[proofplan] We solve the equation by rewriting it as a fixed-point problem in the [Banach space](/page/Banach%20Space) $C([0,T];X)$. The local Lipschitz hypothesis gives both a Lipschitz estimate and a boundedness estimate for $F$ on a ball containing the candidate solution paths, while strong continuity of $S(t)$ at $t=0$ controls the linear term near the constant path $u_0$. Choosing $T>0$ small makes the Duhamel operator map this path ball into itself and makes it a contraction. Uniqueness follows first inside the fixed-point ball and then on all of $C([0,T];X)$ by restarting the mild formula on short subintervals. Continuous dependence and $C^1$ dependence follow from the same contraction estimates, with the linearized Duhamel equation giving the derivative when $F$ is $C^1$. [/proofplan] [step:Fix the path space and the Duhamel operator] The family $(S(t))_{t\geq 0}$ is a strongly continuous semigroup, so $S(0)=I$, $S(t+s)=S(t)S(s)$ for $s,t\geq0$, and $S(t)x\to x$ in $X$ as $t\downarrow0$ for every $x\in X$. Let $E_T$ denote the Banach space $C([0,T];X)$ with norm \begin{align*} \|u\|_{E_T}:=\sup_{t\in[0,T]}\|u(t)\|_X. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$, restricted to subintervals of $[0,T_0]$ when used below. For $a\in X$, define the constant path $c_a\in E_T$ by $c_a(t):=a$ for every $t\in[0,T]$. Define \begin{align*} K:=M e^{\max\{\omega,0\}T_0}. \end{align*} Then $\|S(t)\|_{\mathcal{L}(X)}\leq K$ for every $t\in[0,T_0]$. For a fixed initial value $u_0\in X$ and a time $T\in(0,T_0]$, define the Duhamel operator $\Phi_{u_0,T}:E_T\to E_T$ by \begin{align*} (\Phi_{u_0,T}u)(t):=S(t)u_0+\int_0^t S(t-s)F(u(s))\,d\mathcal{L}^1(s). \end{align*} The integral is the Bochner integral in $X$. If $u\in E_T$, then the map $s\mapsto F(u(s))$ from $[0,T]$ to $X$ is continuous because $u$ is continuous and $F$ is locally Lipschitz, hence continuous. For each fixed $t\in[0,T]$, the map $s\mapsto S(t-s)F(u(s))$ on $[0,t]$ is continuous and therefore Bochner integrable. We now justify continuity of $t\mapsto (\Phi_{u_0,T}u)(t)$. The set $u([0,T])$ is compact in $X$, so $F(u([0,T]))$ is compact and hence bounded; define \begin{align*} C_u:=\sup_{s\in[0,T]}\|F(u(s))\|_X<\infty. \end{align*} Let $\mathbb{1}_I:[0,T]\to\{0,1\}$ denote the indicator function of a subinterval $I\subset[0,T]$, so $\mathbb{1}_I(s)=1$ for $s\in I$ and $\mathbb{1}_I(s)=0$ for $s\notin I$. If $t_n\to t$ in $[0,T]$, extend each integrand by zero outside its natural interval and write it on $[0,T]$ as follows. Define $g_n:[0,T]\to X$ by \begin{align*} g_n(s):=\mathbb{1}_{[0,t_n]}(s)S(t_n-s)F(u(s)). \end{align*} Define $g:[0,T]\to X$ by \begin{align*} g(s):=\mathbb{1}_{[0,t]}(s)S(t-s)F(u(s)). \end{align*} For every $s\in[0,T]$ except possibly $s=t$, strong continuity of the semigroup gives $g_n(s)\to g(s)$ in $X$. Also $\|g_n(s)\|_X\leq KC_u$ for all $n$ and $s$. The Bochner [dominated convergence theorem](/theorems/4) gives \begin{align*} \int_0^{\mathsf T} g_n(s)\,d\mathcal{L}^1(s)\to \int_0^{\mathsf T} g(s)\,d\mathcal{L}^1(s) \end{align*} in $X$. Together with $S(t_n)u_0\to S(t)u_0$, this proves $\Phi_{u_0,T}u\in E_T$. [/step] [step:Choose the time so the Duhamel operator is a contraction on a closed path ball] Fix $r>0$ and set \begin{align*} R:=\|u_0\|_X+r. \end{align*} Let $L_R$ be a Lipschitz constant for $F$ on the closed ball $\{x\in X:\|x\|_X\leq R\}$. Define \begin{align*} B_R:=\|F(0)\|_X+L_R R. \end{align*} Then $\|F(x)\|_X\leq B_R$ whenever $\|x\|_X\leq R$, because \begin{align*} \|F(x)\|_X\leq \|F(0)\|_X+\|F(x)-F(0)\|_X\leq \|F(0)\|_X+L_R\|x\|_X\leq B_R. \end{align*} Let \begin{align*} \mathcal{B}_{T,r}:=\{u\in E_T:\|u-c_{u_0}\|_{E_T}\leq r\}. \end{align*} This is a closed subset of the Banach space $E_T$, hence complete. For every $u\in\mathcal{B}_{T,r}$ and every $s\in[0,T]$, \begin{align*} \|u(s)\|_X\leq \|u_0\|_X+r=R. \end{align*} By strong continuity of the semigroup at $0$, \begin{align*} \delta(T):=\sup_{t\in[0,T]}\|S(t)u_0-u_0\|_X\to 0 \end{align*} as $T\downarrow 0$. Choose $T\in(0,T_0]$ such that \begin{align*} \delta(T)\leq \frac{r}{2},\qquad KTB_R\leq \frac{r}{2},\qquad KTL_R\leq \frac{1}{2}. \end{align*} For $u\in\mathcal{B}_{T,r}$ and $t\in[0,T]$, the triangle inequality and the semigroup bound give \begin{align*} \|(\Phi_{u_0,T}u)(t)-u_0\|_X\leq \|S(t)u_0-u_0\|_X+\int_0^t \|S(t-s)\|_{\mathcal{L}(X)}\|F(u(s))\|_X\,d\mathcal{L}^1(s). \end{align*} Therefore \begin{align*} \|(\Phi_{u_0,T}u)(t)-u_0\|_X\leq \delta(T)+KTB_R\leq r. \end{align*} Thus $\Phi_{u_0,T}$ maps $\mathcal{B}_{T,r}$ into itself. If $u,v\in\mathcal{B}_{T,r}$, then for every $t\in[0,T]$, \begin{align*} \|(\Phi_{u_0,T}u)(t)-(\Phi_{u_0,T}v)(t)\|_X\leq \int_0^t K L_R\|u(s)-v(s)\|_X\,d\mathcal{L}^1(s). \end{align*} Taking the supremum over $t\in[0,T]$ gives \begin{align*} \|\Phi_{u_0,T}u-\Phi_{u_0,T}v\|_{E_T}\leq KTL_R\|u-v\|_{E_T}\leq \frac{1}{2}\|u-v\|_{E_T}. \end{align*} Hence $\Phi_{u_0,T}$ is a contraction on $\mathcal{B}_{T,r}$. [guided] The point of centering the ball at the constant path $c_{u_0}$ is that the semigroup term $S(t)u_0$ is close to $u_0$ for small $t$. We need to control two contributions to $\Phi_{u_0,T}u-u_0$: the linear drift $S(t)u_0-u_0$ and the nonlinear Duhamel integral. First define the uniform semigroup bound \begin{align*} K:=M e^{\max\{\omega,0\}T_0}. \end{align*} This works because for every $t\in[0,T_0]$, \begin{align*} \|S(t)\|_{\mathcal{L}(X)}\leq M e^{\omega t}\leq M e^{\max\{\omega,0\}T_0}=K. \end{align*} Now fix $r>0$ and set $R:=\|u_0\|_X+r$. If $u$ stays within distance $r$ of the constant path $u_0$, then all its values lie in the closed $X$-ball of radius $R$, because \begin{align*} \|u(s)\|_X\leq \|u(s)-u_0\|_X+\|u_0\|_X\leq r+\|u_0\|_X=R. \end{align*} The local Lipschitz hypothesis is now usable on this fixed ball. It gives a constant $L_R$ such that \begin{align*} \|F(x)-F(y)\|_X\leq L_R\|x-y\|_X \end{align*} whenever $\|x\|_X\leq R$ and $\|y\|_X\leq R$. It also gives a bound for $F$ itself on this ball by comparing to $F(0)$. Define \begin{align*} B_R:=\|F(0)\|_X+L_RR. \end{align*} Then for $\|x\|_X\leq R$, \begin{align*} \|F(x)\|_X\leq \|F(0)\|_X+\|F(x)-F(0)\|_X\leq \|F(0)\|_X+L_R\|x\|_X\leq B_R. \end{align*} The remaining choice is the time length. Strong continuity of $S(t)$ means $S(t)u_0\to u_0$ in $X$ as $t\downarrow0$. Therefore \begin{align*} \delta(T):=\sup_{t\in[0,T]}\|S(t)u_0-u_0\|_X \end{align*} satisfies $\delta(T)\to0$ as $T\downarrow0$. We choose $T\in(0,T_0]$ so small that \begin{align*} \delta(T)\leq \frac{r}{2},\qquad KTB_R\leq \frac{r}{2},\qquad KTL_R\leq \frac{1}{2}. \end{align*} The first two inequalities force $\Phi_{u_0,T}$ to preserve the path ball. Indeed, if $u\in\mathcal{B}_{T,r}$, then for every $t\in[0,T]$, \begin{align*} \|(\Phi_{u_0,T}u)(t)-u_0\|_X\leq \|S(t)u_0-u_0\|_X+\int_0^t \|S(t-s)\|_{\mathcal{L}(X)}\|F(u(s))\|_X\,d\mathcal{L}^1(s). \end{align*} Using $\|S(t-s)\|_{\mathcal{L}(X)}\leq K$ and $\|F(u(s))\|_X\leq B_R$ gives \begin{align*} \|(\Phi_{u_0,T}u)(t)-u_0\|_X\leq \delta(T)+KTB_R\leq r. \end{align*} Thus $\Phi_{u_0,T}u$ remains in the same closed ball. The third inequality makes $\Phi_{u_0,T}$ a contraction. For $u,v\in\mathcal{B}_{T,r}$, both $u(s)$ and $v(s)$ lie in the radius-$R$ ball of $X$, so the Lipschitz estimate for $F$ gives \begin{align*} \|F(u(s))-F(v(s))\|_X\leq L_R\|u(s)-v(s)\|_X. \end{align*} Substituting this into the Duhamel difference yields \begin{align*} \|(\Phi_{u_0,T}u)(t)-(\Phi_{u_0,T}v)(t)\|_X\leq \int_0^t K L_R\|u(s)-v(s)\|_X\,d\mathcal{L}^1(s). \end{align*} Since $\|u(s)-v(s)\|_X\leq \|u-v\|_{E_T}$ for every $s$, we obtain \begin{align*} \|\Phi_{u_0,T}u-\Phi_{u_0,T}v\|_{E_T}\leq KTL_R\|u-v\|_{E_T}\leq \frac{1}{2}\|u-v\|_{E_T}. \end{align*} This is exactly the small-time contraction estimate. [/guided] [/step] [step:Construct the mild solution by the contraction principle] We use the following elementary contraction principle. [claim:Contraction principle on a complete metric space] Let $(Y,d)$ be a [complete metric space](/page/Complete%20Metric%20Space) and let $\Psi:Y\to Y$ satisfy $d(\Psi y,\Psi z)\leq qd(y,z)$ for all $y,z\in Y$, where $0\leq q<1$. Then $\Psi$ has a unique fixed point in $Y$. [/claim] [proof] Choose $y_0\in Y$ and define $y_{n+1}:=\Psi y_n$ for every integer $n\geq0$. For $n\geq1$, \begin{align*} d(y_{n+1},y_n)\leq q^n d(y_1,y_0). \end{align*} Thus for $m>n$, \begin{align*} d(y_m,y_n)\leq \sum_{k=n}^{m-1}d(y_{k+1},y_k)\leq \sum_{k=n}^{m-1}q^k d(y_1,y_0)\leq \frac{q^n}{1-q}d(y_1,y_0). \end{align*} Hence $(y_n)$ is Cauchy, so it converges to some $y_\ast\in Y$. Since $\Psi$ is Lipschitz, it is continuous, and therefore \begin{align*} \Psi y_\ast=\Psi\left(\lim_{n\to\infty}y_n\right)=\lim_{n\to\infty}\Psi y_n=\lim_{n\to\infty}y_{n+1}=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 qd(y_\ast,z_\ast). \end{align*} Since $q<1$, this implies $d(y_\ast,z_\ast)=0$, hence $y_\ast=z_\ast$. [/proof] Applying the claim to the complete [metric space](/page/Metric%20Space) $\mathcal{B}_{T,r}$ and the contraction $\Phi_{u_0,T}$ gives a unique $u\in\mathcal{B}_{T,r}$ such that $\Phi_{u_0,T}u=u$. Therefore \begin{align*} u(t)=S(t)u_0+\int_0^t S(t-s)F(u(s))\,d\mathcal{L}^1(s) \end{align*} for every $t\in[0,T]$, so $u$ is a mild solution. [/step] [step:Extend uniqueness from the fixed-point ball to all mild solutions on the same interval] Let $w\in C([0,T];X)$ be any mild solution with initial data $u_0$. Since $u$ and $w$ are continuous on the compact interval $[0,T]$, define \begin{align*} R_\ast:=\sup_{t\in[0,T]}\max\{\|u(t)\|_X,\|w(t)\|_X\}<\infty. \end{align*} Let $L_{R_\ast}$ be a Lipschitz constant for $F$ on the closed ball of radius $R_\ast$ in $X$. Choose $h\in(0,T]$ such that \begin{align*} KhL_{R_\ast}<1. \end{align*} For any $a\in[0,T]$ and $t\in[a,T]$, the semigroup property and the mild formula imply the restarted identity \begin{align*} u(t)=S(t-a)u(a)+\int_a^t S(t-s)F(u(s))\,d\mathcal{L}^1(s), \end{align*} and the same identity holds for $w$. This follows by splitting the integral at $a$ and using $S(t-s)=S(t-a)S(a-s)$ for $0\leq s\leq a\leq t$. On the interval $[0,\min\{h,T\}]$, subtracting the two restarted formulas with $a=0$ gives \begin{align*} \sup_{t\in[0,\min\{h,T\}]}\|u(t)-w(t)\|_X\leq KhL_{R_\ast}\sup_{t\in[0,\min\{h,T\}]}\|u(t)-w(t)\|_X. \end{align*} Since $KhL_{R_\ast}<1$, the supremum is $0$, so $u=w$ on $[0,\min\{h,T\}]$. Repeating the same argument from the endpoint of the previous interval and using the restarted formula covers $[0,T]$ by finitely many intervals of length at most $h$. Hence $u=w$ on all of $[0,T]$. [/step] [step:Prove continuous dependence on the initial data] Fix the constructed solution $u$ for the initial datum $u_0$. Decrease $T$ if necessary so that the contraction constant in the construction is at most $1/2$. Choose $\rho>0$ such that $(K+1)\rho\leq r/4$, and define \begin{align*} R_1:=\|u_0\|_X+\rho+r. \end{align*} For every $a\in X$ with $\|a-u_0\|_X\leq \rho$, the same construction applies on a possibly smaller common time interval $[0,T]$: indeed, \begin{align*} \sup_{t\in[0,T]}\|S(t)a-a\|_X\leq (K+1)\|a-u_0\|_X+\sup_{t\in[0,T]}\|S(t)u_0-u_0\|_X. \end{align*} Then choose $T\in(0,T_0]$ small enough that \begin{align*} \sup_{t\in[0,T]}\|S(t)u_0-u_0\|_X\leq \frac{r}{4}, \qquad KTB_{R_1}\leq \frac{r}{2}, \qquad KTL_{R_1}\leq \frac{1}{2}, \end{align*} where $L_{R_1}$ is a Lipschitz constant for $F$ on the closed $X$-ball of radius $R_1$ and $B_{R_1}:=\|F(0)\|_X+L_{R_1}R_1$. For every $a\in X$ with $\|a-u_0\|_X\leq\rho$, these inequalities give \begin{align*} \sup_{t\in[0,T]}\|S(t)a-a\|_X\leq \frac{r}{2}, \qquad KTB_{R_1}\leq \frac{r}{2}, \qquad KTL_{R_1}\leq \frac{1}{2}. \end{align*} Thus the fixed-point construction above applies on the same interval $[0,T]$ for every such initial datum $a$, with the common radius $R_1$. Let $a,b\in X$ satisfy $\|a-u_0\|_X\leq\rho$ and $\|b-u_0\|_X\leq\rho$, and let $u_a,u_b\in C([0,T];X)$ be the corresponding mild solutions. Both solution paths take values in the radius-$R_1$ ball. Therefore, for every $t\in[0,T]$, \begin{align*} \|u_a(t)-u_b(t)\|_X\leq \|S(t)(a-b)\|_X+\int_0^t K L_{R_1}\|u_a(s)-u_b(s)\|_X\,d\mathcal{L}^1(s). \end{align*} Taking suprema and using $KTL_{R_1}\leq 1/2$ gives \begin{align*} \|u_a-u_b\|_{E_T}\leq K\|a-b\|_X+\frac{1}{2}\|u_a-u_b\|_{E_T}. \end{align*} Hence \begin{align*} \|u_a-u_b\|_{E_T}\leq 2K\|a-b\|_X. \end{align*} Thus the local solution map $\mathcal{S}_T:a\mapsto u_a$ is locally Lipschitz, and in particular continuous, at $u_0$. [/step] [step:Identify the derivative of the solution map when $F$ is $C^1$] Assume now that $F:X\to X$ is Fréchet $C^1$. Decrease $T>0$ and choose a neighbourhood $V\subset X$ of $u_0$ so that all solution paths $\mathcal{S}_T(a)$ with $a\in V$ lie in a common bounded subset of $X$. Define \begin{align*} R_0:=\sup\{\|\mathcal{S}_T(a)(s)\|_X:a\in V,\ s\in[0,T]\}<\infty. \end{align*} Choose a radius $R>R_0$ and let $L_R$ be a Lipschitz constant for $F$ on the closed $X$-ball of radius $R$. Decrease $T$ further, if necessary, so that \begin{align*} KTL_R\leq \frac{1}{2}. \end{align*} For every $x$ in the closed ball of radius $R_0$, the operator norm of $F'(x)$ is bounded by $L_R$: indeed, for each $v\in X$ with $v\neq0$, choose $\tau_0>0$ such that $\|x+\tau v\|_X\leq R$ whenever $0<\tau<\tau_0$. For such $\tau$ the Lipschitz estimate on the radius-$R$ ball gives \begin{align*} \frac{\|F(x+\tau v)-F(x)\|_X}{\tau}\leq L_R\|v\|_X. \end{align*} Letting $\tau\downarrow0$ gives $\|F'(x)v\|_X\leq L_R\|v\|_X$, hence $\|F'(x)\|_{\mathcal{L}(X)}\leq L_R$. For $a\in V$ and $h\in X$, define $z_{a,h}\in E_T$ to be the unique solution of the linearized mild equation \begin{align*} z_{a,h}(t)=S(t)h+\int_0^t S(t-s)F'(\mathcal{S}_T(a)(s))z_{a,h}(s)\,d\mathcal{L}^1(s). \end{align*} This equation has a unique solution by the same contraction argument, because \begin{align*} \left\|\int_0^t S(t-s)F'(\mathcal{S}_T(a)(s))(z_1(s)-z_2(s))\,d\mathcal{L}^1(s)\right\|_X\leq KTL_R\|z_1-z_2\|_{E_T}. \end{align*} The map $h\mapsto z_{a,h}$ is linear, and the estimate \begin{align*} \|z_{a,h}\|_{E_T}\leq 2K\|h\|_X \end{align*} shows that it is bounded. We claim that $D\mathcal{S}_T(a)h=z_{a,h}$. Let $a\in V$, let $h\in X$ be small enough that $a+h\in V$, and set \begin{align*} u:=\mathcal{S}_T(a),\qquad u_h:=\mathcal{S}_T(a+h),\qquad e_h:=u_h-u-z_{a,h}. \end{align*} Subtracting the mild equations for $u_h$, $u$, and $z_{a,h}$ gives \begin{align*} e_h(t)=\int_0^t S(t-s)\left(F(u_h(s))-F(u(s))-F'(u(s))(u_h(s)-u(s))\right)\,d\mathcal{L}^1(s)+\int_0^t S(t-s)F'(u(s))e_h(s)\,d\mathcal{L}^1(s). \end{align*} The first integrand is a Fréchet differentiability remainder. Since $u([0,T])$ is compact in $X$ and $F'$ is continuous, for every $\varepsilon>0$ there exists $\eta>0$ such that whenever $x\in u([0,T])$ and $y\in X$ satisfy $\|y-x\|_X\leq\eta$, one has $\|F'(y)-F'(x)\|_{\mathcal{L}(X)}\leq\varepsilon$. This follows by taking finitely many continuity neighbourhoods covering the compact set $u([0,T])$. For $s\in[0,T]$, define $q_h(s):=u_h(s)-u(s)$. If $\|q_h\|_{E_T}\leq\eta$, then the Banach-space [fundamental theorem of calculus](/theorems/632) applied to the $C^1$ map $\theta\mapsto F(u(s)+\theta q_h(s))$ from $[0,1]$ to $X$ gives \begin{align*} F(u_h(s))-F(u(s))-F'(u(s))q_h(s)=\int_0^1 \left(F'(u(s)+\theta q_h(s))-F'(u(s))\right)q_h(s)\,d\mathcal{L}^1(\theta). \end{align*} The [uniform continuity](/page/Uniform%20Continuity) estimate for $F'$ on the tubular neighbourhood of $u([0,T])$ therefore implies \begin{align*} \|F(u_h(s))-F(u(s))-F'(u(s))q_h(s)\|_X\leq \varepsilon\|q_h(s)\|_X. \end{align*} Since $\|u_h-u\|_{E_T}\leq 2K\|h\|_X$, it follows that \begin{align*} \sup_{s\in[0,T]}\frac{\|F(u_h(s))-F(u(s))-F'(u(s))(u_h(s)-u(s))\|_X}{\|h\|_X}\leq 2K\varepsilon \end{align*} for all sufficiently small $\|h\|_X$. Because $\varepsilon>0$ is arbitrary, the quotient tends to $0$ as $\|h\|_X\to0$. Taking suprema in the equation for $e_h$ and using $KTL_R\leq1/2$ gives \begin{align*} \|e_h\|_{E_T}\leq K T \sup_{s\in[0,T]}\|F(u_h(s))-F(u(s))-F'(u(s))(u_h(s)-u(s))\|_X+\frac{1}{2}\|e_h\|_{E_T}. \end{align*} Therefore \begin{align*} \frac{\|e_h\|_{E_T}}{\|h\|_X}\to0 \end{align*} as $\|h\|_X\to0$. This proves Fréchet differentiability of $\mathcal{S}_T$ at $a$. Finally, the formula for $D\mathcal{S}_T(a)h$ and the same contraction estimate imply continuity of $a\mapsto D\mathcal{S}_T(a)$ in $\mathcal{L}(X,E_T)$. Indeed, if $a_n\to a$ in $V$, then $\mathcal{S}_T(a_n)\to\mathcal{S}_T(a)$ in $E_T$. Since $\mathcal{S}_T(a)([0,T])$ is compact in $X$ and $F'$ is continuous, a finite-cover argument gives uniform continuity of $F'$ along a sufficiently small tubular neighbourhood of this compact path. The [uniform convergence](/page/Uniform%20Convergence) of $\mathcal{S}_T(a_n)$ to $\mathcal{S}_T(a)$ therefore gives \begin{align*} \sup_{s\in[0,T]}\|F'(\mathcal{S}_T(a_n)(s))-F'(\mathcal{S}_T(a)(s))\|_{\mathcal{L}(X)}\to0. \end{align*} Subtracting the two linearized equations and absorbing the term with coefficient at most $1/2$ yields \begin{align*} \|D\mathcal{S}_T(a_n)-D\mathcal{S}_T(a)\|_{\mathcal{L}(X,E_T)}\to0. \end{align*} Thus $\mathcal{S}_T$ is Fréchet $C^1$ after possibly decreasing $T$ and $V$. This completes the proof. [/step]