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Suppose $\mathcal{M}$ is a Riemannian manifold with metric $g$ and with the volume measure $d\mu$ (induced by $g$). Let $f:\mathcal{M}\to \mathbb{R}$ be $C^{\infty}$. Is it true that $$\max_{\mathcal{M}}f-\min_{\mathcal{M}}f\leq\int_{\mathcal{M}}|\nabla f|d\mu \;?$$ Here $|\nabla f|$ is the norm of the gradient of $f$ with respect to the metric $g.$

Edit 1: Following the comments I'd like to know if the following inequalities hold $$(\max_{\mathcal{M}}f-\min_{\mathcal{M}}f)\operatorname{diam}{({\mathcal{M}})}\leq\int_{\mathcal{M}}|\nabla f|d\mu \;?$$ Edit 2:

$$(\max_{\mathcal{M}}f-\min_{\mathcal{M}}f)\operatorname{Vol}{(\mathcal{M})}\leq\int_{\mathcal{M}}|\nabla f|d\mu \;?$$

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A physical intuition is helpful here: your inequality has "the wrong units". Surely you need to multiply $\lvert \nabla f \rvert$ by some kind of length, like $\operatorname{diam}(\mathcal M)$. –  Rahul Aug 15 '11 at 18:41
    
yes I just found it I added it as a comment. –  timhortons Aug 15 '11 at 18:46
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Regarding your edit: You still have the wrong units ($f$ times length on the left, $f$ by length times volume on the right), and the answer is still no. Willie Wong's answer remains a counterexample to your new inequality, because $\operatorname{diam}(M) \approx 1$. I'd conjecture that $\sup_M f - \inf_M f \leq \operatorname{diam}(M) \sup_M \lvert \nabla f \rvert$ instead. –  Rahul Aug 15 '11 at 19:03
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I'd suggest that you stop for a moment and think about what the inequality would actually mean, and why you might or might not expect it to be true, instead of throwing in multiplicative factors willy-nilly hoping for something to stick. –  Rahul Aug 15 '11 at 19:46
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A bit of unsolicited advice: speaking from the point of view of someone who has been through it, I encourage you to pause and try and figure out how I came up with my counterexamples, and think about how @Rahul got his conjecture, before you write your next edit. Sometimes it is this step that helps the most when learning new things. –  Willie Wong Aug 15 '11 at 20:10
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1 Answer 1

up vote 5 down vote accepted

No.

Let $M = (0,1)\times (0,\epsilon)\subset \mathbb{R}^2$ with coordinates $(x,y)$ and with the metric inherited from the Euclidean metric. Take $f(x,y) = x$. $|\nabla x| = 1$. $\sup f - \inf f = 1$ (Please note that in general on an arbitrary set $S$ given a function $f:S\to\mathbb{R}$, the notion $\max_S f$ is generally not defined, since you cannot guarantee that $f$ will attain its maximum on $S$). But $\int_M |\nabla f| d\mu = \epsilon$ which can be made arbitrarily small.


Response to edits 1 & 2

  1. My counter example above is still a counterexample for Edit 1.
  2. For a counter example to edit 2: observe that it is false even in one dimension. Let $f$ be an arbitrary non-negative bump function supported on (-1,1). Consider the manifolds $M_n = (-n,n)$. The RHS stays constant, but the LHS can grow to as large as you want.

Remark: What you are writing down is getting closer and closer to Sobolev/Poincare inequalities. For more details on the scaling/dimensional analysis argument, you may want to read Terry's answer here and his linked blog post.

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thanks for directing me towards Sobolev inequalities that is what i wanted indeed. –  timhortons Aug 15 '11 at 20:26
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