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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.

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