$$\| (x^2-y^2 - ( x_0^2 - y_0^2), 2xy - 2x_0y_0 ) \| < \varepsilon$$

$$\sqrt{\Big[x^2-y^2 - ( x_0^2 - y_0^2) \Big]^2 + \Big[ 2xy - 2x_0y_0 \Big]^2} < \varepsilon$$

$$\Big[x^2-y^2 - ( x_0^2 - y_0^2) \Big]^2 + \Big[ 2xy - 2x_0y_0 \Big]^2 < \varepsilon^2$$

$$\vdots$$

$$\|(x-x_0, y- y_0) \| < \delta$$

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Am I on the right track here? The \vdots are the remaining steps I need to obtain my $\delta$ but it doesn't appear to me that I can end up with $|(x-x_0 , y-y_0) | <$ [expression] without lumping variables over on the righthand side.

Any advice or suggestions are greatly appreciated.

Answer 1

Your idea will absolutely work! In general, to show that a function $f : X \to Y$ is continuous (where $X$ and $Y$ have notions of "distance", say $d_X$ and $d_Y$) we can always show that

$$\forall \epsilon . \exists \delta . d_X(x_1, x_2) < \delta \implies d_Y(f x_1, f x_2) < \epsilon.$$

For $\mathbb{R}^n$ the standard euclidean metric $d_{\mathbb{R}^n}(x_1, x_2) = \lVert x_1 - x_2 \rVert$ works, and we exactly recover your idea.

However, as you've noticed this feels messy. We have to work with a bunch of squares and square roots, and even working with the usual tricks (like trying to get $d(x_1,x_2)^2$ small instead) still leaves us wondering if there's an easier way. The answer, thankfully, is yes!

Theorem: $f : \mathbb{R}^n \to \mathbb{R}^m$ is continuous if and only if each of the coordinate functions are continuous separately.

What does that mean concretely? Well $f = (f_1, f_2, \ldots, f_m)$ where each $f_i : \mathbb{R}^n \to \mathbb{R}$. The theorem tells us that it's enough to check that each of these $f_i$ are continuous! In fact, this generalizes:

Theorem: $f : X \to Y \times Z$ is continuous if and only if the functions $\pi_Y \circ f : X \to Y$ and $\pi_Z \circ c: X \to Z$ are continuous. (Here $\pi_Y (y,z) = y$ and $\pi_Z(y,z) = z$).

This turns out to be related to the categorical notion of "product", but I won't go into that here.

Exercise: Can you prove this theorem? At least in the case of $X = Y = \mathbb{R}^2$? You'll need to do an $\epsilon$-$\delta$ argument using the $\sqrt{x^2 + y^2}$ distance.

Now how does this simplify your life in your concrete example? Well now we just want to show that

• $(x,y) \mapsto x^2 - y^2$
• $(x,y) \mapsto 2xy$

are continuous. But this should be easier. First of all, they're polynomials, so if you know that all polynomials are continuous you're done immediately. But even if you don't know that, it should be easier to use an $\epsilon$-$\delta$ argument to finish this up now that the codomain is one dimensional. If nothing else, that should be much cleaner now that you don't have to worry about square roots, etc.

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I hope this helps ^_^