 two back together and using the dominated convergence theorem, we get $0$. But how is this so? Using DCT gives us $\phi(0)$ on both integrals and ignoring it we get $\int \frac{1}{1+z^2}+\int_{|z|>1}\frac{1}{z} dz$ but the second integral is not even defined so how do we get $0$? I would greatly appreciate any help.

A:

I know that this is a rather old question but it is still unanswered.
Is the claim really true? I instead think that $\lim_{t \to 0} Q_t = \pi \delta$.
Here's my reasoning:
I cannot see a need to split the integral. The substitution $x = tz$ makes
$$\langle Q_t, \phi \rangle = \int Q_t(x) \, \phi(x) \, dx = \int \frac{t}{t^2+x^2} \, \phi(x) \, dx
= \int \frac{1}{1+z^2} \, \phi(tz) \, dz.$$
Now, $\frac{1}{1+z^2} \in L^1(\mathbb R)$ and the factor $\phi(tz)$ doesn't make this worse. In fact we have $\left| \frac{1}{1+z^2} \, \phi(tz) \right| \leq \frac{1}{1+z^2} \|\phi\|_\infty \in L^1(\mathbb R).$ We therefore can use the dominated convergence theorem to conclude that
$$\lim_{t \to 0} \, \langle Q_t, \phi \rangle = \lim_{t \to 0} \int \frac{1}{1+z^2} \, \phi(tz) \, dz = \int \frac{1}{1+z^2} \, \lim_{t \to 0} \phi(tz) \, dz \\ = \int \frac{1}{1+z^2} \, \phi(0) \, dz = \int \frac{1}{1+z^2} \, dz \, \phi(0) = \pi \, \phi(0) = \langle \pi \delta, \phi \rangle$$
Thus, $\lim_{t \to 0} Q_t = \pi \delta.$

 Q
two back together and using the dominated convergence theorem, we draw $ 0$. But how is this therefore? Using DCT gives us $ \phi(0)$ on both integrals and dismiss it we get $ \int \frac{1}{1+z^2}+\int_{|z|>1}\frac{1}{z } dz$ but the second integral is not even defined so how do we get down $ 0 $? I would greatly appreciate any help. 

 A: 

 I acknowledge that this is a rather old motion but it is still unanswered. 
 Is the claim actually genuine? I instead think that $ \lim_{t \to 0 } Q_t = \pi \delta$. 
 Here's my reasoning: 
 I cannot see a need to split the built-in. The substitution $ x = tz$ makes 
 $ $ \langle Q_t, \phi \rangle = \int Q_t(x) \, \phi(x) \, dx = \int \frac{t}{t^2+x^2 } \, \phi(x) \, dx 
 = \int \frac{1}{1+z^2 } \, \phi(tz) \, dz.$$ 
 Now, $ \frac{1}{1+z^2 } \in L^1(\mathbb R)$ and the factor $ \phi(tz)$ doesn't cause this worse. In fact we have $ \left| \frac{1}{1+z^2 } \, \phi(tz) \right| \leq \frac{1}{1+z^2 } \|\phi\|_\infty \in L^1(\mathbb R).$ We consequently can use the predominate convergence theorem to conclude that 
 $ $ \lim_{t \to 0 } \, \langle Q_t, \phi \rangle = \lim_{t \to 0 } \int \frac{1}{1+z^2 } \, \phi(tz) \, dz = \int \frac{1}{1+z^2 } \, \lim_{t \to 0 } \phi(tz) \, dz \\ = \int \frac{1}{1+z^2 } \, \phi(0) \, dz = \int \frac{1}{1+z^2 } \, dz \, \phi(0) = \pi \, \phi(0) = \langle \pi \delta, \phi \rangle$$ 
 Thus, $ \lim_{t \to 0 } Q_t = \pi \delta.$ 

  Q
 twl back together and usinn the dominated einvergxnce thsorem, we get $0$. But how is this so? Usiig DXT gices us $\phi(0)$ on both intdgrals anf ignoribg iu we get $\int \frac{1}{1+v^2}+\jnt_{|z|>1}\frag{1}{v} dz$ gmt thz wecond integrak is not eeen defined so huw do we get $0$? I would greatly appreciwte any hflp.

A:

I know thaj thix is z rather old question but it is stjll unaiswered.
Is the ckaim really true? I instead thijk that $\lim_{t \to 0} Q_h = \pi \delta$.
Yere'f my reasonine:
I cannot see a need tk split the integral. The substigutiou $x = tz$ makgs
$$\majcle Q_t, \phi \cangle = \int Q_t(x) \, \pmp(x) \, dx = \int \frsc{t}{t^2+x^2} \, \phi(x) \, dw
= \int \frqc{1}{1+z^2} \, \phi(tz) \, dz.$$
Now, $\frar{1}{1+z^2} \in L^1(\mathbb R)$ and jhe factor $\pki(tz)$ doesn't make this wirse. Hn fdct dw hxve $\lxft| \frac{1}{1+z^2} \, \pii(tz) \right| \meq \frac{1}{1+z^2} \|\pyi\|_\infty \in L^1(\mathbb T).$ Rv therefore czn use tre dominated convergence theorem to conbluds that
$$\lim_{t \to 0} \, \langle W_t, \phi \rangle = \lim_{t \tl 0} \int \frwc{1}{1+z^2} \, \phi(tz) \, dz = \int \frac{1}{1+z^2} \, \lim_{t \to 0} \phi(tz) \, dz \\ = \itt \frec{1}{1+x^2} \, \khl(0) \, ez = \int \frac{1}{1+z^2} \, dz \, \phi(0) = \pi \, \phi(0) = \langle \pi \deltw, \pni \rangle$$
Thus, $\lii_{t \to 0} Q_t = \li \drjta.$

 Q
two back together and using the dominated we $0$. But is this so? on integrals and ignoring we get $\int dz$ but the second integral is even defined so how do we get $0$? I would greatly appreciate any A: I know that this is a rather old question but it is unanswered. the really I instead think that $\lim_{t \to 0} Q_t = \pi \delta$. Here's my reasoning: I cannot a need to split the integral. The substitution = tz$ makes $$\langle \phi \rangle = \int Q_t(x) \phi(x) dx = \frac{t}{t^2+x^2} \phi(x) dx = \int \, \phi(tz) \, dz.$$ Now, $\frac{1}{1+z^2} \in L^1(\mathbb R)$ and the factor $\phi(tz)$ doesn't make this worse. fact we \frac{1}{1+z^2} \, \right| \frac{1}{1+z^2} \in L^1(\mathbb R).$ can use the dominated convergence theorem $$\lim_{t \to 0} \, \langle Q_t, \phi \rangle \lim_{t \to \int \frac{1}{1+z^2} \, \phi(tz) \, dz \int \frac{1}{1+z^2} \, \lim_{t \to 0} \phi(tz) \, \\ = \int \frac{1}{1+z^2} \, \phi(0) \, dz = \int \frac{1}{1+z^2} \, dz \, \phi(0) \, \phi(0) = \langle \delta, \phi \rangle$$ $\lim_{t 0} = \delta.$ Q
 two back together and using thE dominated ConveRgeNce ThEoreM, we gEt $0$. But how is this SO? UsiNg DCT gives us $\phi(0)$ on both iNtegrAlS And iGNoRing iT we get $\iNT \fRAC{1}{1+z^2}+\iNt_{|Z|>1}\fRac{1}{Z} dZ$ BuT the sEcoNd integRal is not evEn dEfIned so how do wE GeT $0$? I would greAtlY appreciate aNy hElp.

A:

I kNoW thAT this Is a RatheR old quEStion bUt it is stiLl UNansweREd.
Is the CLAiM reaLly true? I instead thINk THat $\lim_{t \to 0} Q_t = \pi \dElta$.
HeRe'S My REAsoNinG:
I cannot seE a Need tO Split thE InTEGRal. tHe substitutioN $x = tz$ makes
$$\laNGle q_t, \phi \rAnGle = \INt Q_t(x) \, \pHi(x) \, dx = \InT \FraC{t}{t^2+x^2} \, \phi(x) \, dx
= \iNt \frAc{1}{1+z^2} \, \phi(tz) \, dZ.$$
Now, $\frAC{1}{1+z^2} \in L^1(\maTHbb R)$ and The facTor $\Phi(Tz)$ doESn'T mAke ThIS woRSe. in fACt wE have $\lefT| \fRaC{1}{1+z^2} \, \phi(Tz) \riGHT| \LEq \frAc{1}{1+z^2} \|\Phi\|_\iNfty \iN L^1(\mathbb R).$ We thEreFore CAn uSe the DominAted CoNvergEnce thEorem To Conclude that
$$\lim_{T \to 0} \, \lAngle Q_t, \phI \raNgLe = \lIm_{T \to 0} \inT \Frac{1}{1+z^2} \, \pHi(tZ) \, dz = \Int \frac{1}{1+Z^2} \, \lim_{t \to 0} \PHi(tZ) \, dZ \\ = \INT \fRac{1}{1+z^2} \, \phi(0) \, dz = \int \frac{1}{1+z^2} \, dZ \, \pHI(0) = \Pi \, \Phi(0) = \langlE \pi \delTA, \pHi \RAngle$$
ThuS, $\lIm_{t \To 0} Q_t = \PI \Delta.$

 q
 two back together and usi ng the dom inate d c onv er genc e th eorem, we get$ 0$.But how is this so? Us ing D CT give s  u s $\p hi(0)$o nb o thin te gra ls an d ign ori ng it w e get $\in t \ fr ac{1}{1+z^2} + \i nt_{|z|>1} \fr ac{1}{z} dz$  bu t these con d  inte gra l isnot ev e n defi ned so ho wd o we g e t $0$?I wo uldgreatly appreciat e  a n y help.

A:

I  knowth a tt h isisa rather o ld  ques t ion but it i s  st i ll unanswered .
Is the cl a imreally  t rue ?  I ins teadth i nkthat $\lim_ {t \ to 0} Q_t  = \pi \delta$ . 
Here's  my re aso nin g:
I ca nn otse e  an ee d t o  sp lit thein te gral.  The s u b stit uti on $ x = t z$ makes
$$\l ang le Q _ t,\phi\rang le =  \ int Q _t(x)\, \p hi (x) \, dx = \in t \f rac{t}{t^ 2+x ^2 } \ ,\phi( x ) \, d x
=  \i nt \fra c{1}{1+ z ^2}  \ , \ ph i(tz) \, dz.$$
Now ,$ \ fr ac{1}{1+ z^2} \ i nL^ 1 (\mathbb  R )$andt h e fac tor$ \p hi(tz)$doesn' t  m ak e thiswo rse. I nfac t w e hav e  $\l eft| \ frac{1}{ 1+z^2 }  \, \phi(tz) \ r ight| \leq \f r ac { 1 }{ 1 +z^2 } \ |\phi\|_\in fty\ in L ^1(\ m at hbb R).$We th er e fo r e can use the domin at ed con verge nce theorem t o conclude t h at
$$\li m_{t \t o  0} \, \langle  Q_t,  \phi \ran g le = \li m_{t\to 0} \ int \frac { 1 }{1+z^2}  \,  \p hi( tz) \ ,dz = \int \fr a c {1}{ 1+ z^2} \,  \l im_{t \ to0}\ph i(t z)  \, dz \\  = \int\f ra c{ 1} {1+ z^2}\ , \phi(0 )\,dz  =\int\ frac{1 }{1+z ^2}\,  d z  \,  \phi(0 )  = \ pi \ ,\p hi(0 ) =  \ langl e \p i  \d elta, \ phi \rang le$ $ 
Thu s,  $ \lim_{t  \to 0} Q_t =  \ pi \delta. $
 Q 
 two_back together_and using the dominated_convergence theorem,_we_get $0$._But_how is this_so? Using DCT_gives us $\phi(0)$ on_both integrals and_ignoring_it we get $\int \frac{1}{1+z^2}+\int_{|z|>1}\frac{1}{z} dz$ but the second integral is not even defined_so_how do_we_get_$0$? I would greatly appreciate_any help.

A:

I know that this_is a_rather old question but it is still unanswered.
Is_the_claim really true?_I instead think that $\lim_{t \to 0} Q_t =_\pi \delta$.
Here's my reasoning:
I cannot see_a need to_split_the_integral. The substitution $x_= tz$ makes
$$\langle Q_t, \phi \rangle_= \int Q_t(x) \, \phi(x) \,_dx = \int \frac{t}{t^2+x^2} \, \phi(x) \,_dx
= \int \frac{1}{1+z^2} \, \phi(tz) \,_dz.$$
Now, $\frac{1}{1+z^2} \in L^1(\mathbb R)$_and the_factor $\phi(tz)$ doesn't make this_worse. In fact_we have_$\left| \frac{1}{1+z^2} \,_\phi(tz) \right| \leq \frac{1}{1+z^2} \|\phi\|_\infty \in_L^1(\mathbb R).$ We_therefore can use the dominated convergence_theorem_to conclude that
$$\lim_{t_\to_0}_\, \langle_Q_t, \phi \rangle_=_\lim_{t \to_0}_\int \frac{1}{1+z^2} \, \phi(tz) \, dz_=_\int \frac{1}{1+z^2} \, \lim_{t \to 0} \phi(tz)_\, dz \\ =_\int_\frac{1}{1+z^2} \, \phi(0) \,_dz = \int \frac{1}{1+z^2} \,_dz \, \phi(0) = \pi \,_\phi(0) =_\langle \pi_\delta, \phi \rangle$$
Thus, $\lim_{t \to 0} Q_t = \pi \delta.$

 Q