2 Variable Continuity

Calculus Level 3

Suppose that $f(x,y)$ satisfies the following conditions:

For all $X$ , the function $f( X, y )$ is continuous in $y$ .
For all $Y$ , the function $f ( x, Y)$ is continuous in $x$ .

Must the function $f(x,y)$ be continuous?

No, it need not be continuous Yes, it must be continuous

1 solution

Brian Moehring
Jun 15, 2018

Consider the functions $g(x) = \begin{cases}x & 0\leq x\leq 1 \\ 2-x & 1 \leq x \leq 2 \\ 0 & \text{otherwise}\end{cases}$ and $f(x,y) = \begin{cases} \frac{g(x/y)}{y} & x,y > 0 \\ 0 & \text{otherwise}\end{cases}$

Then we claim that this $f$ satisfies the two conditions but is not continuous at the origin.

Property 1: Fix some $X$ . If $X\leq 0$ then $f(X,y) = 0$ for all $y$ , which is continuous. Therefore, let's assume $X>0$ . Then $\begin{aligned}f(X,y) &= \begin{cases}g(X/y)/y & y > 0 \\ 0 & \text{otherwise}\end{cases} \\ &= \begin{cases}\frac{X}{y^2} & 0 \leq \frac{X}{y} \leq 1 \\ \frac{2-\frac{X}{y}}{y} & 1\leq \frac{X}{y} \leq 2 \\ 0 & \text{otherwise}\end{cases} \\ &= \begin{cases}\frac{X}{y^2} & y \geq X \\ \frac{2}{y} - \frac{X}{y^2} & \frac{X}{2} \leq y \leq X \\ 0 & \text{otherwise}\end{cases}\end{aligned}$ which is continuous (note that the pieces "fit" together and at the only possible problem point of $y=0$ , the function is locally zero, so continuous)

Property 2: Fix some $Y$ . As before, if $Y \leq 0$ , then $f(x,Y) = 0$ for all $x$ , which is continuous, so we may assume $Y>0$ . Then $f(x,Y) = g(x/Y)/Y = (h \circ g \circ h)(x)$ is just the composition of the continuous function $g$ and the continuous function $h(x) = x/Y$ , hence continuous.

$f$ is not continuous at the origin: To see this, note that $f(0,0) = 0 \\ \lim_{x\to 0^+} f(x,x) = \lim_{x\to 0^+} \frac{g(x/x)}{x} = \lim_{x\to 0^+} \frac{1}{x} = +\infty.$

Your solution complicates the intention of the question. It provides a counter-example, but doesn't explain what's really going on.

The key point of the question is to realize that for the function to be continuous (say at the origin), we require the limit to exist along every possible path to the origin. However, we're only given 2 directions (x-axis, y-axis), and hence have very limited control over what happens anywhere else. This suggests that we could let the limit along the path $y = Kx$ , or the path $y = Kx^2$ , be varying and hence the limit doesn't exist so the function isn't continuous.

For the path of $y = Kx$ , if we naively set $\frac{y}{x}$ , we do get a varying value of $K$ , but end up with the x-axis being undefined. To fix that, we use a counter example of $f(x,y) = \frac{ xy} { x^2 + y^2}$ and $f(0,0) = 0$ . In this case, setting $y = Kx$ gives us $\frac{ K } { 1 + K^2}$ , which is clearly not equal for all values of $K$ .

See if you can come up with a (simple) counter example for the path $y = K x^2$ . Even though this might seem to locally look like $y = 0$ at the origin, it's still possible to create an issue.

Calvin Lin Staff - 3 years ago

I agree my counterexample isn't the prototypical one, and I even agree that if the intention is to teach someone who understands single-variable continuity but not multi-variable continuity, we ought to include such prototypical examples. It's just been so long since I've taken (or even since I've taught) multivariable calculus, I had forgotten all about them.

However, I am unapologetic about my example. I designed it specifically so that $\lim_{x\to 0^+} f(x,kx) = \begin{cases}\infty & 1/2 < k < \infty \\ 0 & \text{otherwise}\end{cases}$ has only those two possibilities. This is more interesting to me because it suggests a method for creating a function $f^*$ whose restriction to a line is discontinuous for exactly one line and the discontinuity is essential.

As for your last comment, of course $f(0,0)=0$ and $f(x,y) = \frac{x^2y}{y^2+x^4}$ for all other points would work.

Brian Moehring - 3 years ago

2 Variable Continuity

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