# Homework Help: Continuity of piecewise function of two variables

1. Sep 22, 2014

### A330NEO

The question looks like this.
Let $f(x, y)$ = 0 if $y\leq 0$ or $y\geq x^4$, and $f(x, y)$ = 1 if $0 < y < x^4$.
(a) show that $f(x, y) \rightarrow 0$ as $(x, y) \rightarrow (0, 0)$ along any path through (0, 0) of the form $y = mx^a$ with $a < 4$.
(b) Despite part (a), show that $f$ is discontinuous at (0, 0)
(c) Show that $f$ is discontinuous on two entire curves.
What I've came to conclusion is that when $x<0, m>0$, and $a$ being an odd number, $y$ becomes smaller then zero, so $f(x, y)$ can't be any larger than zero. But I don't think that's not enough. I think I need to find a way to generalize that $mx^a (a<4)$is larger than $x^4$ or smaller than 0 when $x$ and $y$ is close enough to zero, where I cant' quite get to.
In regarding (b), I know $f(x, y)$ is discontinuous on certain directions, but can't elaborate it in decent form.
In regarding (C), How can I show it?

Last edited: Sep 23, 2014
2. Sep 23, 2014

### pasmith

For (a): The only way $f(x,mx^a)$ can equal 1 is if $0 < mx^a < x^4$. If both $m$ and $x^a$ are positive then this requires $|x| > m^{1/(4 - a)}$. Is that true as $|x| \to 0$, or is there a point beyond which this constraint is violated? What can you say about the case when both $m$ and $x^a$ are negative?

For (b): Can you show that there are points (x,y) arbitrarily close to (0,0) for which f(x,y) = 1, and yet f(0,0) = 0?

For (c): Use the above idea: Let (X,Y) be a point on such a curve; you need to show that there are points (x,y) arbitrarily close to (X,Y) such that $|f(x,y) -f(X,Y)| = 1$.

3. Sep 23, 2014

### A330NEO

I know why the curve is incontinuous except for points on x-axis or y-axis, but I want to know how to prove it elaborately. How can I do that? can it be done by epsilon-delta? If so, then how? If not, what is the 'other way'?

4. Sep 23, 2014

### pasmith

The definition of continuity at a point is:

"$f: \mathbb{R}^2 \to \mathbb{R}$ is continuous at $\mathbf{a} \in \mathbb{R}^2$ if and only if for every $\epsilon > 0$ there exists a $\delta > 0$ such that for all $\mathbb{x} \in \mathbb{R}^2$, if $\|\mathbf{x} - \mathbf{a}\| < \delta$ then $|f(\mathbf{x}) - f(\mathbf{a})| < \epsilon$".

The negation of that is:

"$f: \mathbb{R}^2 \to \mathbb{R}$ is not continuous at $\mathbf{a} \in \mathbb{R}^2$ if and only if there exists an $\epsilon > 0$ such that for all $\delta > 0$ there exists an $\mathbf{x} \in \mathbb{R}^2$ such that $\|\mathbf{x} - \mathbf{a}\| < \delta$ and $|f(\mathbf{x}) - f(\mathbf{a})| \geq \epsilon$."

5. Sep 23, 2014

### A330NEO

Understood. But, if we approach (0, 0) along the y-axis, regardless of if y >0 or y<0, isn't it continuous? Is it considered discontinuous because there is no 'width' of the contacting point?

6. Sep 23, 2014

### pasmith

Any non-constant function from the plane to {0,1} is of necessity discontinuous. If you compose such a function with a path on which the function is constant, then the result is a constant function from (an interval of) the reals to {0,1}, which is continuous. Here the y-axis lies entirely within the region in which f = 0. There are, of course, points arbitrarily close to (0,0) which lie in the region in which f = 1, which is why f is not continuous at (0,0).

If you want an epsilon-delta proof of that, take $\epsilon = \frac12$ and $\delta > 0$ arbitrary. Let $\delta' = \min\{\delta, 2^{4/3}\}$ and consider the point $(\frac12 \delta', \frac{1}{32}\delta'^4)$. This by construction satisfies $$\|(\tfrac12 \delta', \tfrac{1}{32}\delta'^4)\| = \left( \frac14 \delta'^2 + \frac{1}{2^{10}}\delta'^8\right)^{1/2} < \sqrt{\tfrac12 \delta'^2} < \delta$$ and $$f(\tfrac12 \delta', \tfrac{1}{32}\delta'^4) = 1.$$