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Let $X=C([0,T];L^2(\mathbb{R}))$ and define the map suggested by the variation-of-constants formula:The strong continuity of the evolution family and the continuity of $f$ imply that $\Phi$ maps $X$ into itself. Because $U(t,s)$ is unitary and $\Vert f(t,u){\Vert }_2\le C$,

Equip $X$ with the exponentially weighted supremum normThis equivalent norm makes $X$ a Banach space. Using that $f$ is a globally Lipschitz function and that the unitary operator $U(t,s)$ preserves the norm,Choose $\alpha >L$. Then $\Phi$ is a contraction mapping, so the Banach fixed-point theorem gives a unique fixed point $u\in X$. This fixed point is exactly the required mild solution of an abstract Cauchy problem:The weighted-norm argument works on the whole prescribed finite interval, so no subdivision of $[0,T]$ is needed.

SetThen the forced equation is the abstract Cauchy problemFor $Z_0\in \mathcal{H}$ and $F\in C([0,T];\mathcal{H})$, a mild solution of an abstract Cauchy problem is a function $Z\in C([0,T];\mathcal{H})$ satisfying the variation-of-constants formula

Suppose $F(s)\in D(A)$ and both $F$ and $AF$ are continuous. If also $Z_0\in D(A)$, then the closedness of $A$ permits differentiation under the Bochner integral:Thus $Z\in C^1([0,\infty );\mathcal{H})$ and $Z^{\prime }=AZ+F$. The assumption $Z_0\in D(A)$ is necessary here: a unitary group has no smoothing, so the conditions on $F$ alone cannot make $U(t)Z_0$ differentiable for arbitrary $Z_0\in \mathcal{H}$.

For the resulting strong solution, skew symmetry of $A$ gives the energy estimateIntegration yields