Is the squared difference between a $L^{2}$-function and a Non-$L^{2}$-function in $L^{2}$?












1












$begingroup$


Let $(Omega, mathcal{A}, P)$ be a probability space. Furthermore, let $Y,V : Omega rightarrow mathbb{R}^n$ be random vectors and let



$$ int_{Omega}{vert vert Y vert vert ^2}dP < infty$$
and
$$ int_{Omega}{vert vert V vert vert ^2}dP = infty.$$ Is it true that this implies



$$ int_{Omega}{vert vert V-Y vert vert ^2}dP = infty?$$



It seems intuitive, but I have a hard time showing it rigorously.



Thanks for any help!






integration probability-theory lebesgue-integral lp-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    The use of $| cdot |$ is weird. I would use $left| cdot right|$ instead.
    $endgroup$
    – Yanko
    Jan 18 at 20:17










  • $begingroup$
    I used $lVert . rVert$ since the exercise is about random vectors and not scalars.
    $endgroup$
    – Joker123
    Jan 24 at 18:41


















1












$begingroup$


Let $(Omega, mathcal{A}, P)$ be a probability space. Furthermore, let $Y,V : Omega rightarrow mathbb{R}^n$ be random vectors and let



$$ int_{Omega}{vert vert Y vert vert ^2}dP < infty$$
and
$$ int_{Omega}{vert vert V vert vert ^2}dP = infty.$$ Is it true that this implies



$$ int_{Omega}{vert vert V-Y vert vert ^2}dP = infty?$$



It seems intuitive, but I have a hard time showing it rigorously.



Thanks for any help!






integration probability-theory lebesgue-integral lp-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    The use of $| cdot |$ is weird. I would use $left| cdot right|$ instead.
    $endgroup$
    – Yanko
    Jan 18 at 20:17










  • $begingroup$
    I used $lVert . rVert$ since the exercise is about random vectors and not scalars.
    $endgroup$
    – Joker123
    Jan 24 at 18:41
















1












1








1





$begingroup$


Let $(Omega, mathcal{A}, P)$ be a probability space. Furthermore, let $Y,V : Omega rightarrow mathbb{R}^n$ be random vectors and let



$$ int_{Omega}{vert vert Y vert vert ^2}dP < infty$$
and
$$ int_{Omega}{vert vert V vert vert ^2}dP = infty.$$ Is it true that this implies



$$ int_{Omega}{vert vert V-Y vert vert ^2}dP = infty?$$



It seems intuitive, but I have a hard time showing it rigorously.



Thanks for any help!






integration probability-theory lebesgue-integral lp-spaces







share|cite|improve this question









$endgroup$




Let $(Omega, mathcal{A}, P)$ be a probability space. Furthermore, let $Y,V : Omega rightarrow mathbb{R}^n$ be random vectors and let



$$ int_{Omega}{vert vert Y vert vert ^2}dP < infty$$
and
$$ int_{Omega}{vert vert V vert vert ^2}dP = infty.$$ Is it true that this implies



$$ int_{Omega}{vert vert V-Y vert vert ^2}dP = infty?$$



It seems intuitive, but I have a hard time showing it rigorously.



Thanks for any help!






integration probability-theory lebesgue-integral lp-spaces




integration probability-theory lebesgue-integral lp-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 18 at 19:53









Joker123Joker123

527212




527212












  • $begingroup$
    The use of $| cdot |$ is weird. I would use $left| cdot right|$ instead.
    $endgroup$
    – Yanko
    Jan 18 at 20:17










  • $begingroup$
    I used $lVert . rVert$ since the exercise is about random vectors and not scalars.
    $endgroup$
    – Joker123
    Jan 24 at 18:41




















  • $begingroup$
    The use of $| cdot |$ is weird. I would use $left| cdot right|$ instead.
    $endgroup$
    – Yanko
    Jan 18 at 20:17










  • $begingroup$
    I used $lVert . rVert$ since the exercise is about random vectors and not scalars.
    $endgroup$
    – Joker123
    Jan 24 at 18:41


















$begingroup$
The use of $| cdot |$ is weird. I would use $left| cdot right|$ instead.
$endgroup$
– Yanko
Jan 18 at 20:17




$begingroup$
The use of $| cdot |$ is weird. I would use $left| cdot right|$ instead.
$endgroup$
– Yanko
Jan 18 at 20:17












$begingroup$
I used $lVert . rVert$ since the exercise is about random vectors and not scalars.
$endgroup$
– Joker123
Jan 24 at 18:41






$begingroup$
I used $lVert . rVert$ since the exercise is about random vectors and not scalars.
$endgroup$
– Joker123
Jan 24 at 18:41












1 Answer
1






active

oldest

votes


















2












$begingroup$

Notice that $|V| leq |V-Y| + |Y|$ (by the triangle inequality) and so
$$int |V|^2 leq int (|V-Y| + |Y|)^2 leq 4 int |V-Y|^2 + |Y|^2$$
so if $int |V-Y|^2 < infty$ then $int |V|^2 < infty$, a contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Can you elaborate on the second inequality?
    $endgroup$
    – Joker123
    Jan 18 at 22:01










  • $begingroup$
    Because I feel the leading factor should be 2 and not 4.
    $endgroup$
    – Joker123
    Jan 18 at 22:03






  • 1




    $begingroup$
    @Joker123 yeah, it works with $2$. I just stuck the inefficient bound you get by writing $(a+b)^2 leq (2 max(a,b))^2 leq 4 (a^2+b^2)$.
    $endgroup$
    – Rhys Steele
    Jan 18 at 22:56










  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Joker123
    Jan 19 at 15:04











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1 Answer
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1 Answer
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2












$begingroup$

Notice that $|V| leq |V-Y| + |Y|$ (by the triangle inequality) and so
$$int |V|^2 leq int (|V-Y| + |Y|)^2 leq 4 int |V-Y|^2 + |Y|^2$$
so if $int |V-Y|^2 < infty$ then $int |V|^2 < infty$, a contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Can you elaborate on the second inequality?
    $endgroup$
    – Joker123
    Jan 18 at 22:01










  • $begingroup$
    Because I feel the leading factor should be 2 and not 4.
    $endgroup$
    – Joker123
    Jan 18 at 22:03






  • 1




    $begingroup$
    @Joker123 yeah, it works with $2$. I just stuck the inefficient bound you get by writing $(a+b)^2 leq (2 max(a,b))^2 leq 4 (a^2+b^2)$.
    $endgroup$
    – Rhys Steele
    Jan 18 at 22:56










  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Joker123
    Jan 19 at 15:04
















2












$begingroup$

Notice that $|V| leq |V-Y| + |Y|$ (by the triangle inequality) and so
$$int |V|^2 leq int (|V-Y| + |Y|)^2 leq 4 int |V-Y|^2 + |Y|^2$$
so if $int |V-Y|^2 < infty$ then $int |V|^2 < infty$, a contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Can you elaborate on the second inequality?
    $endgroup$
    – Joker123
    Jan 18 at 22:01










  • $begingroup$
    Because I feel the leading factor should be 2 and not 4.
    $endgroup$
    – Joker123
    Jan 18 at 22:03






  • 1




    $begingroup$
    @Joker123 yeah, it works with $2$. I just stuck the inefficient bound you get by writing $(a+b)^2 leq (2 max(a,b))^2 leq 4 (a^2+b^2)$.
    $endgroup$
    – Rhys Steele
    Jan 18 at 22:56










  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Joker123
    Jan 19 at 15:04














2












2








2





$begingroup$

Notice that $|V| leq |V-Y| + |Y|$ (by the triangle inequality) and so
$$int |V|^2 leq int (|V-Y| + |Y|)^2 leq 4 int |V-Y|^2 + |Y|^2$$
so if $int |V-Y|^2 < infty$ then $int |V|^2 < infty$, a contradiction.






share|cite|improve this answer









$endgroup$



Notice that $|V| leq |V-Y| + |Y|$ (by the triangle inequality) and so
$$int |V|^2 leq int (|V-Y| + |Y|)^2 leq 4 int |V-Y|^2 + |Y|^2$$
so if $int |V-Y|^2 < infty$ then $int |V|^2 < infty$, a contradiction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 18 at 19:58









Rhys SteeleRhys Steele

6,5851829




6,5851829












  • $begingroup$
    Thank you! Can you elaborate on the second inequality?
    $endgroup$
    – Joker123
    Jan 18 at 22:01










  • $begingroup$
    Because I feel the leading factor should be 2 and not 4.
    $endgroup$
    – Joker123
    Jan 18 at 22:03






  • 1




    $begingroup$
    @Joker123 yeah, it works with $2$. I just stuck the inefficient bound you get by writing $(a+b)^2 leq (2 max(a,b))^2 leq 4 (a^2+b^2)$.
    $endgroup$
    – Rhys Steele
    Jan 18 at 22:56










  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Joker123
    Jan 19 at 15:04


















  • $begingroup$
    Thank you! Can you elaborate on the second inequality?
    $endgroup$
    – Joker123
    Jan 18 at 22:01










  • $begingroup$
    Because I feel the leading factor should be 2 and not 4.
    $endgroup$
    – Joker123
    Jan 18 at 22:03






  • 1




    $begingroup$
    @Joker123 yeah, it works with $2$. I just stuck the inefficient bound you get by writing $(a+b)^2 leq (2 max(a,b))^2 leq 4 (a^2+b^2)$.
    $endgroup$
    – Rhys Steele
    Jan 18 at 22:56










  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Joker123
    Jan 19 at 15:04
















$begingroup$
Thank you! Can you elaborate on the second inequality?
$endgroup$
– Joker123
Jan 18 at 22:01




$begingroup$
Thank you! Can you elaborate on the second inequality?
$endgroup$
– Joker123
Jan 18 at 22:01












$begingroup$
Because I feel the leading factor should be 2 and not 4.
$endgroup$
– Joker123
Jan 18 at 22:03




$begingroup$
Because I feel the leading factor should be 2 and not 4.
$endgroup$
– Joker123
Jan 18 at 22:03




1




1




$begingroup$
@Joker123 yeah, it works with $2$. I just stuck the inefficient bound you get by writing $(a+b)^2 leq (2 max(a,b))^2 leq 4 (a^2+b^2)$.
$endgroup$
– Rhys Steele
Jan 18 at 22:56




$begingroup$
@Joker123 yeah, it works with $2$. I just stuck the inefficient bound you get by writing $(a+b)^2 leq (2 max(a,b))^2 leq 4 (a^2+b^2)$.
$endgroup$
– Rhys Steele
Jan 18 at 22:56












$begingroup$
Thank you for your help!
$endgroup$
– Joker123
Jan 19 at 15:04




$begingroup$
Thank you for your help!
$endgroup$
– Joker123
Jan 19 at 15:04


















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