Essential supremum of $n sin (n^2x)$
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I have the function $g_n(x): n sin (n^2x)$ for $x in [0,1]$, $n leq 2$ and I would like to calculate the essential supremum, i.e. $||g_n||_infty= $inf${C geq 0 : m(|g_n(x)| > C)=0}$ where $m$ is a Lebesgue measure on $[0,1]$. I think that since $n sin(n^2x) leq n$ then $||g_n||_infty=n$, but I'm not sure if I'm making some mistakes. Can somebody help me?
functional-analysis lebesgue-measure norm
asked Dec 13 '18 at 14:10
user289143user289143
903313
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add a comment |
$begingroup$
I have the function $g_n(x): n sin (n^2x)$ for $x in [0,1]$, $n leq 2$ and I would like to calculate the essential supremum, i.e. $||g_n||_infty= $inf${C geq 0 : m(|g_n(x)| > C)=0}$ where $m$ is a Lebesgue measure on $[0,1]$. I think that since $n sin(n^2x) leq n$ then $||g_n||_infty=n$, but I'm not sure if I'm making some mistakes. Can somebody help me?
functional-analysis lebesgue-measure norm
asked Dec 13 '18 at 14:10
user289143user289143
903313
$endgroup$
3
$begingroup$
Isn't $nsin(n^2x)$ continuous, so the essential supremum equals the maximum?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:13
add a comment |
$begingroup$
I have the function $g_n(x): n sin (n^2x)$ for $x in [0,1]$, $n leq 2$ and I would like to calculate the essential supremum, i.e. $||g_n||_infty= $inf${C geq 0 : m(|g_n(x)| > C)=0}$ where $m$ is a Lebesgue measure on $[0,1]$. I think that since $n sin(n^2x) leq n$ then $||g_n||_infty=n$, but I'm not sure if I'm making some mistakes. Can somebody help me?
functional-analysis lebesgue-measure norm
asked Dec 13 '18 at 14:10
user289143user289143
903313
$endgroup$
I have the function $g_n(x): n sin (n^2x)$ for $x in [0,1]$, $n leq 2$ and I would like to calculate the essential supremum, i.e. $||g_n||_infty= $inf${C geq 0 : m(|g_n(x)| > C)=0}$ where $m$ is a Lebesgue measure on $[0,1]$. I think that since $n sin(n^2x) leq n$ then $||g_n||_infty=n$, but I'm not sure if I'm making some mistakes. Can somebody help me?
functional-analysis lebesgue-measure norm
functional-analysis lebesgue-measure norm
asked Dec 13 '18 at 14:10
user289143user289143
903313
asked Dec 13 '18 at 14:10
user289143user289143
903313
edited Dec 13 '18 at 14:22
user289143
asked Dec 13 '18 at 14:10
user289143user289143
903313
asked Dec 13 '18 at 14:10
user289143user289143
903313
asked Dec 13 '18 at 14:10
user289143user289143
903313
903313
3
$begingroup$
Isn't $nsin(n^2x)$ continuous, so the essential supremum equals the maximum?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:13
add a comment |
3
$begingroup$
Isn't $nsin(n^2x)$ continuous, so the essential supremum equals the maximum?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:13
3
3
$begingroup$
Isn't $nsin(n^2x)$ continuous, so the essential supremum equals the maximum?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:13
$begingroup$
Isn't $nsin(n^2x)$ continuous, so the essential supremum equals the maximum?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:13
add a comment |
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$begingroup$
Isn't $nsin(n^2x)$ continuous, so the essential supremum equals the maximum?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:13