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For a Gaussian random vector $X\sim N(m,\Sigma )$ and an $L$-Lipschitz function $f$ in the Euclidean norm, put $v=L^2\Vert \Sigma {\Vert }_{\mathrm{op}}$. If $v>0$, the Gaussian concentration inequality givesand thereforeWrite $X=m+{\Sigma }^{1/2}G$ with a standard Gaussian random vector $G$. The composed function has Lipschitz constant at most $L\Vert \Sigma {\Vert }_{\mathrm{op}}^{1/2}$. The Gaussian logarithmic Sobolev inequality and Herbst argument give the exponential bound, and the Chernoff bound supplies the tails.^[ref]

For example, $f(x)=\Vert x\Vert$ has $L=1$, so its deviations depend on the largest covariance eigenvalue rather than directly on the number of coordinates. The center $\mathbb{E}\Vert X\Vert$ can still grow with dimension. If $v=0$, $f(X)$ is constant almost surely and every positive deviation has probability zero.

For independent and identically distributed random variables $X,Y$ with a centered multivariate normal distribution, a continuously differentiable function $f$ with bounded gradient, and a convex function $\Psi$ for which the expressions are integrable, the displayed inequality holds. First apply Jensen inequality conditionally to $f(X)-f(Y)$. For the Gaussian rotation of independent copies, the chain rule gives $f(X)-f(Y)={\int }_0^{\pi /2}⟨\nabla f(U_{\theta }),V_{\theta }⟩d\theta$. A second application of Jensen inequality to the uniform measure on $[0,\pi /2]$, followed by the Gaussian rotation of independent copies, proves the inequality. The useful choice $\Psi (v)=e^{\lambda |v|}$ yields a Gaussian concentration inequality through an exponential moment of an absolute standard normal variable.

Let $X_1,\dots ,X_n\in {\mathbb{R}}^p$ be independent centered Gaussian random vectors with common nonzero covariance matrix $\Sigma$, and put $\hat{\Sigma }=n^{-1}{\sum }_i{X}_i{X}_i^T$. There is a universal constant $C>0$ such that for $t\ge 1$, with probability at least $1-e^{-t}$,where $r(\Sigma )=\mathrm{tr}(\Sigma )/\Vert \Sigma {\Vert }_{\mathrm{op}}$ is the effective rank of a covariance matrix.

The bound combines an expected operator norm error with Gaussian concentration inequality. In Euclidean space, the Gaussian effective-dimension parameter in the primary result is comparable to this trace ratio; replacing maxima by sums gives the displayed version.^[ref]

For $\Sigma =I_p$, the covariance effective rank is $p$, and relative error is controlled by $\sqrt{(p+t)/n}+(p+t)/n$. covariance concentrated in fewer directions can have much smaller covariance effective rank. The estimator here is the uncentered empirical second moment for a known zero mean; an unknown mean and empirical centering require the corresponding formulation. The bound remains meaningful when $\Sigma$ is singular but nonzero, since it does not invert $\Sigma$.

The Brownian reflection principle and Brownian scaling givewhere $Z$ has the standard normal distribution. Combine part (c) with the allowed Gaussian tail bound to obtainThis is the Brownian martingale proof of Gaussian concentration. It proves the intended Gaussian concentration inequality. The denominator printed in the initial bound is $2K$, whereas the gradient hypothesis and part (c) give $2K^2$. The stronger printed bound is false in general. For a concrete counterexample, take $d=1$, $K=2$, and $f(x)=2x$. At $r=10$,where the first strict inequality follows by integrating the decreasing normal density over $[5,6]$. The rightmost term is the printed bound. Thus the square on $K$ is a necessary source correction, rather than a change of proof technique.

With the usual variance-proxy convention for a sub-Gaussian random vector, every column inner product is a sub-Gaussian random variable with proxy at most ${\tau }^2$. The two-sided Chernoff bound and union bound implywith probability at least $1-\delta$. No independence between these inner products is required.

The printed coefficient-one bound is false under this convention. Take $k=m=1$, $A=[1]$, and $g=\pm \tau$ with equal probability. This Rademacher random variable has the stipulated variance proxy, but for $\delta =0.99$ the printed threshold is less than $\tau$, whereas $|g|=\tau$ surely. Even a general sub-Gaussian assumption does not imply Gaussian concentration inequality for arbitrary Lipschitz functions. Under the stronger convention $\mathbb{E}{e}^{⟨a,g⟩}\le e^{{\tau }^2\Vert a{\Vert }_2^2/4}$, the same union bound proves the printed constants. That convention, however, differs from the binomial variance proxy used in 1(b).

Let $f(x)=\Vert x{\Vert }_{\infty }$. Away from ties and zero coordinates, its gradient is $\nabla f(x)=\pm e_i$ for a maximizing coordinate $i$. Conditional on $X$, the random variable $⟨\nabla f(X),Y⟩$ is centered Gaussian with variance at most one. The supplied Gaussian concentration inequality therefore givesA Chernoff bound, optimized at $\lambda =4u/{\pi }^2$, yieldsIt remains to bound the mean. For $t>0$, Jensen's inequality and the Gaussian moment-generating function giveTaking $t=\sqrt{2\log (2d)}$ gives $\mathbb{E}\Vert X{\Vert }_{\infty }\le \sqrt{2\log (2d)}$. Consequently