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Let $A$ be the infinitesimal generator of a $C_0$-semigroup $(S(t))_{t \geq 0}$. Now, for every $x_0 \in X$ the map $t \mapsto S(t) x_0$ is a mild solution of

$ \dot{x} = Ax, \quad x(0) = x_0.\tag{*} $

Now, a continuous function $x: [0, \infty) \to X$ is called a mild solution of $\text{(*)}$ if $\int_0^t x(s) \, ds \in D(A)$ where $D(A)$ is the domain of $A$, $x(0) = x_0$ and

$x(t) - x(0) = A \int_0^t x(\tau) \, d\tau \text{ for all $t \geq 0$}.$

Now, I have a proof of this but it uses Hille's theorem, but it is quite involved (needs a few tricks) and Hille's theorem is not elementary, does someone know an elementary proof of the uniqueness?

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    I do not know it, but I see that I can retrieve it from work. I learned semigroups from Lunardi, and I must confess that I never thought of looking there. I am not sure that she does mild solutions actually... thanks for the reference.2019-02-22

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You just need some simple properties of the Banach space-valued integral, namely linearity (the integral commutes with continuous linear operators) substitution for translations and the fundamental theorem of calculus, which says: $\frac{1}{h}\int_0 ^h f(t) d t \to f(0)$ (in the Banach space) for $h \to 0$.

So after writing out all definitions (and using linearity and the semigroup properties), your job is to show that for any $t>0$ we have $\frac{1}{h}\int_0 ^t (S(\tau+h) x_0 - S(\tau)x_0) d \tau \to S(t)x_0 - x_0$ for $h\to 0$ from above. We can assume $t>h$. Then split the first integral into $\int_0 ^{t-h}+\int_{t-h} ^t$ and the second one into $\int_0 ^h+ \int_h ^t$. By translation the integrals $\int_0 ^{t-h}$ and $\int_h ^t$ cancel, and you just have to apply the fundamental theorem of calculus.

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    @Florian: I modified the question.2010-12-04