0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Context needed: what sorts of objects/spaces are $U$ and $f$ in? (The answer will also imply additional tags) $\endgroup$ – jmerry Feb 7 at 5:46
  • 1
    $\begingroup$ It 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. $\endgroup$ – cjackal Feb 7 at 5:47
  • 1
    $\begingroup$ I 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 $\endgroup$ – saz Feb 7 at 8:45
1
$\begingroup$

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 $$Af=\frac{d}{dt}|_{t=0^+} U_t f, \text{ for all } f\in D(A).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.