Relation between Gaussian width and its squared version

For fixed $g$ note that $\sup\limits_{x,y\in T} \langle g,x-y \rangle = \sup\limits_{x,y\in T} |\langle g,x-y \rangle|$, hence

$$\left(\sup\limits_{x,y\in T} \langle g,x-y \rangle \right)^2=\left(\sup\limits_{x,y\in T} |\langle g,x-y \rangle|\right)^2=\sup\limits_{x,y\in T} |\langle g,x-y \rangle|^2=\sup\limits_{x,y\in T} \langle g,x-y \rangle^2$$

Let $F:g\mapsto \sup\limits_{x,y\in T} \langle g,x-y \rangle$. The previous equalities show that $h(T-T)^2=\mathbb E(F(g)^2)$, and of course $w(T-T)=\mathbb E(F(g))$.
Let us prove that $F$ is $\mathrm{diam}(T)$-Lipschitz: for $g,g'\in \mathbb R^n$, $$\langle g,x-y \rangle = \langle g-g',x-y \rangle + \langle g',x-y \rangle \leq \|g-g'\|\mathrm{diam}(T) + F(g')$$ hence $F(g) - F(g')\leq \|g-g'\|\mathrm{diam}(T)$ and the claim is obtained by symmetry.

Gaussian concentration provides an upper bound on $\mathbb V(F(g))$. Indeed $$\mathbb V(F(g)) = \int_0^\infty P(| F(g)- \mathbb E(F(g))|\geq \sqrt t)\leq 2\int_0^\infty e^{-t/(2 \mathrm{diam}(T)^2)} = 4\mathrm{diam}(T)^2$$

Thus $h(T-T)=\sqrt{\mathbb E(F(g)^2)}\leq \sqrt{w(T-T)^2 + 4\mathrm{diam}(T)^2}\leq w(T-T) + 2\mathrm{diam}(T)$.

Using the Gaussian Poincaré inequality one can get the stronger inequality
$$h(T-T)\leq w(T-T) + \mathrm{diam}(T)$$

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Regarding the other inequalities, $w(T-T)\leq h(T-T)$ follows from Jensen's inequality: $$h(T-T)=\sqrt{\mathbb E\left[\left(\sup\limits_{x,y\in T} |\langle g,x-y \rangle| \right)^2\right]}\geq \mathbb E (\sup\limits_{x,y\in T} |\langle g,x-y \rangle|) = w(T-T)$$ The last inequality $w(T-T)+2\mathrm{diam}(T) \leq Cw(T-T)$ follows from Proposition 7.5.2 of the book: $$w(T-T)+2\mathrm{diam}(T)\leq w(T-T)+ 2\sqrt{2\pi}w(T) = \left(1+\sqrt{2\pi} \right)w(T-T)$$ Using the tighter bound on the variance, the last constant can be improved to $1+\sqrt{\frac \pi 2}$.