[proofplan]
We argue by contradiction. If the curvature supremum stayed bounded as $t \uparrow T$, then compactness and smoothness give a uniform curvature bound on all of $M \times [0,T)$. Hamilton's extension criterion for compact Ricci flows then extends the solution smoothly past time $T$, contradicting the assumed maximality of the flow. Therefore the curvature supremum must be unbounded along times approaching $T$.
[/proofplan]
[step:Convert finite limsup into a uniform curvature bound near the final time]
Assume, toward a contradiction, that
\begin{align*}
L:=\limsup_{t\uparrow T}\sup_{x\in M}|\operatorname{Rm}_{g(t)}(x)|_{g(t)}<\infty.
\end{align*}
Define the curvature supremum function $K: [0,T) \to [0,\infty)$ by
\begin{align*}
K(t):=\sup_{x\in M}|\operatorname{Rm}_{g(t)}(x)|_{g(t)}.
\end{align*}
By the definition of the one-sided limsup, there exists $\tau\in[0,T)$ such that
\begin{align*}
K(t)\le L+1
\end{align*}
for every $t\in[\tau,T)$.
[guided]
Assume, for contradiction, that the curvature does not blow up in the sense stated. This means the one-sided limsup is finite, so we may define
\begin{align*}
L:=\limsup_{t\uparrow T}\sup_{x\in M}|\operatorname{Rm}_{g(t)}(x)|_{g(t)}<\infty.
\end{align*}
We package the spatial supremum into a single function of time by defining $K: [0,T) \to [0,\infty)$ as
\begin{align*}
K(t):=\sup_{x\in M}|\operatorname{Rm}_{g(t)}(x)|_{g(t)}.
\end{align*}
The definition of $\limsup_{t\uparrow T}K(t)$ says that, for every number strictly larger than $L$, the values of $K(t)$ are eventually bounded by that number as $t$ approaches $T$ from below. Applying this with $L+1$ gives a time $\tau\in[0,T)$ such that
\begin{align*}
K(t)\le L+1
\end{align*}
for every $t\in[\tau,T)$.
This is the precise point where the contradiction hypothesis is used: finite limsup is not merely a bound along one sequence of times; it gives a uniform upper bound for all sufficiently late times.
[/guided]
[/step]
[step:Extend the late-time bound to all earlier times]
Since $g$ is smooth on $M\times[0,\tau]$ and $M$ is compact, define $F: M\times[0,\tau] \to [0,\infty)$ by
\begin{align*}
F(x,t):=|\operatorname{Rm}_{g(t)}(x)|_{g(t)}.
\end{align*}
This function is continuous on the [compact space](/page/Compact%20Space) $M\times[0,\tau]$. Hence there exists $A_0<\infty$ such that
\begin{align*}
F(x,t)\le A_0
\end{align*}
for every $(x,t)\in M\times[0,\tau]$. Define
\begin{align*}
A:=\max\{A_0,L+1\}.
\end{align*}
Combining the early-time bound with the late-time bound gives
\begin{align*}
\sup_{(x,t)\in M\times[0,T)}|\operatorname{Rm}_{g(t)}(x)|_{g(t)}\le A<\infty.
\end{align*}
[/step]
[step:Apply Hamilton's extension criterion and contradict maximality]
We now invoke Hamilton's extension criterion for compact Ricci flows: if a smooth Ricci flow on a compact manifold exists on $[0,T)$ with $T<\infty$ and satisfies
\begin{align*}
\sup_{(x,t)\in M\times[0,T)}|\operatorname{Rm}_{g(t)}(x)|_{g(t)}<\infty,
\end{align*}
then there exists $\varepsilon>0$ and a smooth Ricci flow
\begin{align*}
\tilde g: [0,T+\varepsilon)\to \Gamma(\operatorname{Sym}^2T^*M)
\end{align*}
such that $\tilde g(t)=g(t)$ for every $t\in[0,T)$. The hypotheses are satisfied: $M$ is compact, $T<\infty$, $g$ is a smooth Ricci flow on $[0,T)$, and the previous step established the required uniform curvature bound. Therefore such an extension $\tilde g$ exists.
This contradicts the maximality assumption in the statement. Hence the assumption $L<\infty$ is false, and therefore
\begin{align*}
\limsup_{t \uparrow T}\sup_{x\in M}|\operatorname{Rm}_{g(t)}(x)|_{g(t)}=\infty.
\end{align*}
[/step]
