theorem relative to closed subspaces of Hilbert spaces.

[**(strict separation relative to subspaces in Hilbert spaces).**]{}\[Proseparate1\] Let $L$ be a closed subspace of a Hilbert space $X$, and let $\Omega\subset L$ be a nonempty convex set with $\bar{x}\in L$ and $\bar{x}\not\in\Bar{\Omega}$. Then there exists a nonzero vector $u\in L$ such that $$\sup\big\{\langle u,x\rangle\;\big|\;x\in\Omega\big\}<\langle u,\bar{x}\rangle.$$

[**Proof.**]{} Since $\ox\notin\Bar{\Omega}$, we see that the sets $\{\ox\}$ and $\Omega$ are strictly separated in $X$, which means that there exists a vector $h\in X$ such that $$\label{eqlm1}
\sup\big\{\la h,x\ra\;\big|\;x\in\Omega\big\}<\la h,\ox\ra.$$ It is well known that the space $X$ can be represented as the direct sum $X=L\oplus L^{\bot}$ where $$L^{\bot}:=\big\{w\in X\;\big|\;\la w,x\ra=0\;\text{ for all }\;x\in L\big\}.$$ If $h\in L^{\bot}$, then immediately gives us a contradiction. Thus $h\in X$ can be represented as $h=u+w$ with $0\ne u\in L$ and $w\in L^{\bot}$. This implies that for each $x\in\Omega\subset L$ we have $$\begin{aligned}
\la u,x\ra &=\la u,x\ra+\la w,x\ra=\la h,x\ra\le\sup\big\{\la h,x\ra\;\big|\;x\in\Omega\big\}<\la h,\ox\ra\\
&=\la u+w,\ox\ra=\la u,\ox\ra,
\end{aligned}$$ which shows that $\sup\{\la u,x\ra\;|\;x\in\Omega\}<\la u,\ox\ra$ with $u\ne 0$. $\h$

Now we are ready to present sufficient conditions for quasi-regularity of convex sets.

[**(sufficient conditions for quasi-regularity).**]{}\[qreg\] Let $\Omega\subset X$ be a nonempty convex set. Then it is quasi-regular under each of the following 