Continuity of piecewise function of two variables

[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?

 

[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.

 

[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'?

 

[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."

 

[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?

 

[pasmith]

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?

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 &gt; 0 arbitrary. Let \delta&#039; = \min\{\delta, 2^{4/3}\} and consider the point (\frac12 \delta&#039;, \frac{1}{32}\delta&#039;^4). This by construction satisfies <br /> \|(\tfrac12 \delta&#039;, \tfrac{1}{32}\delta&#039;^4)\| = \left( \frac14 \delta&#039;^2 + \frac{1}{2^{10}}\delta&#039;^8\right)^{1/2} &lt; \sqrt{\tfrac12 \delta&#039;^2} &lt; \delta<br /> and f(\tfrac12 \delta&#039;, \tfrac{1}{32}\delta&#039;^4) = 1.