I believe my question is on some very elementary computations (i.e., this is probably answerable even if you don't know what is the subgaussian norm $\| \cdot \|_{\psi_2}$ etc. However, do let me know if you want clarifications on the definitions. I will update the post if needed). Nevertheless, I will post the statement and the proof that I am trying to understand here for context. This is the statement of the proof:
This is the proof that I am trying to understand (I will state my specific question after this picture):
In particular, I am confused on how to check the very last inequality $\| X_{uv} - X_{wz} \|_{\psi_2} \leq C \| Y_{uv} - Y_{wz} \|_2$ for some constant $C$ large enough.
Attempts: Through computation (squaring both sides of the $\| X_{uv} - X_{wz} \|_{\psi_2}$ inequality and expand the quadratic terms on the right hand side), I noticed that it suffices to show $$ \| u - w \|_2 \| v - z \|_2 \operatorname{rad}(S)\operatorname{rad}(T) \leq C \| u - w \|_2^2 \operatorname{rad}(S)^2 + \| v - z \|_2^2 \operatorname{rad}(T)^2 = C\| Y_{uv} - Y_{wz} \|_2^2 $$ for some $C$ large uniformly for all $u,w,v,z$. However, I do not see why this would be true. Perhaps we can do something with the boundedness of the set $S$ and $T$? I am not entirely sure how to proceed from here.
Since $$\| X_{uv} - X_{wz} \|_{\psi_2} \leq \| u - w \|_2 \operatorname{rad}(S) + \| v - z \|_2 \operatorname{rad}(T)$$ and $$\| Y_{uv} - Y_{wz} \|_2^2 = \| u - w \|_2^2 \operatorname{rad}(S)^2 + \| v - z \|_2^2 \operatorname{rad}(T)^2,$$
it suffices to show that $$\| u - w \|_2 \operatorname{rad}(S) + \| v - z \|_2 \operatorname{rad}(T) \lesssim \Big[\| u - w \|_2^2 \operatorname{rad}(S)^2 + \| v - z \|_2^2 \operatorname{rad}(T)^2\Big]^{1/2}.$$
This follows from the standard estimate $(a+b)^2\leq 2a^2 + 2b^2$ applied with $a = \| u - w \|_2 \operatorname{rad}(S)$ and $b=\| v - z \|_2 \operatorname{rad}(T)$.