1. ## Surfaces and Smoothness

Well, here's my problem. I've been at it all night and have made zero progress:

Consider the real function

$\displaystyle f(x,y) = xy(x^2 + y^2)^{-N}$, in the respective cases N = 2, 1, and $\displaystyle \frac{1}{2}$.

a.) Show that in each case the function is differentiable ($\displaystyle \mathcal{C}^{\omega}$) with respect to x, for any fixed y-value (and that the same holds with the roles of x and y reversed).

Nevertheless, f is not smooth as a function of the pair (x,y).

b.) Show this in the case N = 2 by demonstrating that the function is not even bounded in the neighborhood of the origin (0,0).

c.) Show this in the case N = 1 by demonstrating that the function though bounded is not actually continuous as a function of (x,y).

d.) Show this in the case N = $\displaystyle \frac{1}{2}$ by showing that though the function is now continuous, it is not smooth along the line x = y.

Thanks for any and all help. It's much appreciated.

2. Originally Posted by Aryth
$\displaystyle f(x,y) = xy(x^2 + y^2)^{-N}$, in the respective cases N = 2, 1, and $\displaystyle \frac{1}{2}$.

a.) Show that in each case the function is differentiable ($\displaystyle C^{\omega}$) with respect to x, for any fixed y-value (and that the same holds with the roles of x and y reversed).
You end up with $\displaystyle xy_0(x^2+y_0)^{-N}$. Thus, as long as $\displaystyle y_0\not = 0$ it would mean $\displaystyle xy_0(x^2+y_0)^{-N}$ is defined eveywhere. And you can compute the derivatives $\displaystyle g^{(n)}(x)$ where $\displaystyle g(x) = f(x,y_0)$ or all orders $\displaystyle n$. Thus, $\displaystyle g$ is $\displaystyle \mathcal{C}^{\infty}$.
[quote]

Nevertheless, f is not smooth as a function of the pair (x,y).

b.) Show this in the case N = 2 by demonstrating that the function is not even bounded in the neighborhood of the origin (0,0).
You need to show that for any $\displaystyle A>0$ we can find $\displaystyle \epsilon > 0$ so that there is a point $\displaystyle \bold{x}\in \{ 0 < \sqrt{x^2+y^2} < \epsilon \}$ such that $\displaystyle f(\bold{x}) > A$. To show this think along the x-axis. Therefore, there is no way to define the function $\displaystyle f$ to even a continous function at $\displaystyle (0,0)$.

c.) Show this in the case N = 1 by demonstrating that the function though bounded is not actually continuous as a function of (x,y).
Show $\displaystyle \lim_{(x,y)\to (0,0)} f(x,y)$ does not exist.

d.) Show this in the case N = $\displaystyle \frac{1}{2}$ by showing that though the function is now continuous, it is not smooth along the line x = y.
Defining $\displaystyle f(0,0) = 0$ will make the function continous. It is continous for $\displaystyle |f(x,y) - 0| = \left| \tfrac{xy}{\sqrt{x^2+y^2}} \right| \leq \tfrac{|xy|}{\sqrt{2|xy|}} = \tfrac{1}{\sqrt{2}} \cdot \sqrt{|xy|} \leq \tfrac{1}{2\sqrt{2}}\left( \sqrt{x^2+y^2} \right) < \epsilon$ if $\displaystyle 0 < \sqrt{x^2+y^2} < \epsilon$. Thus, the function is continous. To show it is not differenciable it is sufficient to show $\displaystyle x^2\sqrt{x^2+x^2}$ is not differenciable at $\displaystyle 0$ (this is along the path $\displaystyle y=x$).

3. I just had a classic case of "thinking WAY too hard".

That was much simpler than I intended...

Thanks a lot.