The Gaussian Poincare inequality says that if $q:R^n\to R$ is Lipschitz (for simplicity you may additionally assume smooth with compact support), then $Var[f(X)] \le L^2$
for $X\sim N(0,I\_q)$.


Now instead, $X\in R^{n\times d}$ is a matrix with iid entries and $f:R^{n\times d}\to R$ is again $L$-Lipschitz, but with respect to the operator norm $\|X\|\_{op} = \sup\_{y\in R^d: \|y\|=1} \|Xy\|\_2$, i.e.,
\begin{equation}
|f(X) - f(\tilde X)|\le L \|X-\tilde X\|\_{op}.
\tag{1}
\end{equation}
Then $Var[f(X)]\le L^2$ again holds by identifying $R^{n\times d}= R^{nd}$, using the loose upper bound
$\|X-\tilde X\|\_{op}\le \|X-\tilde X\|\_F$ for $\|M\|\_F=(\sum\_{ik}M\_{ik}^2)^{1/2}$ the Frobenius norm, and applying the Gaussian Poincare inequality of the first paragraph for $q=nd$.


The operator norm can be much smaller than the Frobenius norm, and one can hope to get a smaller variance bound if (1) holds.



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> Prove or disprove whether the inequality $Var[f(X)]\le L^2$ can be improved if the Lipschitz condition holds for the operator norm, and by how much for given dimensions $n,d$.
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Lipschitz condition with respect to operator norm of a Gaussian matrix with iid entries. Improved Gaussian Poincare Inequality?
You cannot improve the bound $L^2$ on $Var\,f(X)$ unless an additional condition on $f$ is assumed.


Indeed, let $f(x)\equiv Lx\_{11}$, where $x\_{11}$ is the first diagonal entry of a matrix $x\in R^{n\times d}$. Then $L$ is the Lipschitz constant for $f$ with respect to the operator norm, but $Var\,f(X)=L^2$ if $X\_{11}\sim N(0,1)$.


