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Infinitesimal generator why $limlimits_sto 0P_tleft(fracP_sf-fsright)=P_tLf$?

Does finiteness of $limlimits_xtoinftyf(x)$ and $limlimits_xtoinftyf'(x)$ imply $limlimits_xtoinftyf'(x)=0$?A confusion on Almost Everywhere ConvergenceShow $lim_m to infty ,n to infty f(fracleftlfloor mx rightrfloor m,fracleftlfloor ny rightrfloor n) = f(x,y)$Showing that $ Pleft(limlimits_n to inftyfracS_nn text exists in mathbb Rright) = 0. $Why is $limlimits_ntoinfty sumlimits_k=1^n frac1k - log (n) = sumlimits_k=1^infty frac1k-log(frack+1k)$Why $left|int f f^q-1right|=|f|_q^q$?Show that $f$ is uni. cont. on $(a,b)$ if and only if it is continuous $(a,b)$ and $limlimits_xto a^+f(x)$ and $limlimits_xto b^-f(x)$ existShow there is $Cin [0,1)$ s.t. $|f(x)-f(y)|leq C|x-y|$ when $|x-y|geq a$.Why is $mu(E_1) + limlimits_N rightarrow infty sumlimits_n=1^N mu(E_n+1 -E_n) = limlimits_N rightarrow infty mu(E_N)$If $limlimits_nto infty|x_n|= 0$ then $suplimits_ngeq 1|x_n|<infty$

1

$begingroup$

Let $(P_t)_tgeq 0$ a semi group and $fin D(L)=leftfin mathcal C_0mid limlimits_tto 0fracP_tf-fttext existright$ and $mathcal C_0$ is the set of continuous function s.t. $limlimits_tto infty f(t)=0$.

Set $$Lf=lim_tto 0fracP_t f-ft.$$

Suppose $fin D(L)$. We have $$lim_sto 0fracP_sP_tf-P_tft=lim_sto 0P_tfracP_sf-fs=P_tLf.$$

I don't understand the last equality, i.e. why $$lim_sto 0P_tfracP_sf-fs=P_tlim_sto 0fracP_sf-fs ?$$

real-analysis

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edited Mar 12 at 18:56

rtybase

11.5k31534

asked Mar 12 at 14:27

PierrePierre

6611

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the $P_t$'s? Continuous linear transformations?
  $endgroup$
  – Berci
  Mar 12 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $P_t$ is s.t. $P_0=id$, $P_t+s=P_tcirc P_s$ and $lim_tto 0|P_tf-f|=0$ and $|P_t|leq 1$ for a certain norm $|cdot |$, but it's not specified. @Berci
  $endgroup$
  – Pierre
  Mar 12 at 14:37
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $(P_t)_tgeq 0$ a semi group and $fin D(L)=leftfin mathcal C_0mid limlimits_tto 0fracP_tf-fttext existright$ and $mathcal C_0$ is the set of continuous function s.t. $limlimits_tto infty f(t)=0$.

Set $$Lf=lim_tto 0fracP_t f-ft.$$

Suppose $fin D(L)$. We have $$lim_sto 0fracP_sP_tf-P_tft=lim_sto 0P_tfracP_sf-fs=P_tLf.$$

I don't understand the last equality, i.e. why $$lim_sto 0P_tfracP_sf-fs=P_tlim_sto 0fracP_sf-fs ?$$

real-analysis

share|cite|improve this question

edited Mar 12 at 18:56

rtybase

11.5k31534

asked Mar 12 at 14:27

PierrePierre

6611

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the $P_t$'s? Continuous linear transformations?
  $endgroup$
  – Berci
  Mar 12 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $P_t$ is s.t. $P_0=id$, $P_t+s=P_tcirc P_s$ and $lim_tto 0|P_tf-f|=0$ and $|P_t|leq 1$ for a certain norm $|cdot |$, but it's not specified. @Berci
  $endgroup$
  – Pierre
  Mar 12 at 14:37
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $(P_t)_tgeq 0$ a semi group and $fin D(L)=leftfin mathcal C_0mid limlimits_tto 0fracP_tf-fttext existright$ and $mathcal C_0$ is the set of continuous function s.t. $limlimits_tto infty f(t)=0$.

Set $$Lf=lim_tto 0fracP_t f-ft.$$

Suppose $fin D(L)$. We have $$lim_sto 0fracP_sP_tf-P_tft=lim_sto 0P_tfracP_sf-fs=P_tLf.$$

I don't understand the last equality, i.e. why $$lim_sto 0P_tfracP_sf-fs=P_tlim_sto 0fracP_sf-fs ?$$

real-analysis

share|cite|improve this question

edited Mar 12 at 18:56

rtybase

11.5k31534

asked Mar 12 at 14:27

PierrePierre

6611

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 12 at 18:56

rtybase

11.5k31534

asked Mar 12 at 14:27

PierrePierre

6611

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 12 at 18:56

rtybase

11.5k31534

asked Mar 12 at 14:27

PierrePierre

6611

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 12 at 18:56

rtybase

11.5k31534

edited Mar 12 at 18:56

rtybase

11.5k31534

asked Mar 12 at 14:27

PierrePierre

6611

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 12 at 14:27

PierrePierre

6611