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Suppose $X$ is a Banach space, and $T(t)$ $t\ge0$ is a strongly continuous semigroup with generator $A$. Assume $\frac{T(t)-I}{t}x$ weakly converges to $y\in X$ when $t\to 0$, then I need to prove $x\in D(A)$ and $y=Ax$. Could you provide some help?

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    $\begingroup$ This is proved for example in Pazy's "Semigroups of Linear Operators and Applications to Partial Differential Equations", section 2.1 "Weak Equals Strong". $\endgroup$
    – unwissen
    Commented Jul 8 at 6:45
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    $\begingroup$ @unwissen Thanks! $\endgroup$
    – Richard
    Commented Jul 8 at 9:26

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