$f in L^1, g in L^1$ s.t. $f cdot g not in L^1$












-1












$begingroup$


For what functions for instance stands: $f in L^1, g in L^1$ but $f cdot g$ are not in $L^1$? Thank you






real-analysis functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
    $endgroup$
    – twnly
    Jan 18 at 9:22












  • $begingroup$
    I am asking for multiplication. It is connected with definition of convolution
    $endgroup$
    – Lukáš Altman
    Jan 18 at 9:24
















-1












$begingroup$


For what functions for instance stands: $f in L^1, g in L^1$ but $f cdot g$ are not in $L^1$? Thank you






real-analysis functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
    $endgroup$
    – twnly
    Jan 18 at 9:22












  • $begingroup$
    I am asking for multiplication. It is connected with definition of convolution
    $endgroup$
    – Lukáš Altman
    Jan 18 at 9:24














-1












-1








-1





$begingroup$


For what functions for instance stands: $f in L^1, g in L^1$ but $f cdot g$ are not in $L^1$? Thank you






real-analysis functions







share|cite|improve this question











$endgroup$




For what functions for instance stands: $f in L^1, g in L^1$ but $f cdot g$ are not in $L^1$? Thank you






real-analysis functions




real-analysis functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 12:18









Andrews

6741318




6741318










asked Jan 18 at 9:10









Lukáš AltmanLukáš Altman

11




11












  • $begingroup$
    The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
    $endgroup$
    – twnly
    Jan 18 at 9:22












  • $begingroup$
    I am asking for multiplication. It is connected with definition of convolution
    $endgroup$
    – Lukáš Altman
    Jan 18 at 9:24


















  • $begingroup$
    The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
    $endgroup$
    – twnly
    Jan 18 at 9:22












  • $begingroup$
    I am asking for multiplication. It is connected with definition of convolution
    $endgroup$
    – Lukáš Altman
    Jan 18 at 9:24
















$begingroup$
The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
$endgroup$
– twnly
Jan 18 at 9:22






$begingroup$
The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
$endgroup$
– twnly
Jan 18 at 9:22














$begingroup$
I am asking for multiplication. It is connected with definition of convolution
$endgroup$
– Lukáš Altman
Jan 18 at 9:24




$begingroup$
I am asking for multiplication. It is connected with definition of convolution
$endgroup$
– Lukáš Altman
Jan 18 at 9:24










1 Answer
1






active

oldest

votes


















2












$begingroup$

$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
    $endgroup$
    – Lukáš Altman
    Jan 18 at 9:20












  • $begingroup$
    You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
    $endgroup$
    – Klaus
    Jan 18 at 9:39










  • $begingroup$
    Thank you very much
    $endgroup$
    – Lukáš Altman
    Jan 18 at 10:58











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
    $endgroup$
    – Lukáš Altman
    Jan 18 at 9:20












  • $begingroup$
    You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
    $endgroup$
    – Klaus
    Jan 18 at 9:39










  • $begingroup$
    Thank you very much
    $endgroup$
    – Lukáš Altman
    Jan 18 at 10:58
















2












$begingroup$

$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
    $endgroup$
    – Lukáš Altman
    Jan 18 at 9:20












  • $begingroup$
    You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
    $endgroup$
    – Klaus
    Jan 18 at 9:39










  • $begingroup$
    Thank you very much
    $endgroup$
    – Lukáš Altman
    Jan 18 at 10:58














2












2








2





$begingroup$

$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$






share|cite|improve this answer









$endgroup$



$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 18 at 9:12









KlausKlaus

2,0099




2,0099












  • $begingroup$
    Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
    $endgroup$
    – Lukáš Altman
    Jan 18 at 9:20












  • $begingroup$
    You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
    $endgroup$
    – Klaus
    Jan 18 at 9:39










  • $begingroup$
    Thank you very much
    $endgroup$
    – Lukáš Altman
    Jan 18 at 10:58


















  • $begingroup$
    Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
    $endgroup$
    – Lukáš Altman
    Jan 18 at 9:20












  • $begingroup$
    You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
    $endgroup$
    – Klaus
    Jan 18 at 9:39










  • $begingroup$
    Thank you very much
    $endgroup$
    – Lukáš Altman
    Jan 18 at 10:58
















$begingroup$
Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
$endgroup$
– Lukáš Altman
Jan 18 at 9:20






$begingroup$
Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
$endgroup$
– Lukáš Altman
Jan 18 at 9:20














$begingroup$
You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
$endgroup$
– Klaus
Jan 18 at 9:39




$begingroup$
You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
$endgroup$
– Klaus
Jan 18 at 9:39












$begingroup$
Thank you very much
$endgroup$
– Lukáš Altman
Jan 18 at 10:58




$begingroup$
Thank you very much
$endgroup$
– Lukáš Altman
Jan 18 at 10:58


















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