Relation of absolute values inside integrals
Clash Royale CLAN TAG#URR8PPP
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I have to prove the following:
$$||gu||_2 leq ||u||_2$$
where $g(w)=(2n^2/omega^2)(1-cos(omega/n)), n in mathbbN$, and $u(omega)$ a general function $in$ $L^2$.
Can I state the following?
$$int_-infty^+infty|(2n^2/omega^2)(1-cos(omega/n))u(omega)|^2domega leq int_-infty^+infty|(2n^2/omega^2)u(omega)|^2domega leq int_-infty^+infty|u(omega)|^2domega$$
The first passage is pretty obvious because $|(1-cos(omega/n))| leq 1$, but the second one is actually true when $omega rightarrow infty$. If not, how could I prove it?
integration absolute-value
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up vote
0
down vote
favorite
I have to prove the following:
$$||gu||_2 leq ||u||_2$$
where $g(w)=(2n^2/omega^2)(1-cos(omega/n)), n in mathbbN$, and $u(omega)$ a general function $in$ $L^2$.
Can I state the following?
$$int_-infty^+infty|(2n^2/omega^2)(1-cos(omega/n))u(omega)|^2domega leq int_-infty^+infty|(2n^2/omega^2)u(omega)|^2domega leq int_-infty^+infty|u(omega)|^2domega$$
The first passage is pretty obvious because $|(1-cos(omega/n))| leq 1$, but the second one is actually true when $omega rightarrow infty$. If not, how could I prove it?
integration absolute-value
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to prove the following:
$$||gu||_2 leq ||u||_2$$
where $g(w)=(2n^2/omega^2)(1-cos(omega/n)), n in mathbbN$, and $u(omega)$ a general function $in$ $L^2$.
Can I state the following?
$$int_-infty^+infty|(2n^2/omega^2)(1-cos(omega/n))u(omega)|^2domega leq int_-infty^+infty|(2n^2/omega^2)u(omega)|^2domega leq int_-infty^+infty|u(omega)|^2domega$$
The first passage is pretty obvious because $|(1-cos(omega/n))| leq 1$, but the second one is actually true when $omega rightarrow infty$. If not, how could I prove it?
integration absolute-value
I have to prove the following:
$$||gu||_2 leq ||u||_2$$
where $g(w)=(2n^2/omega^2)(1-cos(omega/n)), n in mathbbN$, and $u(omega)$ a general function $in$ $L^2$.
Can I state the following?
$$int_-infty^+infty|(2n^2/omega^2)(1-cos(omega/n))u(omega)|^2domega leq int_-infty^+infty|(2n^2/omega^2)u(omega)|^2domega leq int_-infty^+infty|u(omega)|^2domega$$
The first passage is pretty obvious because $|(1-cos(omega/n))| leq 1$, but the second one is actually true when $omega rightarrow infty$. If not, how could I prove it?
integration absolute-value
asked Jul 24 at 8:49
giovanni_13
566
566
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1 Answer
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$1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.
oh thank you very much!
– giovanni_13
Jul 24 at 8:57
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.
oh thank you very much!
– giovanni_13
Jul 24 at 8:57
add a comment |Â
up vote
0
down vote
accepted
$1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.
oh thank you very much!
– giovanni_13
Jul 24 at 8:57
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.
$1-cos t leq frac t^2 2$. This gives the desired inequality. In your argument both steps are wrong. $1-cos t$ can take the value $2$ so the first step is wrong. The second step is also wrong.
answered Jul 24 at 8:54


Kavi Rama Murthy
20.2k2829
20.2k2829
oh thank you very much!
– giovanni_13
Jul 24 at 8:57
add a comment |Â
oh thank you very much!
– giovanni_13
Jul 24 at 8:57
oh thank you very much!
– giovanni_13
Jul 24 at 8:57
oh thank you very much!
– giovanni_13
Jul 24 at 8:57
add a comment |Â
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