答案1
像这样?
\documentclass{article}
\usepackage[most]{tcolorbox}
\usepackage{mathtools}
\usepackage{lipsum}
\newtcbtheorem[auto counter]{PeterTheorem}{Theorem}%
{theorem style=plain, enhanced, breakable,
empty,
coltitle=black,
fonttitle=\upshape\scshape,
boxrule=0.0pt,
overlay unbroken={\fill[gray](frame.north west)rectangle ([xshift=1.5mm]frame.south west);},
overlay first={\fill[top color=gray, bottom color=white](frame.north west)rectangle ([xshift=1.5mm]frame.south west);},
overlay last={\fill[top color=white, bottom color=gray](frame.north west)rectangle ([xshift=1.5mm]frame.south west);},
overlay middle={\fill[top color=white, bottom color=white, middle color=gray](frame.north west)rectangle ([xshift=1.5mm]frame.south west);},
}{theo}
\begin{document}
\lipsum[1-4]
\begin{PeterTheorem}{}{theo1-3}%
Consider the n.n.f. Insing model on $Z^d,\ d\geq 1$. There exists acritical inverse temperature $\beta_c=\beta_c(d) \in[0,\infty]$, such that the Ising undergoes a \emph{sharp} ferromagnetic phase transistion at $\beta_c$:
\begin{equation*}
(\sigma_0)^+_{\beta;\Lambda_n}
\begin{cases}\leq e^{-cn} & $for$\ \beta<\beta_c\ $with$\ c=c(\beta)>0\\
\geq \sqrt{1-(\beta_c/\beta)^2} & $for$\ \beta\geq \beta_c \end{cases}
\end{equation*}
uniformly in $n\geq0$, where $\Lambda_n:=Z^d\cap[-n,n]^d$ denotes the box of size $n$ around the origin. Moreover, for\ $d\geq2$, the phase transition is \emph{non-trivial}, meaning that $0<\beta_c<\infty$.
\end{PeterTheorem}
\end{document}