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For a differentiable real-valued function on $\mathbb{R}^n$, denoting $\partial_i f$ for the $i$th partial derivative, we can define the functional $$ T_n(f) = \sum_{i=1}^n \frac{1}{1 + \log(\|\partial_i f\|_2/\|\partial_i f\|_1)} \|\partial_i f\|_2^2. $$ Above, $\|\cdot\|_p$ denotes the usual $L^p$ norm under the standard Gaussian law $\gamma_n$ on $\mathbb{R}^n$.

The $L^1-L^2$ inequality implies that there exists a constant $C > 0$ such that for each $n \geq 1$ and any suitably regular $f \colon \mathbb{R}^n \to \mathbb{R}$, it holds that $$ \mathrm{Var}_{\gamma_n}(f) \leq C \, T_n(f), $$ where the variance is taken again with respect to standard Gaussian law on $\mathbb{R}^n$. We take $C$ above to be the smallest such constant: $$ C = \sup_{n \geq 1} \sup_{f} \frac{\mathrm{Var}_{\gamma_n}{f}}{T_n(f)}. $$

By considering linear functionals, it is clear that $C \geq 1$. On the other hand it is known that $C \leq 4$ (see, e.g., Corollary 5 in [1]).

Is the value of $C$ known? (In particular, are there any examples where $\mathrm{Var}(f) > T_n(f)$?)

[1] https://arxiv.org/abs/1105.4533

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  • $\begingroup$ I don't understand the supremum over the dimension $n$ in the main inequality. $\endgroup$ Commented Dec 5 at 23:13
  • $\begingroup$ The constant could in principle depend on $n$, right? I am asking for the smallest that holds uniformly in the dimension. $\endgroup$ Commented Dec 5 at 23:27
  • $\begingroup$ But what happens to the supremum over $f$ as $n$ changes ? I would rather fix $n$ and look for the optimal $C_{n}$. $\endgroup$ Commented Dec 5 at 23:48
  • $\begingroup$ We know that $C_n \in [1, 4]$ for all $n$, so my interest is primarily in determining what the largest it can be is. Of course, if we can even determine $C_n$ then this is even better. It all depends on what level of accuracy you hope for. $\endgroup$ Commented Dec 5 at 23:49
  • $\begingroup$ Remark: I believe that one can select $\lambda > 0$ such that $f_\lambda(x) = x + \lambda (x^2 -1)$ makes it such that $\mathrm{Var}(f_\lambda) > T(f_\lambda)$. $\endgroup$ Commented Dec 5 at 23:57

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