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Introduction Let $\mathcal{M}$ be a compact $m$ - dimensional Riemmanian manifold with normalized measure $\mu$ (derived from the metric). It is know that in this setting we have concentration inequalities for Lipschitz functions on $\mathcal{M}$ (see here, page 30):

$\mu( \left(x\in\mathcal{M} |\ |f(x)-\mathbb{E}(f)|\geq \epsilon\right ) \leq 2 \exp(-\frac{\epsilon^2}{2CL^2})$,

where $L$ is the Lipschitz constant of $\mathcal{C}^\infty$ function $f$, $C$ is the constant subject to ineqiality:

$\lambda_1\geq\frac{1}{C}\geq \frac{m}{m-1}R$,

where $\lambda_1$ is first nonzero eigenvalue of Laplace -Beltlami operator on $\mathcal{M}$ ($\Delta f = -div(grad (f))$ ) and $R$ is any lower bound for the Ricci curvature of $\mathcal{M}$.

If follows that when $\lambda_1$ is "large" (thhink of the $n$ dimensional sphere as $n\rightarrow \infty $) then we have effective concentration of measure on $\mathcal{M}$.

Question: Are there any geometric conditions on $\mathcal{M}$ that would imply the lack of effective measure concentration? By lack of effective measure concentration I mean the inequality of the type:

$\mu( \left(x\in\mathcal{M} |\ |f(x)-\mathbb{E}(f)|\geq \epsilon\right ) \geq 2 A(\mathcal{M},L,\epsilon)$,

where $A(\mathcal{M},L,\epsilon)$ is some estimate depending upon the relevant parameters in the problem (obviously$A(\mathcal{M},L,\epsilon)$ would be a decreasing function of $\epsilon$.

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    $\begingroup$ The kind of inequality you want cannot be true for all $L$-Lipschitz functions, even if $L$ is assumed to be the best Lipschitz constant of $f$, could you precise what you mean (in particular the quantifier on $f, \mathcal{M}, \dots$? $\endgroup$ Commented Jun 16, 2012 at 13:59

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