If $varphiin C_c^infty(mathbb R)$, then $left|varphiright|lesup_{xinmathbb R}left|varphi'(x)right|$ [closed]
Let $varphiin C_c^infty(mathbb R)$ and $K:=operatorname{supp}varphi$. By the mean value theorem, $$forall-infty<a<b<infty:exists cin(a,b):varphi'(c)=frac{varphi(b)-varphi(a)}{b-a}tag1$$ and hence $$left|varphi(a)-varphi(b)right|lesup_{cinmathbb R}left|varphi'(c)right|left|a-bright|;;;text{for all }a,binmathbb Rtag2.$$ So, $$left|varphi(a)right|leleft|varphi(0)right|+sup_{cinmathbb R}left|varphi'(c)right|sup_{xin K}left|xright|;;;text{for all }ainmathbb Rtag3.$$
However, I've read that $left|varphiright|$ is actually bounded by $sup_{xinmathbb R}left|varphi'(x)right|$. How can we show that?
calculus analysis distribution-theory
asked Dec 12 '18 at 22:40
0xbadf00d0xbadf00d
1,77741430
closed as unclear what you're asking by Did, BigbearZzz, Davide Giraudo, Joel Reyes Noche, kjetil b halvorsen Dec 16 '18 at 14:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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Let $varphiin C_c^infty(mathbb R)$ and $K:=operatorname{supp}varphi$. By the mean value theorem, $$forall-infty<a<b<infty:exists cin(a,b):varphi'(c)=frac{varphi(b)-varphi(a)}{b-a}tag1$$ and hence $$left|varphi(a)-varphi(b)right|lesup_{cinmathbb R}left|varphi'(c)right|left|a-bright|;;;text{for all }a,binmathbb Rtag2.$$ So, $$left|varphi(a)right|leleft|varphi(0)right|+sup_{cinmathbb R}left|varphi'(c)right|sup_{xin K}left|xright|;;;text{for all }ainmathbb Rtag3.$$
However, I've read that $left|varphiright|$ is actually bounded by $sup_{xinmathbb R}left|varphi'(x)right|$. How can we show that?
calculus analysis distribution-theory
asked Dec 12 '18 at 22:40
0xbadf00d0xbadf00d
1,77741430
closed as unclear what you're asking by Did, BigbearZzz, Davide Giraudo, Joel Reyes Noche, kjetil b halvorsen Dec 16 '18 at 14:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
Where did you read this? I think the statement is false.
– Kavi Rama Murthy
Dec 12 '18 at 23:20
2
It is slightly sad that you come here to make us check an obviously wrong statement, but choose to vanish into thin air when asked about the source of said statement.
– Did
Dec 16 '18 at 11:27
add a comment |
Let $varphiin C_c^infty(mathbb R)$ and $K:=operatorname{supp}varphi$. By the mean value theorem, $$forall-infty<a<b<infty:exists cin(a,b):varphi'(c)=frac{varphi(b)-varphi(a)}{b-a}tag1$$ and hence $$left|varphi(a)-varphi(b)right|lesup_{cinmathbb R}left|varphi'(c)right|left|a-bright|;;;text{for all }a,binmathbb Rtag2.$$ So, $$left|varphi(a)right|leleft|varphi(0)right|+sup_{cinmathbb R}left|varphi'(c)right|sup_{xin K}left|xright|;;;text{for all }ainmathbb Rtag3.$$
However, I've read that $left|varphiright|$ is actually bounded by $sup_{xinmathbb R}left|varphi'(x)right|$. How can we show that?
calculus analysis distribution-theory
asked Dec 12 '18 at 22:40
0xbadf00d0xbadf00d
1,77741430
Let $varphiin C_c^infty(mathbb R)$ and $K:=operatorname{supp}varphi$. By the mean value theorem, $$forall-infty<a<b<infty:exists cin(a,b):varphi'(c)=frac{varphi(b)-varphi(a)}{b-a}tag1$$ and hence $$left|varphi(a)-varphi(b)right|lesup_{cinmathbb R}left|varphi'(c)right|left|a-bright|;;;text{for all }a,binmathbb Rtag2.$$ So, $$left|varphi(a)right|leleft|varphi(0)right|+sup_{cinmathbb R}left|varphi'(c)right|sup_{xin K}left|xright|;;;text{for all }ainmathbb Rtag3.$$
However, I've read that $left|varphiright|$ is actually bounded by $sup_{xinmathbb R}left|varphi'(x)right|$. How can we show that?
calculus analysis distribution-theory
calculus analysis distribution-theory
asked Dec 12 '18 at 22:40
0xbadf00d0xbadf00d
1,77741430
asked Dec 12 '18 at 22:40
0xbadf00d0xbadf00d
1,77741430
asked Dec 12 '18 at 22:40
0xbadf00d0xbadf00d
1,77741430
asked Dec 12 '18 at 22:40
0xbadf00d0xbadf00d
1,77741430
asked Dec 12 '18 at 22:40
0xbadf00d0xbadf00d
1,77741430
1,77741430
closed as unclear what you're asking by Did, BigbearZzz, Davide Giraudo, Joel Reyes Noche, kjetil b halvorsen Dec 16 '18 at 14:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, BigbearZzz, Davide Giraudo, Joel Reyes Noche, kjetil b halvorsen Dec 16 '18 at 14:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
Where did you read this? I think the statement is false.
– Kavi Rama Murthy
Dec 12 '18 at 23:20
2
It is slightly sad that you come here to make us check an obviously wrong statement, but choose to vanish into thin air when asked about the source of said statement.
– Did
Dec 16 '18 at 11:27
add a comment |
4
Where did you read this? I think the statement is false.
– Kavi Rama Murthy
Dec 12 '18 at 23:20
2
It is slightly sad that you come here to make us check an obviously wrong statement, but choose to vanish into thin air when asked about the source of said statement.
– Did
Dec 16 '18 at 11:27
4
4
Where did you read this? I think the statement is false.
– Kavi Rama Murthy
Dec 12 '18 at 23:20
Where did you read this? I think the statement is false.
– Kavi Rama Murthy
Dec 12 '18 at 23:20
2
2
It is slightly sad that you come here to make us check an obviously wrong statement, but choose to vanish into thin air when asked about the source of said statement.
– Did
Dec 16 '18 at 11:27
It is slightly sad that you come here to make us check an obviously wrong statement, but choose to vanish into thin air when asked about the source of said statement.
– Did
Dec 16 '18 at 11:27
add a comment |
1 Answer
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I am pretty sure that you need to factor in the size of the support somehow.
Else, given some function $phi$, you can consider $phi_r:x longmapsto phi(rx)$ where $r>0$ is very small. Every $phi_r$ is smooth compactly supported and has max-norm the max-norm of $phi$; the max-norm of $phi_r’$ is $r$ times the max-norm of $phi’$ so can be arbitrarily small.
Now, if $phi$ has support in $[-1,1]$, and if $|a| leq 1$, there exists $r$ such that $|a-r| leq 1$ and $phi(r)=0$. So we know that $phi(a)-phi(r)$ can be written as $(a-r)phi’(c)$ for some $c$, so that $|phi(a)| leq 1*|phi’(c)| leq |phi’|_{infty}$.
answered Dec 12 '18 at 22:55
MindlackMindlack
2,33217
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I am pretty sure that you need to factor in the size of the support somehow.
Else, given some function $phi$, you can consider $phi_r:x longmapsto phi(rx)$ where $r>0$ is very small. Every $phi_r$ is smooth compactly supported and has max-norm the max-norm of $phi$; the max-norm of $phi_r’$ is $r$ times the max-norm of $phi’$ so can be arbitrarily small.
Now, if $phi$ has support in $[-1,1]$, and if $|a| leq 1$, there exists $r$ such that $|a-r| leq 1$ and $phi(r)=0$. So we know that $phi(a)-phi(r)$ can be written as $(a-r)phi’(c)$ for some $c$, so that $|phi(a)| leq 1*|phi’(c)| leq |phi’|_{infty}$.
answered Dec 12 '18 at 22:55
MindlackMindlack
2,33217
add a comment |
I am pretty sure that you need to factor in the size of the support somehow.
Else, given some function $phi$, you can consider $phi_r:x longmapsto phi(rx)$ where $r>0$ is very small. Every $phi_r$ is smooth compactly supported and has max-norm the max-norm of $phi$; the max-norm of $phi_r’$ is $r$ times the max-norm of $phi’$ so can be arbitrarily small.
Now, if $phi$ has support in $[-1,1]$, and if $|a| leq 1$, there exists $r$ such that $|a-r| leq 1$ and $phi(r)=0$. So we know that $phi(a)-phi(r)$ can be written as $(a-r)phi’(c)$ for some $c$, so that $|phi(a)| leq 1*|phi’(c)| leq |phi’|_{infty}$.
answered Dec 12 '18 at 22:55
MindlackMindlack
2,33217
add a comment |
I am pretty sure that you need to factor in the size of the support somehow.
Else, given some function $phi$, you can consider $phi_r:x longmapsto phi(rx)$ where $r>0$ is very small. Every $phi_r$ is smooth compactly supported and has max-norm the max-norm of $phi$; the max-norm of $phi_r’$ is $r$ times the max-norm of $phi’$ so can be arbitrarily small.
Now, if $phi$ has support in $[-1,1]$, and if $|a| leq 1$, there exists $r$ such that $|a-r| leq 1$ and $phi(r)=0$. So we know that $phi(a)-phi(r)$ can be written as $(a-r)phi’(c)$ for some $c$, so that $|phi(a)| leq 1*|phi’(c)| leq |phi’|_{infty}$.
answered Dec 12 '18 at 22:55
MindlackMindlack
2,33217
I am pretty sure that you need to factor in the size of the support somehow.
Else, given some function $phi$, you can consider $phi_r:x longmapsto phi(rx)$ where $r>0$ is very small. Every $phi_r$ is smooth compactly supported and has max-norm the max-norm of $phi$; the max-norm of $phi_r’$ is $r$ times the max-norm of $phi’$ so can be arbitrarily small.
Now, if $phi$ has support in $[-1,1]$, and if $|a| leq 1$, there exists $r$ such that $|a-r| leq 1$ and $phi(r)=0$. So we know that $phi(a)-phi(r)$ can be written as $(a-r)phi’(c)$ for some $c$, so that $|phi(a)| leq 1*|phi’(c)| leq |phi’|_{infty}$.
answered Dec 12 '18 at 22:55
MindlackMindlack
2,33217
answered Dec 12 '18 at 22:55
MindlackMindlack
2,33217
answered Dec 12 '18 at 22:55
MindlackMindlack
2,33217
answered Dec 12 '18 at 22:55
MindlackMindlack
2,33217
2,33217
add a comment |
add a comment |
4
Where did you read this? I think the statement is false.
– Kavi Rama Murthy
Dec 12 '18 at 23:20
2
It is slightly sad that you come here to make us check an obviously wrong statement, but choose to vanish into thin air when asked about the source of said statement.
– Did
Dec 16 '18 at 11:27