www.fixkar.com/articles/kb/1150/ and changed my site collection administrator but it didn't change anything in the log file...
If you have a piece of clue it will be so great!
Thank you by advance and sorry for my bad english

A:

Assuming that you ran the Powershell shell run as Administrator, what happened if you specified the -force parameter? Did anything in the target list get created?

 Q:

Map of Pre-Hilbert Space is affine if it is an isometry

Let $(H, <\cdot,\cdot>)   $ be a real pre-Hilbert space and let $T: H \to H$ be an isometry so that $||T(x)-T(y)|| = ||x - y||$ for all $x, y \in H$. Prove $T$ is affine.  
How do I prove this?

A:

Replace $T$ by $S(x) = T(x) - T(0)$.  $S$ is an isometry with
 $S(0) = 0$, and  the task is to prove that $S$ is linear.  
Now $S$ preserves both distances and norms.  Writing out
$$
||S(a)-S(b)||^2 = d(S(a), S(b))^2 = d(a, b)^2  = ||a-b||^2,
$$
and expanding gives $$\langle S(a), S(b)\rangle  = \langle a, b \rangle.$$
  Using this, compute 
$$
\langle S(a + \lambda b) - S(a) - \lambda S(b), 
 S(a + \lambda b) - S(a) - \lambda S(b)\rangle = 0.
$$
Hence $S(a + \lambda b) - S(a) - \lambda S(b) = 0$, and $S$ is linear.
Addendum:  Maybe the last step is more palatable in the following form:
For all elements $S(z)$ in the range of $S$, one has
$$
\langle S(a + \lambda b) - S(a) - \lambda S(b), S(z) \rangle = 0.
$$
Hence for all $y
www.fixkar.com/articles/kb/1150/ and changed my site collection administrator but it didn't change anything in the logarithm file... 
 If you take a piece of clue it will be therefore capital! 
 Thank you by advance and sorry for my regretful english 

 A: 

 Assuming that you ran the Powershell shell test as Administrator, what happened if you specified the -force argument? Did anything in the target list get created? 

  Q: 

 Map of Pre - Hilbert Space is affine if it is an isometry 

 lease $ (H, < \cdot,\cdot >)    $ be a real pre - Hilbert space and let $ T: H \to H$ be an isometry so that $ ||T(x)-T(y)|| = ||x - y||$ for all $ x, y \in H$. testify $ T$ is affine.  
 How do I prove this? 

 A: 

 Replace $ T$ by $ S(x) = T(x) - T(0)$.   $ S$ is an isometry with 
  $ S(0) = 0 $, and   the task is to raise that $ S$ is linear.  
 Now $ S$ preserves both distances and norm.   Writing out 
 $ $ 
 ||S(a)-S(b)||^2 = d(S(a), S(b))^2 = d(a, b)^2   = ||a - b||^2, 
 $ $ 
 and expanding gives $ $ \langle S(a), S(b)\rangle   = \langle a, b \rangle.$$ 
   Using this, compute 
 $ $ 
 \langle S(a + \lambda b) - S(a) - \lambda S(b), 
  S(a + \lambda b) - S(a) - \lambda S(b)\rangle = 0. 
 $ $ 
 therefore $ S(a + \lambda b) - S(a) - \lambda S(b) = 0 $, and $ S$ is linear. 
 Addendum:   Maybe the last step is more palatable in the following form: 
 For all elements $ S(z)$ in the compass of $ S$, one has 
 $ $ 
 \langle S(a + \lambda b) - S(a) - \lambda S(b), S(z) \rangle = 0. 
 $ $ 
 Hence for all $ y
www.vixkar.com/articles/kb/1150/ and ghanged my site eillectmon admjnistratur but it didn't change anythmng un tht log file...
If you haxe a piecv of clue it xill be so great!
Vgank yom by zfvanee and sorry for my bad enclish

A:

Assuming tfac you ran the Powershell shell run af Adminoshrator, what hakpenec if gou specified the -force parameter? Sid anyuhing in the targey list get created?

 Q:

Map of Pre-Jilbert Space is avfine if it is wb isometry

Leg $(H, <\cdot,\cdot>)   $ be a rgal pre-Hilbert space and let $T: H \to H$ be an isonerry vo that $||T(x)-T(b)|| = ||x - j||$ for all $x, y \in H$. Prmve $T$ ix affine.  
How co M prive this?

A:

Replace $T$ by $S(x) = T(x) - T(0)$.  $S$ is ag isometrf cith
 $S(0) = 0$, and  the tasj us to proee tfqt $R$ ia kihear.  
Jow $S$ preservss both disrances and norms.  Wtiepmg out
$$
||S(a)-S(b)||^2 = s(S(a), S(b))^2 = q(a, b)^2  = ||a-b||^2,
$$
and expanding gives $$\langle S(a), V(b)\rzngle  = \langle a, b \rangoe.$$
  Using this, computg 
$$
\langle S(w + \lambda b) - S(a) - \lambda S(b), 
 S(a + \lambda b) - S(a) - \lamtda S(u)\rxngoe = 0.
$$
Yejce $S(a + \lambda b) - S(a) - \lambda S(b) = 0$, and $S$ is ligsat.
Acdendum:  Maybe the last syeo of more palatacle in ths following form:
Fog all ejemenrs $S(z)$ in uhe rsnge of $S$, one has
$$
\langle S(a + \lambda b) - S(c) - \oambda S(b), S(z) \ranglz = 0.
$$
Hence for all $y
www.fixkar.com/articles/kb/1150/ and changed my site collection administrator didn't anything in log file... If clue will be so Thank you by and sorry for my bad english Assuming that you ran the Powershell shell run as Administrator, what happened if specified the -force parameter? Did anything in the target list get created? Q: of Space affine it is an isometry Let $(H, <\cdot,\cdot>) $ be a real pre-Hilbert space and let $T: \to H$ be an isometry so that $||T(x)-T(y)|| ||x - y||$ for $x, y \in H$. Prove is How do prove A: $T$ by $S(x) T(x) - T(0)$. $S$ is an isometry with $S(0) = 0$, and the task is to prove $S$ is $S$ preserves distances norms. out $$ ||S(a)-S(b)||^2 S(b))^2 = d(a, b)^2 = ||a-b||^2, gives $$\langle S(a), S(b)\rangle = \langle a, b Using this, $$ \langle S(a + \lambda b) S(a) - \lambda S(b), S(a + \lambda b) S(a) - \lambda S(b)\rangle = 0. $$ Hence $S(a + \lambda b) - S(a) - = 0$, and $S$ linear. Addendum: Maybe last is palatable the following For all elements $S(z)$ in the range of $S$, one has \langle S(a + \lambda b) - S(a) - \lambda S(b), = $$ Hence for $y
www.fixkar.com/articles/kb/1150/ and Changed my sIte coLleCtiOn AdmiNistRator but it didn'T ChanGe anything in the log file...
if you HaVE a piECe Of cluE it will BE sO GReaT!
THaNk yOu BY aDvancE anD sorry fOr my bad engLisH

A:

assuming that YOu Ran the PoweRshEll shell run aS AdMinistRaTor, WHat haPpeNed if You speCIfied tHe -force paRaMEter? DiD AnythinG IN tHe taRget list get createD?

 q:

MAP of Pre-Hilbert SPace is AfFInE IF it Is aN isometry

LEt $(h, <\cdot,\CDot>)   $ be a rEAl PRE-hilBErt space and leT $T: H \to H$ be an iSOmeTry so tHaT $||T(x)-t(Y)|| = ||x - y||$ for All $x, y \In h$. proVe $T$ is affine.  
how dO I prove thIs?

A:

RepLAce $T$ by $S(X) = t(x) - T(0)$.  $S$ is aN isomeTry WitH
 $S(0) = 0$, anD  ThE tAsk Is TO prOVe ThaT $s$ is Linear.  
NoW $S$ PrEservEs boTH DIStanCes And nOrms.  WRiting out
$$
||S(a)-S(b)||^2 = D(S(a), s(b))^2 = d(a, B)^2  = ||A-b||^2,
$$
aNd expAndinG givEs $$\LanglE S(a), S(b)\rAngle  = \LaNgle a, b \rangle.$$
  UsiNg thIs, compute 
$$
\LanGlE S(a + \LaMbda b) - s(A) - \lambdA S(b), 
 s(a + \lAmbda b) - S(A) - \lambda s(B)\raNgLE = 0.
$$
hEnCe $S(a + \lambda b) - S(a) - \lambdA S(B) = 0$, ANd $s$ is lineaR.
AddenDUm:  maYBe the lasT sTep Is moRE PalatAble IN tHe followIng forM:
foR aLl elemeNtS $S(z)$ in tHe RanGe oF $S$, one HAs
$$
\laNgle S(a + \Lambda b) - S(A) - \lambDA S(b), S(z) \rangle = 0.
$$
HenCE for all $y
www.fixkar.com/articles/kb /1150/ and  chan ged  my  s itecoll ection adminis t rato r but it didn't change  anyt hi n g in th e log  file.. . 
I f you  h av e a  p i ec e ofclu e it wi ll be so g rea t! 
Thank you b y  a dvance and  so rry for my b adenglis h
A: 
Assu min g tha t your an the  Powershe ll shellr un as A d m in istr ator, what happen e di f you specifie d the-f o rc e par ame ter? Did a ny thing in thet ar g e t  li s t get created ?

 Q:

Map ofPre-Hi lb ert Spaceis af fi n e i f it is anisom etry

Let  $(H,< \cdot,\ c dot>)  $ bea r eal  pre - Hi lb ert  s p ace an d l e t $ T: H \to  H $be an  iso m e t r y so  th at $ ||T(x )-T(y)|| = || x -  y|| $  fo r all  $x,y \i nH$. P rove $ T$ is  a ffine.  
How do  I p rove this ?
A: 

R ep lace$ T$ by$S( x)= T(x)- T(0)$ .   $ S$ i s  a n isometry with
 $ S( 0 )  =  0$, and   thet as ki s to pro ve  th at $ S $  is l inea r . 
Now $S $ pres e rv es  both d is tances  a ndnor ms. W riti ng out 
$$
||S( a)-S( b )||^2 = d(S(a) ,  S(b))^2 = d( a ,b ) ^2  = | |a- b||^2,
$$
a nd e x pand ingg iv es$ $\lan gle S (a ) ,S (b)\rangle  = \lang le  a, b\rang le.$$
  Using  this, com p u t e 
$$
\l angl e  S ( a + \lambda b)  - S( a) - \lamb d a S(b),
 S(a  + \lamb da b) - S ( a ) - \lam bda  S( b)\ ran g l e= 0.
$$
Hence $ S(a+\lambda  b)  - S(a)  -\la mbd a S (b ) = 0$, a nd $S$ i sli ne ar .
A ddend u m:  Mayb ethe  l ast  step is mor e pal atab le  i n  th e follo w in g form :Fo r al l e le ments  $S( z )$in therange of$S$ ,  one  h as 
$$
\la ngle S(a + \l am bda b) - S (a ) -  \lamb d a  S(b), S (z) \rangle = 0.
$$
Hen c e for a ll$y
www.fixkar.com/articles/kb/1150/ and_changed my_site collection administrator but_it didn't_change_anything in_the_log file...
If you_have a piece_of clue it will_be so great!
Thank_you_by advance and sorry for my bad english

A:

Assuming that you ran the Powershell shell_run_as Administrator,_what_happened_if you specified the -force_parameter? Did anything in the_target list_get created?

 Q:

Map of Pre-Hilbert Space is affine_if_it is an_isometry

Let $(H, <\cdot,\cdot>)   $ be a real_pre-Hilbert space and let $T: H_\to H$ be_an_isometry_so that $||T(x)-T(y)|| =_||x - y||$ for all $x,_y \in H$. Prove $T$ is_affine.  
How do I prove this?

A:

Replace_$T$ by $S(x) = T(x) -_T(0)$.  $S$ is an_isometry with
_$S(0) = 0$, and _the task is_to prove_that $S$ is_linear.  
Now $S$ preserves both_distances and norms._ Writing out
$$
||S(a)-S(b)||^2 = d(S(a), S(b))^2_=_d(a, b)^2 _=_||a-b||^2,
$$
and_expanding gives_$$\langle S(a), S(b)\rangle__= \langle_a,_b \rangle.$$
  Using this, compute_
$$
\langle_S(a + \lambda b) - S(a) -_\lambda S(b), 
 S(a_+_\lambda b) - S(a)_- \lambda S(b)\rangle = 0.
$$
Hence_$S(a + \lambda b) - S(a)_- \lambda_S(b) =_0$, and $S$ is linear.
Addendum:  Maybe the last step is_more palatable in the following form:
For_all elements $S(z)$ in_the range_of_$S$, one has
$$
\langle_S(a_+ \lambda_b) - S(a) - \lambda S(b), S(z)_\rangle =_0.
$$
Hence for all $y