I have seen such expression:
the generator of the semigroup is defined by
$$ [Uf] = \lim_{t\rightarrow 0} \frac{U^t f - f}{t}. $$
However, I don't understand, isn't the $[Uf] = df/dt$ that simple?
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Context needed: what sorts of objects/spaces are $U$ and $f$ in? (The answer will also imply additional tags) – jmerry Feb 07 '19 at 05:46
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1It depends on the context and notations, but the notation $df/dt$ usually implies that $f$ is a function on the variable $t$, and $df/dt$ is the derivative taken with respect to the argument $t$ of the function $f$. On the other hand, it is implied in the notation of $[Uf]$ that the semigroup containing $U$(thus $U^t$) acts on the space that contains $f$ (a function space if $f$ is a function) and we take the derivative with respect to this action. – cjackal Feb 07 '19 at 05:47
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1I agree with the previous comment. For instance if $(U_t)$ is the semigroup associated with one-dimensional brownian motion, then $Uf = \frac{1}{2} f''$. You might want to take a look at this question – saz Feb 07 '19 at 08:45
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Assuming that $(U_t)_{t\geq 0}$ is a semigroup of linear operators on a Banach space $X$. The infinitesimal generator $A$ of $U_t$ is defined by $$D(A)=\{f\in X \colon \lim_{t \to 0}\frac{U_t f - f}{t} \text{ exists in } X\},$$ and $$Af=\lim_{t \to 0}\frac{U_t f - f}{t}, \text{ for all } f\in D(A).$$ It is clear from the first semigroup property $U_0 =I_X$ (The identity operator) that $$Af=\frac{d}{dt}|_{t=0^+} U_t f, \text{ for all } f\in D(A).$$
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