Bounded and conitnuous

[mario22]

Assume that functions f, g : R[sup:9qmuzlsq]2[/sup:9qmuzlsq] --> R[sup:9qmuzlsq]2[/sup:9qmuzlsq] satisfy f(0, 0) = g(0, 0) = 0
and for all (x, y)? R[sup:9qmuzlsq]2[/sup:9qmuzlsq] \ {(0, 0)}
f(x, y) = xy[sup:9qmuzlsq]2[/sup:9qmuzlsq](x[sup:9qmuzlsq]2[/sup:9qmuzlsq] + y[sup:9qmuzlsq]4[/sup:9qmuzlsq])[sup:9qmuzlsq]-1[/sup:9qmuzlsq];
g(x, y) = xy[sup:9qmuzlsq]2[/sup:9qmuzlsq](x[sup:9qmuzlsq]2[/sup:9qmuzlsq] + y[sup:9qmuzlsq]6[/sup:9qmuzlsq])[sup:9qmuzlsq]-1[/sup:9qmuzlsq];

Show that
1) f is bounded on R[sup:9qmuzlsq]2[/sup:9qmuzlsq]
2) g is not bounded in any neighborhood of (0, 0)
3) f is not conitnuous at (0,0)

regarding 3 - I wrote that the lim of f when x,y --> 0 is +-?, but f(0,0)=0 - so f is not conitnuous.

who can help wuth 1,2?

 

[daon]

As written, your reasoning for 3) contradicts 1). You are trying to show that the limit, if it exists, is not 0. The problem is that the limit does not exist... Show that by approaching (0,0) from two different ways gives you contradictory answers. Try x=y^2

For 1, you need to show there is an M for which |f(x,y)| <= M for all x,y. As either x,y go to infinity, f -> 0 (note f is continuous except possibly at 0, so it attains its maximum on any closed region). You need only check that f is bounded in some neighborhood of 0. For then it attains its maximum on some region with part of its boundary ("the inside of the annulus") being |(x,y)|=delta for some delta > 0 and outside the region |f(x,y)|<epsilon by the second sentence. Since f is bounded on |(x,y)| >=k for all k>0, you may assume |x|, |y| < k. But justify that.

For 2, show that for any epsilon > 0, there is an M such that |(x,y)|<epsilon imply |g(x,y)| > M. Hint: let x=1. Find an inequality that lets you choose an appropriate y (in terms of M) that gives g(x,y)>=M.