[proofplan] We prove the contrapositive. If the mild solution stays bounded before a finite endpoint, then the fixed-point construction for the mild equation can be restarted with a lifespan bounded below uniformly over all restart times near the endpoint. This uniform lifespan comes from using balls around the linear orbit $s\mapsto S(s)x$, not balls around the initial value $x$, so strong continuity of the semigroup causes no uniform-continuity difficulty on bounded sets. Restarting sufficiently close to $T_{\max}$ then gives a mild solution past $T_{\max}$, contradicting maximality. [/proofplan]

[step:Assume boundedness up to a finite maximal time] Assume, toward a contradiction, that $T_{\max}<\infty$ and \begin{align*} \limsup_{t\uparrow T_{\max}}\|u(t)\|_X<\infty. \end{align*} Then there exist $\delta\in(0,T_{\max})$ and $R_1>0$ such that $\|u(t)\|_X\leq R_1$ for every $t\in[T_{\max}-\delta,T_{\max})$. Since $u:[0,T_{\max}-\delta]\to X$ is continuous on a compact interval, there exists $R_0>0$ such that $\|u(t)\|_X\leq R_0$ for every $t\in[0,T_{\max}-\delta]$. Define \begin{align*} R:=\max\{R_0,R_1\}. \end{align*} Then \begin{align*} \|u(t)\|_X\leq R \end{align*} for every $t\in[0,T_{\max})$. [/step]

[step:Construct a uniform local solution for every restart value in the bounded orbit]By the local [boundedness theorem](/theorems/181) for strongly continuous semigroups, namely the result that every strongly continuous semigroup is uniformly bounded on compact time intervals, $(S(t))_{t\geq 0}$ is locally bounded. Define \begin{align*} M:=\sup_{0\leq s\leq 1}\|S(s)\|_{\mathcal{L}(X)}<\infty. \end{align*} Define the radius \begin{align*} \rho:=MR+1. \end{align*} Define the closed ball in $X$ of radius $\rho$ centered at $0$ by \begin{align*} \overline{B}_X(0,\rho):=\{y\in X:\|y\|_X\leq \rho\}. \end{align*} Because $F$ is Lipschitz on bounded subsets, there exists $L_\rho\geq 0$ such that $F$ is $L_\rho$-Lipschitz on $\overline{B}_X(0,\rho)$. Also define \begin{align*} B_\rho:=\|F(0)\|_X+L_\rho\rho. \end{align*} Then $\|F(y)\|_X\leq B_\rho$ whenever $\|y\|_X\leq \rho$. Choose \begin{align*} \tau:=\min\left\{1,\frac{1}{M(B_\rho+1)},\frac{1}{2M(L_\rho+1)}\right\}. \end{align*} For each $x\in X$ with $\|x\|_X\leq R$, let $Y_\tau:=C([0,\tau];X)$ with the norm \begin{align*} \|v\|_{Y_\tau}:=\sup_{0\leq s\leq\tau}\|v(s)\|_X. \end{align*} Define the closed subset \begin{align*} K_x:=\{v\in Y_\tau:\sup_{0\leq s\leq\tau}\|v(s)-S(s)x\|_X\leq 1\}. \end{align*} For $v\in K_x$, define $\Phi_x(v):[0,\tau]\to X$ by \begin{align*} \Phi_x(v)(s):=S(s)x+\int_0^s S(s-\sigma)F(v(\sigma))\,d\mathcal{L}^1(\sigma). \end{align*} This is a well-defined element of $Y_\tau$. Indeed, $v:[0,\tau]\to X$ is continuous, $F$ is continuous on the bounded set containing $v([0,\tau])$, and therefore the map $g_v:[0,\tau]\to X$ defined by $g_v(\sigma):=F(v(\sigma))$ is continuous. Define the integral term $I_v:[0,\tau]\to X$ by \begin{align*} I_v(s):=\int_0^s S(s-\sigma)g_v(\sigma)\,d\mathcal{L}^1(\sigma). \end{align*} For $0\leq s\leq s'\leq\tau$, the difference splits as the sum of \begin{align*} \int_0^s [S(s'-\sigma)-S(s-\sigma)]g_v(\sigma)\,d\mathcal{L}^1(\sigma) \end{align*} and \begin{align*} \int_s^{s'} S(s'-\sigma)g_v(\sigma)\,d\mathcal{L}^1(\sigma). \end{align*} The first term tends to $0$ as $s'\to s$ by strong continuity of the semigroup and dominated convergence, using the domination $2M B_\rho$ on $[0,s]$. The second term has norm at most $M B_\rho\mathcal{L}^1([s,s'])$, which tends to $0$ as $s'\to s$. The case $s'\leq s$ is identical with the roles of $s$ and $s'$ reversed. Hence $I_v$ is continuous, so $\Phi_x(v)\in C([0,\tau];X)=Y_\tau$. If $v\in K_x$ and $0\leq s\leq\tau$, then \begin{align*} \|v(s)\|_X\leq \|S(s)x\|_X+\|v(s)-S(s)x\|_X\leq MR+1=\rho. \end{align*} Thus $\|F(v(s))\|_X\leq B_\rho$. Hence, for every $s\in[0,\tau]$, \begin{align*} \|\Phi_x(v)(s)-S(s)x\|_X\leq \int_0^s \|S(s-\sigma)\|_{\mathcal{L}(X)}\|F(v(\sigma))\|_X\,d\mathcal{L}^1(\sigma)\leq M\tau B_\rho\leq 1. \end{align*} Therefore $\Phi_x(K_x)\subset K_x$. If $v,w\in K_x$, then for every $s\in[0,\tau]$, \begin{align*} \|\Phi_x(v)(s)-\Phi_x(w)(s)\|_X\leq \int_0^s M L_\rho\|v(\sigma)-w(\sigma)\|_X\,d\mathcal{L}^1(\sigma)\leq M L_\rho\tau\|v-w\|_{Y_\tau}\leq \frac{1}{2}\|v-w\|_{Y_\tau}. \end{align*} Thus $\Phi_x:K_x\to K_x$ is a contraction with contraction constant at most $1/2$. [claim:Each restart value has a unique fixed point on the uniform time interval] For every $x\in X$ with $\|x\|_X\leq R$, there exists a unique $v_x\in K_x$ such that $\Phi_x(v_x)=v_x$. [/claim] [proof] Choose any $v_{x,0}\in K_x$, for instance $v_{x,0}:[0,\tau]\to X$ defined by $v_{x,0}(s):=S(s)x$. Define a sequence $(v_{x,k})_{k\geq 0}$ in $K_x$ recursively by \begin{align*} v_{x,k+1}:=\Phi_x(v_{x,k}). \end{align*} Since $\Phi_x$ is a contraction with constant at most $1/2$, \begin{align*} \|v_{x,k+1}-v_{x,k}\|_{Y_\tau}\leq 2^{-k}\|v_{x,1}-v_{x,0}\|_{Y_\tau} \end{align*} for every $k\geq 0$. Therefore the series of successive differences is summable in $Y_\tau$, and $(v_{x,k})_{k\geq 0}$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in the [Banach space](/page/Banach%20Space) $Y_\tau$. Since $K_x$ is closed in $Y_\tau$, there exists $v_x\in K_x$ such that $v_{x,k}\to v_x$ in $Y_\tau$. The contraction estimate gives the explicit continuity bound \begin{align*} \|\Phi_x(v_{x,k})-\Phi_x(v_x)\|_{Y_\tau}\leq \frac{1}{2}\|v_{x,k}-v_x\|_{Y_\tau}. \end{align*} Thus $\Phi_x(v_{x,k})\to \Phi_x(v_x)$ in $Y_\tau$, and passing to the limit in $v_{x,k+1}=\Phi_x(v_{x,k})$ gives $v_x=\Phi_x(v_x)$. If $z_x\in K_x$ is another fixed point, then \begin{align*} \|v_x-z_x\|_{Y_\tau}=\|\Phi_x(v_x)-\Phi_x(z_x)\|_{Y_\tau}\leq \frac{1}{2}\|v_x-z_x\|_{Y_\tau}. \end{align*} Subtracting the right-hand side gives $\|v_x-z_x\|_{Y_\tau}=0$, so $v_x=z_x$. [/proof] The fixed point $v_x$ satisfies $v_x(0)=x$ and \begin{align*} v_x(s)=S(s)x+\int_0^s S(s-\sigma)F(v_x(\sigma))\,d\mathcal{L}^1(\sigma) \end{align*} for every $s\in[0,\tau]$. Thus $v_x$ is a mild solution on $[0,\tau]$ with initial value $x$.[/step]

[guided]The goal of this step is to obtain a local existence time that does not depend on the particular restart value $x$, as long as $\|x\|_X\leq R$. The delicate point is that for a general strongly continuous semigroup, $S(s)x\to x$ as $s\downarrow 0$ holds for each fixed $x$, but not uniformly over the whole bounded ball $\{x:\|x\|_X\leq R\}$ in operator norm. For this reason we build the fixed-point ball around the linear orbit $s\mapsto S(s)x$, not around the constant path $s\mapsto x$. Strong continuity of the semigroup implies local boundedness, so \begin{align*} M:=\sup_{0\leq s\leq 1}\|S(s)\|_{\mathcal{L}(X)}<\infty. \end{align*} We set \begin{align*} \rho:=MR+1. \end{align*} Define the closed ball in $X$ of radius $\rho$ centered at $0$ by \begin{align*} \overline{B}_X(0,\rho):=\{y\in X:\|y\|_X\leq \rho\}. \end{align*} This is the ball in $X$ that will contain all paths in the fixed-point construction. Since $F$ is Lipschitz on bounded subsets of $X$, there exists $L_\rho\geq 0$ such that \begin{align*} \|F(y)-F(z)\|_X\leq L_\rho\|y-z\|_X \end{align*} whenever $\|y\|_X\leq \rho$ and $\|z\|_X\leq \rho$. In particular, taking $z=0$ gives \begin{align*} \|F(y)\|_X\leq \|F(0)\|_X+L_\rho\|y\|_X\leq \|F(0)\|_X+L_\rho\rho. \end{align*} Define \begin{align*} B_\rho:=\|F(0)\|_X+L_\rho\rho. \end{align*} Thus $\|F(y)\|_X\leq B_\rho$ for all $\|y\|_X\leq \rho$. Now choose \begin{align*} \tau:=\min\left\{1,\frac{1}{M(B_\rho+1)},\frac{1}{2M(L_\rho+1)}\right\}. \end{align*} This choice enforces two estimates at once: $M\tau B_\rho\leq 1$, which keeps the fixed-point map inside the path ball, and $M L_\rho\tau\leq 1/2$, which makes the fixed-point map a contraction. Fix $x\in X$ with $\|x\|_X\leq R$. Let $Y_\tau=C([0,\tau];X)$ with norm \begin{align*} \|v\|_{Y_\tau}:=\sup_{0\leq s\leq\tau}\|v(s)\|_X. \end{align*} Define \begin{align*} K_x:=\{v\in Y_\tau:\sup_{0\leq s\leq\tau}\|v(s)-S(s)x\|_X\leq 1\}. \end{align*} A path in $K_x$ remains within distance $1$ of the linear orbit. Since $\|S(s)x\|_X\leq MR$ for $0\leq s\leq\tau\leq 1$, every $v\in K_x$ satisfies \begin{align*} \|v(s)\|_X\leq \|S(s)x\|_X+\|v(s)-S(s)x\|_X\leq MR+1=\rho. \end{align*} This is exactly why $\rho$ was chosen as $MR+1$: it gives a uniform ball on which both the Lipschitz constant and the bound for $F$ are valid. For $v\in K_x$, define the mild-solution operator $\Phi_x(v):[0,\tau]\to X$ by \begin{align*} \Phi_x(v)(s):=S(s)x+\int_0^s S(s-\sigma)F(v(\sigma))\,d\mathcal{L}^1(\sigma). \end{align*} This formula defines an element of $Y_\tau$. To check the only non-immediate point, define $g_v:[0,\tau]\to X$ by $g_v(\sigma):=F(v(\sigma))$ and define $I_v:[0,\tau]\to X$ by \begin{align*} I_v(s):=\int_0^s S(s-\sigma)g_v(\sigma)\,d\mathcal{L}^1(\sigma). \end{align*} The map $g_v$ is continuous because $v$ is continuous and $F$ is continuous on the bounded ball containing $v([0,\tau])$. If $0\leq s\leq s'\leq\tau$, then $I_v(s')-I_v(s)$ is the sum of \begin{align*} \int_0^s [S(s'-\sigma)-S(s-\sigma)]g_v(\sigma)\,d\mathcal{L}^1(\sigma) \end{align*} and \begin{align*} \int_s^{s'} S(s'-\sigma)g_v(\sigma)\,d\mathcal{L}^1(\sigma). \end{align*} For the first term, strong continuity of $S$ gives pointwise convergence of the integrand to $0$ as $s'\to s$, and the uniform bound $\|S(r)\|_{\mathcal{L}(X)}\leq M$ on $0\leq r\leq1$ gives domination by $2M B_\rho$. Dominated convergence therefore sends the first term to $0$. The second term has norm at most $M B_\rho\mathcal{L}^1([s,s'])$, so it also tends to $0$. The case $s'\leq s$ is handled by reversing the roles of $s$ and $s'$. Thus $I_v$ is continuous, and hence $\Phi_x(v)\in Y_\tau$. We first show that $\Phi_x$ maps $K_x$ into itself. Since $\|v(\sigma)\|_X\leq\rho$, we have $\|F(v(\sigma))\|_X\leq B_\rho$. Therefore, for every $s\in[0,\tau]$, \begin{align*} \|\Phi_x(v)(s)-S(s)x\|_X\leq \int_0^s \|S(s-\sigma)\|_{\mathcal{L}(X)}\|F(v(\sigma))\|_X\,d\mathcal{L}^1(\sigma)\leq M\tau B_\rho\leq 1. \end{align*} Thus $\Phi_x(v)\in K_x$. Next we prove contraction. If $v,w\in K_x$, then both paths take values in $\overline{B}_X(0,\rho)$, so the $L_\rho$-Lipschitz estimate applies pointwise: \begin{align*} \|F(v(\sigma))-F(w(\sigma))\|_X\leq L_\rho\|v(\sigma)-w(\sigma)\|_X. \end{align*} Using this inside the integral gives, for every $s\in[0,\tau]$, \begin{align*} \|\Phi_x(v)(s)-\Phi_x(w)(s)\|_X\leq \int_0^s M L_\rho\|v(\sigma)-w(\sigma)\|_X\,d\mathcal{L}^1(\sigma)\leq M L_\rho\tau\|v-w\|_{Y_\tau}\leq \frac{1}{2}\|v-w\|_{Y_\tau}. \end{align*} Taking the supremum over $s\in[0,\tau]$ shows that $\Phi_x:K_x\to K_x$ is a contraction. For completeness, we spell out the Banach fixed-point argument rather than only citing the [contraction mapping principle](/page/Contraction%20Mapping%20Principle). Start with $v_{x,0}:[0,\tau]\to X$ given by $v_{x,0}(s):=S(s)x$, and define $v_{x,k+1}:=\Phi_x(v_{x,k})$. The contraction estimate implies \begin{align*} \|v_{x,k+1}-v_{x,k}\|_{Y_\tau}\leq 2^{-k}\|v_{x,1}-v_{x,0}\|_{Y_\tau}. \end{align*} Hence $(v_{x,k})_{k\geq 0}$ is Cauchy in $Y_\tau$. Since $Y_\tau$ is complete and $K_x$ is closed, the sequence converges to some $v_x\in K_x$. The explicit estimate \begin{align*} \|\Phi_x(v_{x,k})-\Phi_x(v_x)\|_{Y_\tau}\leq \frac{1}{2}\|v_{x,k}-v_x\|_{Y_\tau} \end{align*} shows that $\Phi_x(v_{x,k})\to\Phi_x(v_x)$ in $Y_\tau$, so passing to the limit in $v_{x,k+1}=\Phi_x(v_{x,k})$ gives $v_x=\Phi_x(v_x)$. If $z_x\in K_x$ is another fixed point, then \begin{align*} \|v_x-z_x\|_{Y_\tau}\leq \frac{1}{2}\|v_x-z_x\|_{Y_\tau}, \end{align*} so $v_x=z_x$. Thus for every restart value $x$ with $\|x\|_X\leq R$, there is a unique mild solution on the same interval $[0,\tau]$.[/guided]

[step:Restart the solution close enough to the finite endpoint] Choose $t_0\in[0,T_{\max})$ such that \begin{align*} T_{\max}-\tau<t_0<T_{\max}. \end{align*} Set $x_0:=u(t_0)$. By the boundedness established above, $\|x_0\|_X\leq R$. Let $v_{x_0}:[0,\tau]\to X$ be the fixed point constructed in the previous step. Define $\widetilde{u}:[0,t_0+\tau]\to X$ by \begin{align*} \widetilde{u}(t):=u(t) \text{ for } 0\leq t\leq t_0, \qquad \widetilde{u}(t):=v_{x_0}(t-t_0) \text{ for } t_0\leq t\leq t_0+\tau. \end{align*} The two definitions agree at $t=t_0$, because $v_{x_0}(0)=x_0=u(t_0)$. It remains to check that this concatenated path is a mild solution. For $t\leq t_0$, this is the original mild equation for $u$. For $t\in[t_0,t_0+\tau]$, write $s:=t-t_0$. Since $v_{x_0}$ solves the restarted mild equation, \begin{align*} \widetilde{u}(t)=v_{x_0}(s)=S(s)u(t_0)+\int_0^s S(s-\sigma)F(v_{x_0}(\sigma))\,d\mathcal{L}^1(\sigma). \end{align*} Using the original mild equation at time $t_0$ and the semigroup identity $S(s)S(t_0-r)=S(t-r)$, we obtain \begin{align*} S(s)u(t_0)=S(t)u_0+\int_0^{t_0}S(t-r)F(u(r))\,d\mathcal{L}^1(r). \end{align*} In the restarted integral, substitute $r=t_0+\sigma$. Since $d\mathcal{L}^1(r)=d\mathcal{L}^1(\sigma)$ and $\sigma\in[0,s]$ corresponds exactly to $r\in[t_0,t]$, we get \begin{align*} \int_0^s S(s-\sigma)F(v_{x_0}(\sigma))\,d\mathcal{L}^1(\sigma)=\int_{t_0}^t S(t-r)F(\widetilde{u}(r))\,d\mathcal{L}^1(r). \end{align*} Combining the last two identities gives \begin{align*} \widetilde{u}(t)=S(t)u_0+\int_0^t S(t-r)F(\widetilde{u}(r))\,d\mathcal{L}^1(r). \end{align*} Thus $\widetilde{u}$ is a mild solution on $[0,t_0+\tau]$. [/step]

[step:Contradict maximality and conclude blow-up] By the choice of $t_0$, \begin{align*} t_0+\tau>T_{\max}. \end{align*} The function $\widetilde{u}:[0,t_0+\tau]\to X$ is a mild solution with initial value $u_0$ and agrees with the original solution $u$ on $[0,t_0]$. It remains to prove agreement on the overlap $[t_0,T_{\max})$. Let $a\in(0,T_{\max}-t_0)$, and define $q_a:[0,a]\to X$ by $q_a(s):=u(t_0+s)$. The mild equation for $u$ on $[0,T_{\max})$, the semigroup identity, and the substitution $r=t_0+\sigma$ show that \begin{align*} q_a(s)=S(s)x_0+\int_0^s S(s-\sigma)F(q_a(\sigma))\,d\mathcal{L}^1(\sigma) \end{align*} for every $s\in[0,a]$. Also $\|q_a(s)\|_X\leq R\leq\rho$ for $0\leq s\leq a$, while $\|v_{x_0}(s)\|_X\leq\rho$ for $0\leq s\leq a$ because $v_{x_0}\in K_{x_0}$. Subtracting the restarted mild equations and using the $L_\rho$-Lipschitz estimate for $F$ on $\overline{B}_X(0,\rho)$ gives \begin{align*} \sup_{0\leq s\leq a}\|q_a(s)-v_{x_0}(s)\|_X\leq M L_\rho a\sup_{0\leq s\leq a}\|q_a(s)-v_{x_0}(s)\|_X. \end{align*} Since $a<T_{\max}-t_0<\tau$ and $M L_\rho\tau\leq 1/2$, the coefficient on the right is strictly less than $1$. Hence the supremum is $0$, so $u(t_0+s)=v_{x_0}(s)$ for every $s\in[0,a]$. Because $a<T_{\max}-t_0$ was arbitrary, $\widetilde{u}=u$ on all of $[t_0,T_{\max})$. Thus $\widetilde{u}$ is a genuine extension of $u$ beyond $T_{\max}$. This contradicts the maximality of $u:[0,T_{\max})\to X$. Therefore the assumption \begin{align*} \limsup_{t\uparrow T_{\max}}\|u(t)\|_X<\infty \end{align*} is false. Consequently, if $T_{\max}<\infty$, then \begin{align*} \limsup_{t\uparrow T_{\max}}\|u(t)\|_X=\infty. \end{align*} [/step]