Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Fix a metric $g$ on a smooth, closed manifold $\mathcal{M}$. Take a finite subcover of the manifold from its atlas. Is it true that any smooth partition of unity subordinate to this cover has uniformly bounded derivatives in $L^p (\mathcal{M})$ for each $p \geq 1$ ?

share|improve this question
    
Have you tried to figure this out yourself? What difficulty did you run into? Also, note that you need to clarify what you mean by "uniformly bounded derivatives in $L^p(M)$". –  Deane Yang Feb 6 '12 at 19:46
    
I mean that $|| \nabla ^k \phi _j ||_{ L^p (\mathcal{M})} \leq K$ for each $k \geq 0$, where $K$ can depend on $j$ but is independent of $k$. Here, $\phi _j$ is one function in the partition of unity. In terms of trying it for myself, I have given it some thought but was hoping I was missing something trivial. –  T-' Feb 6 '12 at 20:19
    
@Michael, something is not competely clear to me: you want a bound on the $L^p$ norms of the derivatives, which is uniform on which of the following: $p\ge1$; any partition of unity subordinated to the subcover $\mathcal{U}$; at least some partition of unity subordinated to the subcover $\mathcal{U}$; the subcover $\mathcal{U}$ itself ? Thanks. –  Pietro Majer Feb 6 '12 at 20:24
    
It appears that he wants it to be uniform in the number of derivatives. I don't have an answer to this, but I think it suffices to find a single compactly supported smooth function on $R^n$ with a uniformly bound for any number of derivatives in $L^p$. –  Deane Yang Feb 6 '12 at 20:44
    
I doubt it. Here's why: I would guess that you can reduce this to a partition of unity for a closed interval. Now the derivatives of a given element of the "standard explicit" partition (see, e.g., Dubrovin, Fomenko, and Novikov's construction) are going to oscillate more as the order of the derivative increases. Looking at tanh (which is similarish) I don't see why the extrema of these derivatives would be bounded (though I don't have ready access to software to check this numerically at the moment). –  Steve Huntsman Feb 6 '12 at 21:02

1 Answer 1

up vote 4 down vote accepted

A counterexample: Let $M$ be the unit circle. Let the two charts be the arcs $A = (0,2\pi)$ and $B = (\pi/2, 5\pi/2)$. For each $n$, consider the partition of unity subordinate to $\{A,B\}$ given by

$$ \psi_{A,n} = \sin^2 ( (2n+1) \theta ) $$

and

$$ \psi_{B,n} = \cos^2 ( (2n+1) \theta ) $$

Check that $\psi_{A,n} (0) = 0 = \psi_{B,n} (\pi/2)$, and clearly $\psi_{A,n} + \psi_{B,n} = 1$.

By the scaling property it is easy to check that

$$ \| \partial^k\psi_{A,n} \| = (2n+1)^k \| \partial^k\psi_{A,1} \| $$

And using that

$$ \sin^2(\theta) = \frac12 (1-\cos 2\theta) $$

you see immediately that your desired uniform bound is impossible.

share|improve this answer
    
Thanks for this! –  T-' Mar 1 '12 at 16:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.