# Lipschitz condition with respect to operator norm of a Gaussian matrix with iid entries. Improved Gaussian Poincare Inequality?

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.

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 $$n,d$$.

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 $$X_{11}\sim N(0,1)$$.