Hi @Daniel. Cool problem I've been giving the kids I'm working with ... still don't have the maximally elegant proof :) Prove that any $C^\infty$ function $f\colon\Bbb R^2\to\Bbb R$ that vanishes on the coordinate axes can be written as $f(x,y)=xyg(x,y)$ for a smooth function $g$. :)
Well, I had them prove the so-called $C^\infty$ trick: A $C^\infty$ $f\colon\Bbb R^2\to\Bbb R$ with $f(0)=0$ can be written as $f(x,y)=xg_1(x,y)+y g_2(x,y)$ for smooth functions $g_i$. That uses an integral. What do you mean?
Using my approach, @Daniel, it works but smoothness along the axes takes an argument. Your approach is, as I expect of you, much cleaner. We'll see if my undergrads think of it :)
They're both quite talented, and we've been doing multivariable analysis/manifolds (mostly from my book) to get ready for the differential topology class this fall.
One has taken 4 classes from me, the other just diff geo this past spring. I was getting him ready for diff top. He sadly missed out on the tough multivariable math class because undergrads these days are very docile about studying the course catalog to see what might be good for them.
And advisers are worthless ... generally speaking.
@DanielF, forgive me for asking a personal question (which you can choose not to answer, of course). Since I observe that you're here 'round the clock, are you now retired?
You should do the problem I posed earlier. But don't scroll up. Suppose $f\colon \Bbb R^2\to\Bbb R$ is smooth and vanishes on the coordinate axes. Prove that $f(x,y)=xyg(x,y)$ for some smooth $g$.
@robjohn No. It was that way from the beginning. This is just a hard version. I remember you evaluated the version without square (you gave an answer to one of my questions).
@PedroTamaroff Uniformly continuous means it has a continuous extension to $\overline{B(0,1)}$, which is compact (if $X$ is not yet complete, the extension has codomain $\tilde{X}$).
@BalarkaSen, consider the polynomial $f(x) = x^3 + x + 1$. Is $\sqrt{-31} \in \mathbb{Q}[x]/\langle f(x) \rangle$? Is $\sqrt{-31}$ an element of the splitting field?
@DanielFischer This one was easy too. Suppose $X$ is a connected unbounded metric space. Then for each $a\in X$ and $r>0$ there is $x$ with $d(a,x)=r$.
@DanielFischer Hm, another one was showing that if $f:X\times Y\to Z$ is such that each $f(x,-)$ is continuous and the family of $f(-,y)$ is equicontinuous, then $f$ is continuous.
Easy too. Then it asked about the case when each $f(x,-)$ is UC and the family of $f(-,y)$ is unif. equicontinuous. I couldn't think of a counterexample, then usual spirit d'escalier and a friend's comment gave $f(x,y)=xy$, $\Bbb R^2\to\Bbb R$.
I figured this doesn't have enough content for a full question so I came here to ask: Does anyone have a hint as to how to prove that for $1 < p < q < \infty$ we have $$ \lVert f \rVert_p \le \mu(\Omega) \lVert f \rVert_q $$ where $\Omega$ is the space you're working over and doesn't contain any set of arbitrarily large measure. I know that I have to use Hölder's Inequality but I'm not sure how.
$\mu(\Omega)^{1/p-1/q}$*, sorry I tried to edit it above but apparently it was too late (and I have a horrid inet connection ATM so that's probably why)
