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The support of a real-valued functions

Continuity of measureCharacteristic functions (Statistics)Support of measurable function regular?Which of the following is not uniformly continuous?Support of a discrete measureProving that $f$ functions of compact support are linearly independant.$maxf_1,dots,f_n$ continuous for continuous functions $f_i:X to mathbbR$Write trinomials as nonnegative linear combinationLet $F$ the set of all continuous real functions with domain $[0,a]$. Which of the following are metrics on $F$?Proof check: continuous function with compact support is $lambda^d$ integrable

1

$begingroup$

Find the supp($f_i$) for:

1) $f_1=mathbb1_mathbbQ(x)$. The rational numbers are dense in $mathbbR$ so the support will be $mathbbR$.

2)$f_2 =x$ . So $f_2(x)=0$ iff $x=0$ so $x in mathbbR:f_2(x)neq0 = [-infty,0) cup (0,+infty]$ and the closure of this set is $mathbbR$.

3) $f_3 =$ max$0,x$. So $x in mathbbR:f_3(x)neq0 = (0,+infty]$ and the closure of this set is $mathbbR^+$

4) $f_4 = e^frac1x^2-1cdotmathbb1_(-1,1)(x)$. Then $x in mathbbR:f_3(x)neq0 = (-1,1)$ and the closure is $[-1,1]$

Are my toughts on this correct?

real-analysis general-topology measure-theory

share|cite|improve this question

asked Mar 13 at 9:28

KingDingelingKingDingeling

1837

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If by $mathbbR^+$ you mean $[0, +infty)$, everything is correct.
  $endgroup$
  – mechanodroid
  Mar 13 at 9:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, should be correct
  $endgroup$
  – supinf
  Mar 13 at 9:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot guys for your help and quick answer! :)
  $endgroup$
  – KingDingeling
  Mar 13 at 9:36
  

  
  
  
  

add a comment | 

1

$begingroup$

Find the supp($f_i$) for:

1) $f_1=mathbb1_mathbbQ(x)$. The rational numbers are dense in $mathbbR$ so the support will be $mathbbR$.

2)$f_2 =x$ . So $f_2(x)=0$ iff $x=0$ so $x in mathbbR:f_2(x)neq0 = [-infty,0) cup (0,+infty]$ and the closure of this set is $mathbbR$.

3) $f_3 =$ max$0,x$. So $x in mathbbR:f_3(x)neq0 = (0,+infty]$ and the closure of this set is $mathbbR^+$

4) $f_4 = e^frac1x^2-1cdotmathbb1_(-1,1)(x)$. Then $x in mathbbR:f_3(x)neq0 = (-1,1)$ and the closure is $[-1,1]$

Are my toughts on this correct?

real-analysis general-topology measure-theory

share|cite|improve this question

asked Mar 13 at 9:28

KingDingelingKingDingeling

1837

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If by $mathbbR^+$ you mean $[0, +infty)$, everything is correct.
  $endgroup$
  – mechanodroid
  Mar 13 at 9:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, should be correct
  $endgroup$
  – supinf
  Mar 13 at 9:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot guys for your help and quick answer! :)
  $endgroup$
  – KingDingeling
  Mar 13 at 9:36
  

  
  
  
  

add a comment | 

$begingroup$

Find the supp($f_i$) for:

1) $f_1=mathbb1_mathbbQ(x)$. The rational numbers are dense in $mathbbR$ so the support will be $mathbbR$.

2)$f_2 =x$ . So $f_2(x)=0$ iff $x=0$ so $x in mathbbR:f_2(x)neq0 = [-infty,0) cup (0,+infty]$ and the closure of this set is $mathbbR$.

3) $f_3 =$ max$0,x$. So $x in mathbbR:f_3(x)neq0 = (0,+infty]$ and the closure of this set is $mathbbR^+$

4) $f_4 = e^frac1x^2-1cdotmathbb1_(-1,1)(x)$. Then $x in mathbbR:f_3(x)neq0 = (-1,1)$ and the closure is $[-1,1]$

Are my toughts on this correct?

real-analysis general-topology measure-theory

share|cite|improve this question

asked Mar 13 at 9:28

KingDingelingKingDingeling

1837

$endgroup$

share|cite|improve this question

asked Mar 13 at 9:28

KingDingelingKingDingeling

1837

share|cite|improve this question

asked Mar 13 at 9:28

KingDingelingKingDingeling

1837

asked Mar 13 at 9:28

KingDingelingKingDingeling

1837

asked Mar 13 at 9:28

KingDingelingKingDingeling

1837