[proofplan] We solve the nonlinear evolution equation directly in a weighted space of globally defined continuous functions. The exponential decay of the semigroup controls the linear term, while the quadratic Lipschitz estimate on $N$ makes the Duhamel operator a strict contraction on a sufficiently small weighted ball. The fixed point gives a global mild solution with exponential decay, and uniqueness on finite time intervals shows that every local mild solution with the same initial datum agrees with it and therefore extends globally. [/proofplan]

[step:Choose the weighted space and the smallness scale] Let $(T(t))_{t\geq 0}$ denote the strongly continuous semigroup on $X$ generated by $L$. Let $M\geq 1$ and $\omega>0$ be constants witnessing exponential stability of the semigroup, so that \begin{align*} \|T(t)\|_{\mathcal{L}(X)}\leq Me^{-\omega t} \end{align*} for every $t\geq 0$. Fix $\omega_0\in(0,\omega)$, and let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,\infty)$ and its subintervals. Define \begin{align*} a:=\omega-\omega_0>0. \end{align*} Let $E$ be the [Banach space](/page/Banach%20Space) \begin{align*} E:=\left\{u\in C([0,\infty);X):\|u\|_E:=\sup_{t\geq 0} e^{\omega_0 t}\|u(t)\|_X<\infty\right\}. \end{align*} Here $C_b([0,\infty);X)$ denotes the Banach space of bounded continuous maps $f:[0,\infty)\to X$ with norm $\|f\|_{C_b}:=\sup_{t\geq 0}\|f(t)\|_X$. Completeness follows because the map $u\mapsto (t\mapsto e^{\omega_0t}u(t))$ is an isometric isomorphism from $E$ onto $C_b([0,\infty);X)$. If $C=0$, replace $C$ by $1$ in the estimates below; the hypothesis with constant $0$ implies the same hypothesis with constant $1$ on $B_X(0,r)$. Thus we may fix a constant, still denoted $C$, with $C>0$. Choose \begin{align*} \rho:=\min\left\{\frac{r}{2},\frac{a}{4MC}\right\}>0, \end{align*} and define \begin{align*} \delta:=\frac{\rho}{2M}. \end{align*} For a fixed $v_0\in X$ with $\|v_0\|_X<\delta$, let \begin{align*} \mathcal{B}_\rho:=\{u\in E:\|u\|_E\leq \rho\}. \end{align*} If $u\in\mathcal{B}_\rho$, then $\|u(t)\|_X\leq \rho\leq r/2<r$ for every $t\geq 0$, so $u(t)\in B_X(0,r)$ and $N(u(t))$ is defined for every $t\geq 0$. [/step]

[step:Show that the Duhamel operator maps the weighted ball into itself]Define the Duhamel operator $\Phi:\mathcal{B}_\rho\to E$ by \begin{align*} (\Phi u)(t):=T(t)v_0+\int_0^t T(t-s)N(u(s))\,d\mathcal{L}^1(s). \end{align*} The integrand $s\mapsto T(t-s)N(u(s))$ is continuous from $[0,t]$ to $X$, because $T$ is strongly continuous, $u$ is continuous, and $N$ is continuous on $B_X(0,r)$ by its Lipschitz estimate. Moreover, the map \begin{align*} t\mapsto \int_0^t T(t-s)N(u(s))\,d\mathcal{L}^1(s) \end{align*} is continuous from $[0,\infty)$ to $X$. To verify this, fix $T_\ast>0$ and set $F:[0,T_\ast]\to X$ by $F(s):=N(u(s))$. The map $F$ is continuous, hence uniformly continuous and bounded on $[0,T_\ast]$. For $0\leq t\leq T_\ast$ and $h$ with $t+h\in[0,T_\ast]$, split the difference of the two convolutions into the integral over the common interval $[0,t]$ and the remaining interval between $t$ and $t+h$. Let $I_h:=[0,t]\cap[0,t+h]=[0,\min\{t,t+h\}]$ denote the common interval on which both $T(t-s)$ and $T(t+h-s)$ are defined. The remaining interval, namely the symmetric difference of $[0,t]$ and $[0,t+h]$, has $\mathcal{L}^1$-measure at most $|h|$ and contributes norm at most $M e^{\omega T_\ast}\|F\|_{C([0,T_\ast];X)}|h|$. On $I_h$, strong continuity of $T$ gives pointwise convergence of $(T(t+h-s)-T(t-s))F(s)$ to $0$ as $h\to 0$, and the same bound $2M e^{\omega T_\ast}\|F\|_{C([0,T_\ast];X)}$ is integrable on $[0,T_\ast]$ with respect to $\mathcal{L}^1$. The [dominated convergence theorem](/theorems/4) for Bochner integrals therefore gives continuity on $[0,T_\ast]$, and since $T_\ast$ was arbitrary, on $[0,\infty)$. Hence $\Phi u\in C([0,\infty);X)$ once its weighted norm is finite. Since $N(0)=0$, the Lipschitz hypothesis with $w=0$ gives \begin{align*} \|N(z)\|_X\leq C\|z\|_X^2 \end{align*} for every $z\in B_X(0,r)$. Hence, for $u\in\mathcal{B}_\rho$ and $t\geq 0$, \begin{align*} e^{\omega_0t}\|(\Phi u)(t)\|_X\leq M\|v_0\|_X+MC\|u\|_E^2\int_0^t e^{-(\omega-\omega_0)(t-s)}e^{-\omega_0s}\,d\mathcal{L}^1(s). \end{align*} Since $e^{-\omega_0s}\leq 1$ for $s\geq 0$, we have \begin{align*} \int_0^t e^{-a(t-s)}e^{-\omega_0s}\,d\mathcal{L}^1(s)\leq \int_0^t e^{-a(t-s)}\,d\mathcal{L}^1(s)\leq \frac{1}{a}. \end{align*} Therefore \begin{align*} \|\Phi u\|_E\leq M\|v_0\|_X+\frac{MC}{a}\rho^2. \end{align*} By \begin{align*} \|v_0\|_X<\frac{\rho}{2M} \end{align*} and \begin{align*} \rho\leq \frac{a}{4MC}, \end{align*} this gives \begin{align*} \|\Phi u\|_E\leq \frac{\rho}{2}+\frac{\rho}{4}<\rho. \end{align*} Thus $\Phi$ maps $\mathcal{B}_\rho$ into itself.[/step]

[guided]The point of the weighted norm is that decay is built into the space. If $u\in\mathcal{B}_\rho$, then the estimate $\|u\|_E\leq\rho$ means \begin{align*} \|u(t)\|_X\leq \rho e^{-\omega_0t} \end{align*} for every $t\geq 0$. In particular $\|u(t)\|_X\leq\rho\leq r/2<r$, so $u(t)\in B_X(0,r)$ and the nonlinear expression $N(u(t))$ is legitimate for all times. We define the candidate solution operator by the mild formula: \begin{align*} (\Phi u)(t):=T(t)v_0+\int_0^t T(t-s)N(u(s))\,d\mathcal{L}^1(s). \end{align*} The integral is a Bochner integral over $[0,t]$ with respect to one-dimensional Lebesgue measure. For fixed $t\geq 0$, the integrand $s\mapsto T(t-s)N(u(s))$ is continuous from $[0,t]$ to $X$: the map $u$ is continuous, the Lipschitz estimate makes $N$ continuous on $B_X(0,r)$, and the semigroup is strongly continuous. We still have to prove that $\Phi u$ is a [continuous function](/page/Continuous%20Function) of $t$, not only that each fixed-time integral is well-defined. Fix $T_\ast>0$ and define $F:[0,T_\ast]\to X$ by $F(s):=N(u(s))$. Since $u$ is continuous and $N$ is continuous on $B_X(0,r)$, the map $F$ is continuous. Hence $F$ is uniformly continuous and bounded on $[0,T_\ast]$; denote its supremum norm by \begin{align*} \|F\|_{C([0,T_\ast];X)}:=\sup_{0\leq s\leq T_\ast}\|F(s)\|_X. \end{align*} For $0\leq t\leq T_\ast$ and $h$ with $t+h\in[0,T_\ast]$, compare the two convolution terms at times $t$ and $t+h$. Let \begin{align*} I_h:=[0,t]\cap[0,t+h]=[0,\min\{t,t+h\}] \end{align*} denote the common part of the two integration intervals. The remaining part is the symmetric difference of $[0,t]$ and $[0,t+h]$, whose $\mathcal{L}^1$-measure is at most $|h|$. On that remaining part, the semigroup bound gives \begin{align*} \|T(\tau)F(s)\|_X\leq Me^{\omega T_\ast}\|F\|_{C([0,T_\ast];X)} \end{align*} for the relevant nonnegative times $\tau\leq T_\ast$, so the remaining interval contributes at most \begin{align*} Me^{\omega T_\ast}\|F\|_{C([0,T_\ast];X)}|h|. \end{align*} On the common interval $I_h$, the difference of the integrands is \begin{align*} (T(t+h-s)-T(t-s))F(s). \end{align*} For each fixed $s\in I_h$, strong continuity of the semigroup gives convergence of this expression to $0$ as $h\to 0$. Moreover, for all sufficiently small $h$, its norm is bounded by \begin{align*} 2Me^{\omega T_\ast}\|F\|_{C([0,T_\ast];X)}, \end{align*} which is integrable on $[0,T_\ast]$ with respect to $\mathcal{L}^1$. The dominated convergence theorem for Bochner integrals therefore sends the integral over $I_h$ to $0$, while the estimate on the remaining interval also tends to $0$. Thus \begin{align*} t\mapsto \int_0^t T(t-s)N(u(s))\,d\mathcal{L}^1(s) \end{align*} is continuous on $[0,T_\ast]$. Since $T_\ast>0$ was arbitrary and $t\mapsto T(t)v_0$ is continuous by strong continuity, we have $\Phi u\in C([0,\infty);X)$ once the weighted norm is finite. We now estimate the weighted norm. Since $N(0)=0$, applying the Lipschitz hypothesis with $w=0$ gives \begin{align*} \|N(z)\|_X=\|N(z)-N(0)\|_X\leq C(\|z\|_X+\|0\|_X)\|z-0\|_X=C\|z\|_X^2 \end{align*} for every $z\in B_X(0,r)$. For $z=u(s)$ this becomes \begin{align*} \|N(u(s))\|_X\leq C\|u(s)\|_X^2\leq C\|u\|_E^2e^{-2\omega_0s}. \end{align*} Using the semigroup decay estimate, we obtain \begin{align*} e^{\omega_0t}\|(\Phi u)(t)\|_X\leq M\|v_0\|_X+MC\|u\|_E^2\int_0^t e^{\omega_0t}e^{-\omega(t-s)}e^{-2\omega_0s}\,d\mathcal{L}^1(s). \end{align*} The exponential factor simplifies as \begin{align*} e^{\omega_0t}e^{-\omega(t-s)}e^{-2\omega_0s}=e^{-(\omega-\omega_0)(t-s)}e^{-\omega_0s}. \end{align*} Since $a=\omega-\omega_0>0$ and $e^{-\omega_0s}\leq 1$, the integral is bounded by \begin{align*} \int_0^t e^{-a(t-s)}e^{-\omega_0s}\,d\mathcal{L}^1(s)\leq \int_0^t e^{-a(t-s)}\,d\mathcal{L}^1(s)\leq \frac{1}{a}. \end{align*} Thus \begin{align*} \|\Phi u\|_E\leq M\|v_0\|_X+\frac{MC}{a}\|u\|_E^2\leq M\|v_0\|_X+\frac{MC}{a}\rho^2. \end{align*} The choices $\|v_0\|_X<\rho/(2M)$ and $\rho\leq a/(4MC)$ give \begin{align*} \|\Phi u\|_E\leq \frac{\rho}{2}+\frac{\rho}{4}<\rho. \end{align*} So $\Phi u\in E$ and $\Phi u\in\mathcal{B}_\rho$. This proves that the Duhamel operator sends the closed weighted ball $\mathcal{B}_\rho$ into itself.[/guided]

[step:Prove that the Duhamel operator is a strict contraction]Let $u,w\in\mathcal{B}_\rho$. For every $s\geq 0$, \begin{align*} \|N(u(s))-N(w(s))\|_X\leq C(\|u(s)\|_X+\|w(s)\|_X)\|u(s)-w(s)\|_X. \end{align*} Using $\|u(s)\|_X+\|w(s)\|_X\leq 2\rho e^{-\omega_0s}$ and $\|u(s)-w(s)\|_X\leq e^{-\omega_0s}\|u-w\|_E$, we get \begin{align*} \|N(u(s))-N(w(s))\|_X\leq 2C\rho e^{-2\omega_0s}\|u-w\|_E. \end{align*} Therefore \begin{align*} e^{\omega_0t}\|(\Phi u)(t)-(\Phi w)(t)\|_X\leq 2MC\rho\|u-w\|_E\int_0^t e^{-a(t-s)}e^{-\omega_0s}\,d\mathcal{L}^1(s). \end{align*} As above, the integral is at most $1/a$, so \begin{align*} \|\Phi u-\Phi w\|_E\leq \frac{2MC\rho}{a}\|u-w\|_E. \end{align*} Since $\rho\leq a/(4MC)$, we have \begin{align*} \|\Phi u-\Phi w\|_E\leq \frac{1}{2}\|u-w\|_E. \end{align*} Thus $\Phi$ is a strict contraction on $\mathcal{B}_\rho$.[/step]

[guided]The contraction estimate uses the same weighted decay mechanism as the mapping estimate, but now applied to the difference of two candidate solutions. Let $u,w\in\mathcal{B}_\rho$. Since $u(s),w(s)\in B_X(0,r)$ for every $s\geq 0$, the Lipschitz hypothesis on $N$ applies and gives \begin{align*} \|N(u(s))-N(w(s))\|_X\leq C(\|u(s)\|_X+\|w(s)\|_X)\|u(s)-w(s)\|_X. \end{align*} The ball condition gives $\|u(s)\|_X+\|w(s)\|_X\leq 2\rho e^{-\omega_0s}$, while the definition of the weighted norm gives $\|u(s)-w(s)\|_X\leq e^{-\omega_0s}\|u-w\|_E$. Substituting both estimates yields \begin{align*} \|N(u(s))-N(w(s))\|_X\leq 2C\rho e^{-2\omega_0s}\|u-w\|_E. \end{align*} Using the semigroup decay estimate in the Duhamel difference, we obtain \begin{align*} e^{\omega_0t}\|(\Phi u)(t)-(\Phi w)(t)\|_X\leq 2MC\rho\|u-w\|_E\int_0^t e^{-a(t-s)}e^{-\omega_0s}\,d\mathcal{L}^1(s). \end{align*} As in the preceding step, the integral is at most $1/a$. Therefore \begin{align*} \|\Phi u-\Phi w\|_E\leq \frac{2MC\rho}{a}\|u-w\|_E. \end{align*} The choice $\rho\leq a/(4MC)$ makes the contraction constant at most $1/2$, so \begin{align*} \|\Phi u-\Phi w\|_E\leq \frac{1}{2}\|u-w\|_E. \end{align*} Thus the nonlinear Duhamel map is a strict contraction on the closed weighted ball.[/guided]

[step:Construct the global fixed point by the contraction argument][claim:Contraction maps on complete metric spaces have unique fixed points] Let $(Y,d)$ be a [complete metric space](/page/Complete%20Metric%20Space), and let $\Psi:Y\to Y$ satisfy $d(\Psi y,\Psi z)\leq \kappa d(y,z)$ for all $y,z\in Y$, where $0\leq\kappa<1$. Then $\Psi$ has a unique fixed point in $Y$. [/claim] [proof] Choose $y_0\in Y$, and define a sequence $(y_n)_{n\geq 0}$ in $Y$ by $y_{n+1}:=\Psi y_n$. For every $n\geq 0$, \begin{align*} d(y_{n+1},y_n)\leq \kappa^n d(y_1,y_0). \end{align*} Hence, for integers $m>n$, \begin{align*} d(y_m,y_n)\leq \sum_{j=n}^{m-1}d(y_{j+1},y_j)\leq \frac{\kappa^n}{1-\kappa}d(y_1,y_0). \end{align*} Thus $(y_n)$ is Cauchy, and completeness gives a point $y_\ast\in Y$ with $y_n\to y_\ast$. Since $\Psi$ is Lipschitz continuous, $\Psi y_n\to \Psi y_\ast$. But $\Psi y_n=y_{n+1}\to y_\ast$, so $\Psi y_\ast=y_\ast$. If $z_\ast$ is another fixed point, then \begin{align*} d(y_\ast,z_\ast)=d(\Psi y_\ast,\Psi z_\ast)\leq \kappa d(y_\ast,z_\ast). \end{align*} Since $\kappa<1$, this implies $d(y_\ast,z_\ast)=0$, so $y_\ast=z_\ast$. [/proof] The set $\mathcal{B}_\rho$ is closed in the Banach space $E$, hence complete. Applying the claim to $\Phi:\mathcal{B}_\rho\to\mathcal{B}_\rho$, there exists a unique $v\in\mathcal{B}_\rho$ such that $\Phi v=v$. Therefore, for every $t\geq 0$, \begin{align*} v(t)=T(t)v_0+\int_0^t T(t-s)N(v(s))\,d\mathcal{L}^1(s). \end{align*} Thus $v$ is a global mild solution.[/step]

[guided]We first recall why the fixed point claim applies. Let $(Y,d)$ be a complete [metric space](/page/Metric%20Space), and let $\Psi:Y\to Y$ satisfy \begin{align*} d(\Psi y,\Psi z)\leq \kappa d(y,z) \end{align*} for all $y,z\in Y$, where $0\leq\kappa<1$. Choose $y_0\in Y$, and define a sequence $(y_n)_{n\geq 0}$ in $Y$ by \begin{align*} y_{n+1}:=\Psi y_n. \end{align*} Iterating the contraction estimate gives, for every $n\geq 0$, \begin{align*} d(y_{n+1},y_n)\leq \kappa^n d(y_1,y_0). \end{align*} Hence, for integers $m>n$, the triangle inequality and the geometric-series estimate give \begin{align*} d(y_m,y_n)\leq \sum_{j=n}^{m-1}d(y_{j+1},y_j)\leq \frac{\kappa^n}{1-\kappa}d(y_1,y_0). \end{align*} Since $0\leq\kappa<1$, the right-hand side tends to $0$ as $n\to\infty$, so $(y_n)$ is Cauchy. Completeness of $Y$ gives a point $y_\ast\in Y$ with $y_n\to y_\ast$. The map $\Psi$ is Lipschitz continuous because it is a contraction, so $\Psi y_n\to\Psi y_\ast$. But $\Psi y_n=y_{n+1}$ and $y_{n+1}\to y_\ast$, hence \begin{align*} \Psi y_\ast=y_\ast. \end{align*} If $z_\ast$ is another fixed point, then \begin{align*} d(y_\ast,z_\ast)=d(\Psi y_\ast,\Psi z_\ast)\leq \kappa d(y_\ast,z_\ast). \end{align*} Since $\kappa<1$, this implies $d(y_\ast,z_\ast)=0$, and therefore $y_\ast=z_\ast$. We now check these hypotheses for the present metric space. The set $\mathcal{B}_\rho$ is closed in the Banach space $E$ because it is a closed norm ball, hence $\mathcal{B}_\rho$ is complete with the metric \begin{align*} d(u,w):=\|u-w\|_E. \end{align*} The preceding two steps prove that $\Phi:\mathcal{B}_\rho\to\mathcal{B}_\rho$ and that, for all $u,w\in\mathcal{B}_\rho$, \begin{align*} \|\Phi u-\Phi w\|_E\leq \frac{1}{2}\|u-w\|_E. \end{align*} Thus the fixed point claim applies with \begin{align*} Y=\mathcal{B}_\rho, \end{align*} \begin{align*} d(u,w)=\|u-w\|_E, \end{align*} \begin{align*} \Psi=\Phi, \end{align*} and \begin{align*} \kappa=\frac{1}{2}. \end{align*} Consequently there is a unique $v\in\mathcal{B}_\rho$ such that \begin{align*} \Phi v=v. \end{align*} Expanding this fixed point equation gives \begin{align*} v(t)=T(t)v_0+\int_0^t T(t-s)N(v(s))\,d\mathcal{L}^1(s) \end{align*} for every $t\geq 0$. This is exactly the mild formulation of the nonlinear evolution equation on the whole interval $[0,\infty)$, so the fixed point is a global mild solution.[/guided]

[step:Extract the sharp decay bound from the fixed point equation]Let \begin{align*} A:=\|v\|_E. \end{align*} Repeating the mapping estimate with $u=v$ and using $\Phi v=v$ gives \begin{align*} A\leq M\|v_0\|_X+\frac{MC}{a}A^2. \end{align*} Since $v\in\mathcal{B}_\rho$, we have $A\leq\rho$, and since $\rho\leq a/(4MC)$, \begin{align*} \frac{MC}{a}A\leq \frac{1}{4}. \end{align*} Therefore \begin{align*} A\leq M\|v_0\|_X+\frac{1}{4}A. \end{align*} Subtracting $\frac{1}{4}A$ from both sides gives \begin{align*} A\leq \frac{4M}{3}\|v_0\|_X\leq 2M\|v_0\|_X. \end{align*} By the definition of $\|\cdot\|_E$, this implies \begin{align*} \|v(t)\|_X\leq 2M e^{-\omega_0t}\|v_0\|_X \end{align*} for every $t\geq 0$.[/step]

[guided]The fixed point already gives decay because it lies in $\mathcal{B}_\rho$, but the preceding bound would only give the coefficient $\rho/\|v_0\|_X$, which is not uniform as $v_0\to 0$. To extract a uniform linear estimate in $\|v_0\|_X$, define \begin{align*} A:=\|v\|_E. \end{align*} Since $v=\Phi v$, the mapping estimate with $u=v$ gives \begin{align*} A\leq M\|v_0\|_X+\frac{MC}{a}A^2. \end{align*} Because $v\in\mathcal{B}_\rho$, we have $A\leq\rho$. The smallness choice $\rho\leq a/(4MC)$ therefore implies \begin{align*} \frac{MC}{a}A\leq \frac{1}{4}. \end{align*} Substituting this into the quadratic term gives \begin{align*} A\leq M\|v_0\|_X+\frac{1}{4}A. \end{align*} Subtracting $A/4$ from both sides yields \begin{align*} A\leq \frac{4M}{3}\|v_0\|_X\leq 2M\|v_0\|_X. \end{align*} Finally, the definition of the weighted norm says $e^{\omega_0t}\|v(t)\|_X\leq A$ for every $t\geq 0$. Hence \begin{align*} \|v(t)\|_X\leq 2M e^{-\omega_0t}\|v_0\|_X \end{align*} for every $t\geq 0$.[/guided]

[step:Identify every local mild solution with the global fixed point]Let $u\in C([0,T];X)$ and $w\in C([0,T];X)$ be two mild solutions on a finite interval $[0,T]$ with the same initial datum $v_0$, and suppose $u(t),w(t)\in B_X(0,r)$ for every $t\in[0,T]$. Define \begin{align*} D_T:=\sup_{0\leq t\leq T}\|u(t)-w(t)\|_X. \end{align*} For $t\in[0,T]$, subtracting the two mild formulas gives \begin{align*} u(t)-w(t)=\int_0^t T(t-s)(N(u(s))-N(w(s)))\,d\mathcal{L}^1(s). \end{align*} If $R_T:=\sup_{0\leq s\leq T}(\|u(s)\|_X+\|w(s)\|_X)$, then $R_T\leq 2r$. If $R_T=0$, then $u(t)=w(t)=0$ for every $t\in[0,T]$, so uniqueness is immediate. Hence assume $R_T>0$. The Lipschitz estimate gives \begin{align*} \|u(t)-w(t)\|_X\leq MCR_T\int_0^t \|u(s)-w(s)\|_X\,d\mathcal{L}^1(s). \end{align*} Choose a partition $0=t_0<t_1<\cdots<t_m=T$ such that $t_j-t_{j-1}< (MCR_T)^{-1}$ for every $j$. On $[0,t_1]$, taking the supremum over $0\leq t\leq t_1$ in the preceding estimate gives \begin{align*} \sup_{0\leq t\leq t_1}\|u(t)-w(t)\|_X\leq MCR_Tt_1\sup_{0\leq t\leq t_1}\|u(t)-w(t)\|_X. \end{align*} Since $MCR_Tt_1<1$, this supremum is zero, so $u=w$ on $[0,t_1]$. Suppose now that $u=w$ on $[0,t_{j-1}]$. For $t\in[t_{j-1},t_j]$, the mild formula restarted at time $t_{j-1}$ follows from the original mild formula by splitting the Duhamel integral over $[0,t_{j-1}]$ and $[t_{j-1},t]$, then using the semigroup property $T(t-s)=T(t-t_{j-1})T(t_{j-1}-s)$ on the first integral. Thus both restarted formulas have the same initial value $u(t_{j-1})=w(t_{j-1})$, and the same estimate gives \begin{align*} \sup_{t_{j-1}\leq t\leq t_j}\|u(t)-w(t)\|_X\leq MCR_T(t_j-t_{j-1})\sup_{t_{j-1}\leq t\leq t_j}\|u(t)-w(t)\|_X. \end{align*} Because $MCR_T(t_j-t_{j-1})<1$, the supremum is zero. Induction over $j$ gives $u=w$ on all of $[0,T]$. Now let $\tilde v$ be any local mild solution with initial datum $v_0$, defined on an interval $[0,\tau)$ with $\tau>0$. Since \begin{align*} \|v_0\|_X<\delta\leq \frac{\rho}{2M}\leq \rho\leq \frac{r}{2}<r, \end{align*} continuity gives a nontrivial interval on which $\tilde v(t)\in B_X(0,r)$. Let \begin{align*} \sigma:=\sup\left\{s\in[0,\tau):\tilde v(t)\in B_X(0,r)\text{ for every }0\leq t\leq s\right\}. \end{align*} For every $S<\sigma$, the preceding uniqueness argument applies on $[0,S]$ to $\tilde v$ and to the global fixed point $v$, so $\tilde v=v$ on $[0,S]$. Hence \begin{align*} \|\tilde v(t)\|_X=\|v(t)\|_X\leq 2Me^{-\omega_0t}\|v_0\|_X<2M\delta=\rho\leq \frac{r}{2} \end{align*} for every $t<\sigma$. If $\sigma<\tau$, continuity of $\tilde v$ at $\sigma$ gives $\|\tilde v(\sigma)\|_X\leq r/2$, and then continuity gives some $\varepsilon>0$ such that $\tilde v(t)\in B_X(0,r)$ for $0\leq t\leq \sigma+\varepsilon<\tau$, contradicting the definition of $\sigma$. Thus $\sigma=\tau$, and $\tilde v=v$ on every compact subinterval of $[0,\tau)$. If $\tilde v$ is a maximal local mild solution and $\tau<\infty$, then the equality $\tilde v(t)=v(t)$ on $[0,\tau)$ gives $\tilde v(t)\to v(\tau)$ in $X$ as $t\uparrow\tau$. The global fixed point $v$ provides the mild continuation beyond $\tau$, contradicting maximality. Therefore every maximal local mild solution is global and equals $v$. The decay estimate from the previous step completes the proof.[/step]

[guided]It remains to connect the constructed global fixed point with arbitrary local mild solutions. Let $u,w\in C([0,T];X)$ be two mild solutions on a finite interval $[0,T]$ with the same initial datum $v_0$, and assume $u(t),w(t)\in B_X(0,r)$ for every $t\in[0,T]$. Define \begin{align*} R_T:=\sup_{0\leq s\leq T}(\|u(s)\|_X+\|w(s)\|_X). \end{align*} If $R_T=0$, then both solutions are identically zero and uniqueness holds. If $R_T>0$, subtracting the two mild formulas gives \begin{align*} u(t)-w(t)=\int_0^t T(t-s)(N(u(s))-N(w(s)))\,d\mathcal{L}^1(s). \end{align*} The semigroup bound and the Lipschitz estimate for $N$ imply \begin{align*} \|u(t)-w(t)\|_X\leq MCR_T\int_0^t \|u(s)-w(s)\|_X\,d\mathcal{L}^1(s). \end{align*} Choose a partition $0=t_0<t_1<\cdots<t_m=T$ with $t_j-t_{j-1}<(MCR_T)^{-1}$. On the first interval, taking the supremum gives \begin{align*} \sup_{0\leq t\leq t_1}\|u(t)-w(t)\|_X\leq MCR_Tt_1\sup_{0\leq t\leq t_1}\|u(t)-w(t)\|_X. \end{align*} Since $MCR_Tt_1<1$, the supremum is zero, so $u=w$ on $[0,t_1]$. Repeating the same estimate on each later interval is valid after restarting the mild formula at $t_{j-1}$; the restart follows by splitting the Duhamel integral and using the semigroup property. Induction gives $u=w$ on $[0,T]$. Now let $\tilde v$ be any local mild solution with initial datum $v_0$, defined on $[0,\tau)$. The point that requires care is that the Lipschitz hypothesis for $N$ is only assumed on $B_X(0,r)$. We therefore prove that no small solution can exit that ball before it has been identified with the global fixed point. Because \begin{align*} \|v_0\|_X<\delta\leq \frac{\rho}{2M}\leq \rho\leq \frac{r}{2}<r, \end{align*} continuity gives an initial interval on which $\tilde v$ lies in $B_X(0,r)$. Define \begin{align*} \sigma:=\sup\left\{s\in[0,\tau):\tilde v(t)\in B_X(0,r)\text{ for every }0\leq t\leq s\right\}. \end{align*} For each $S<\sigma$, both $\tilde v$ and the fixed point $v$ take values in $B_X(0,r)$ on $[0,S]$, so the finite-interval uniqueness argument gives $\tilde v=v$ on $[0,S]$. The decay estimate for the fixed point then implies \begin{align*} \|\tilde v(t)\|_X=\|v(t)\|_X\leq 2Me^{-\omega_0t}\|v_0\|_X<2M\delta=\rho\leq \frac{r}{2} \end{align*} for every $t<\sigma$. If $\sigma<\tau$, continuity at $\sigma$ keeps $\tilde v$ strictly inside $B_X(0,r)$ for a little longer, contradicting the definition of $\sigma$. Therefore $\sigma=\tau$, and $\tilde v=v$ throughout its local interval. Finally suppose $\tilde v$ is maximal and $\tau<\infty$. Since $\tilde v(t)=v(t)$ for $t<\tau$ and $v$ is continuous on $[0,\infty)$, the local solution has the limit $v(\tau)$ as $t\uparrow\tau$. The already constructed global fixed point supplies a mild continuation past $\tau$, contradicting maximality. Hence every maximal local mild solution is global, equals $v$, and satisfies the decay estimate proved above.[/guided]