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Consider a strongly continuous semigroup of bounded linear operators $S(t):X\to X$. The infinitesimal generator of $S(t)$ is the linear operator $A:D(A)\subseteq X \to X$ defined by $$D(A):=\bigg\lbrace x\in X \ | \ \exists \lim_{t\to 0^+} \frac{S(t)x-x}{t} \text{ in } X\bigg\rbrace$$ and $$A\,x := \lim_{t\to 0^+} \frac{S(t)x-x}{t} \text{ in } X, \ \forall x\in D(A)$$ The operator $A$ is densly defined, closed but not necessarily bounded.

However, a corollary of the uniform boundedness principle states that the pointwise limit of a sequence of bounded operators is bounded. So since all of the $S(t)$'s are bounded, I would expect $A$ to be bounded as well.

Can anyone point out the flaw in my reasoning? (I belive it has to do with the fact that $A$ is in general not everywhere-defined).

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Your argument would be valid for bounded limit of bounded operators. In the definition of the generator, you divide with $t$ as $t\to 0$, and you cannot guarantee it to remain bounded.

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Yes $A$ is not defined everywhere, but we can still ask the question whether $A$ is bounded on its domain and thus $A$ would have a bounded extension in the whole space $X$.

The operators $\dfrac{T(t)-I}{t}$ are bounded on $D(A)$ and converges pointwise to $A$. However $D(A)$ is not a Banach space (because it is not closed in $X$ in general), so you can't apply the the uniform boundedness principle.

Note: While $D(A)$ is not in general a Banach space with the inherited norm, it is a Banach space if we consider the following norm: $|x|_A:=|x|+|Ax|$ and I think this is equivalent to $A$ being a closed operator.

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