Extending a function defined in $mathbb{R}^nsetminus{0}$ to a continuous function defined in $mathbb{R}^{n}$.
$begingroup$
Let $g: mathbb{R}^nsetminus{0} to mathbb{R}$ be a function of class $C^{1}$ and suppose that there is $M > 0$ such that
$$left|frac{partial}{partial x_{i}}g(x)right| leq M.$$
Prove that if $n geq 2$ then $g$ can be extended to a continuous function defined in $mathbb{R}^n$. Show that if $n = 1$ the statement is false.
My attempt.
I define the extension $bar{g}: mathbb{R}^n to mathbb{R}$ by $bar{g}(x) = g(x)$ if $x in mathbb{R}^{n}setminus{0$} and $displaystyle bar{g}(0) = lim_{x to 0}g(x)$. Thus
$$lim_{x to 0} bar{g}(x) = lim_{x to 0}g(x) = bar{g}(0),$$
so $bar{g}$ is continuous. So the question is reduced to proving that $displaystyle lim_{x to 0}g(x)$ exists. Thus, I must show that
$$forall epsilon > 0, exists delta > 0text{ s.t. } Vert X Vert < delta Longrightarrow |g(x)|<epsilon.$$
The hypothesis
$$left|frac{partial}{partial x_{i}}g(x)right| leq M$$
seems necessary for getting
$$|g(x)-g(y)| leq M|x-y|$$
using the Mean Value Inequality. This almost solves the problem, because if $g(0) = 0$ we can write
$$|g(x)| leq M|x|.$$
But $g(0) = 0$ doesn't make sense. I'm stuck here.
Also, I cannot see why it is necessary that $n geq 2$, where I use this in the demonstration, and why it fails when $n=1$.
real-analysis multivariable-calculus derivatives continuity
edited Jan 20 at 18:49
J. W. Tanner
2,1471117
asked Jan 20 at 18:37
Lucas CorrêaLucas Corrêa
1,5711321
$endgroup$
|
show 6 more comments
$begingroup$
Let $g: mathbb{R}^nsetminus{0} to mathbb{R}$ be a function of class $C^{1}$ and suppose that there is $M > 0$ such that
$$left|frac{partial}{partial x_{i}}g(x)right| leq M.$$
Prove that if $n geq 2$ then $g$ can be extended to a continuous function defined in $mathbb{R}^n$. Show that if $n = 1$ the statement is false.
My attempt.
I define the extension $bar{g}: mathbb{R}^n to mathbb{R}$ by $bar{g}(x) = g(x)$ if $x in mathbb{R}^{n}setminus{0$} and $displaystyle bar{g}(0) = lim_{x to 0}g(x)$. Thus
$$lim_{x to 0} bar{g}(x) = lim_{x to 0}g(x) = bar{g}(0),$$
so $bar{g}$ is continuous. So the question is reduced to proving that $displaystyle lim_{x to 0}g(x)$ exists. Thus, I must show that
$$forall epsilon > 0, exists delta > 0text{ s.t. } Vert X Vert < delta Longrightarrow |g(x)|<epsilon.$$
The hypothesis
$$left|frac{partial}{partial x_{i}}g(x)right| leq M$$
seems necessary for getting
$$|g(x)-g(y)| leq M|x-y|$$
using the Mean Value Inequality. This almost solves the problem, because if $g(0) = 0$ we can write
$$|g(x)| leq M|x|.$$
But $g(0) = 0$ doesn't make sense. I'm stuck here.
Also, I cannot see why it is necessary that $n geq 2$, where I use this in the demonstration, and why it fails when $n=1$.
real-analysis multivariable-calculus derivatives continuity
edited Jan 20 at 18:49
J. W. Tanner
2,1471117
asked Jan 20 at 18:37
Lucas CorrêaLucas Corrêa
1,5711321
$endgroup$
1
$begingroup$
$f(x) = x/|x|$ would be a counter-example for $n=1$. I assume that the connectedness of $mathbb{R}^nsetminus{0} $ for $n ge 2$ plays a role.
$endgroup$
– Martin R
Jan 20 at 18:44
2
$begingroup$
The problem is to show that the limit exist... To show abstracly that the limit exists, use sequential caracterisation and show the sequence is a cauchy sequence with your mean value inequality. If n=1 the mean value inequality fails. Do you see why ?
$endgroup$
– DLeMeur
Jan 20 at 18:44
1
$begingroup$
@DLeMeur, I think fails because $mathbb{R}setminus{0}$ is disconnected, as Martin said. We can define $f(x) = -1$ for $x<0$ and $f(x)=1$ for $x>0$. Then we can find $x,y$ such that $|f(x) - f(y)| = 2$ but $f'(x) = 0$ for all $x$.
$endgroup$
– Lucas Corrêa
Jan 20 at 18:52
1
$begingroup$
Yes. But you can t conclude directly the convergence of $g $.You must also show that L does not depend of $(x_n)$.
$endgroup$
– DLeMeur
Jan 20 at 19:06
1
$begingroup$
To have the MVI apply to $|f(x)-f(y)|,$ you need to know $[x,y]$ does not go through $0.$
$endgroup$
– zhw.
Jan 20 at 19:33
|
show 6 more comments
$begingroup$
Let $g: mathbb{R}^nsetminus{0} to mathbb{R}$ be a function of class $C^{1}$ and suppose that there is $M > 0$ such that
$$left|frac{partial}{partial x_{i}}g(x)right| leq M.$$
Prove that if $n geq 2$ then $g$ can be extended to a continuous function defined in $mathbb{R}^n$. Show that if $n = 1$ the statement is false.
My attempt.
I define the extension $bar{g}: mathbb{R}^n to mathbb{R}$ by $bar{g}(x) = g(x)$ if $x in mathbb{R}^{n}setminus{0$} and $displaystyle bar{g}(0) = lim_{x to 0}g(x)$. Thus
$$lim_{x to 0} bar{g}(x) = lim_{x to 0}g(x) = bar{g}(0),$$
so $bar{g}$ is continuous. So the question is reduced to proving that $displaystyle lim_{x to 0}g(x)$ exists. Thus, I must show that
$$forall epsilon > 0, exists delta > 0text{ s.t. } Vert X Vert < delta Longrightarrow |g(x)|<epsilon.$$
The hypothesis
$$left|frac{partial}{partial x_{i}}g(x)right| leq M$$
seems necessary for getting
$$|g(x)-g(y)| leq M|x-y|$$
using the Mean Value Inequality. This almost solves the problem, because if $g(0) = 0$ we can write
$$|g(x)| leq M|x|.$$
But $g(0) = 0$ doesn't make sense. I'm stuck here.
Also, I cannot see why it is necessary that $n geq 2$, where I use this in the demonstration, and why it fails when $n=1$.
real-analysis multivariable-calculus derivatives continuity
edited Jan 20 at 18:49
J. W. Tanner
2,1471117
asked Jan 20 at 18:37
Lucas CorrêaLucas Corrêa
1,5711321
$endgroup$
Let $g: mathbb{R}^nsetminus{0} to mathbb{R}$ be a function of class $C^{1}$ and suppose that there is $M > 0$ such that
$$left|frac{partial}{partial x_{i}}g(x)right| leq M.$$
Prove that if $n geq 2$ then $g$ can be extended to a continuous function defined in $mathbb{R}^n$. Show that if $n = 1$ the statement is false.
My attempt.
I define the extension $bar{g}: mathbb{R}^n to mathbb{R}$ by $bar{g}(x) = g(x)$ if $x in mathbb{R}^{n}setminus{0$} and $displaystyle bar{g}(0) = lim_{x to 0}g(x)$. Thus
$$lim_{x to 0} bar{g}(x) = lim_{x to 0}g(x) = bar{g}(0),$$
so $bar{g}$ is continuous. So the question is reduced to proving that $displaystyle lim_{x to 0}g(x)$ exists. Thus, I must show that
$$forall epsilon > 0, exists delta > 0text{ s.t. } Vert X Vert < delta Longrightarrow |g(x)|<epsilon.$$
The hypothesis
$$left|frac{partial}{partial x_{i}}g(x)right| leq M$$
seems necessary for getting
$$|g(x)-g(y)| leq M|x-y|$$
using the Mean Value Inequality. This almost solves the problem, because if $g(0) = 0$ we can write
$$|g(x)| leq M|x|.$$
But $g(0) = 0$ doesn't make sense. I'm stuck here.
Also, I cannot see why it is necessary that $n geq 2$, where I use this in the demonstration, and why it fails when $n=1$.
real-analysis multivariable-calculus derivatives continuity
real-analysis multivariable-calculus derivatives continuity
edited Jan 20 at 18:49
J. W. Tanner
2,1471117
asked Jan 20 at 18:37
Lucas CorrêaLucas Corrêa
1,5711321
edited Jan 20 at 18:49
J. W. Tanner
2,1471117
asked Jan 20 at 18:37
Lucas CorrêaLucas Corrêa
1,5711321
edited Jan 20 at 18:49
J. W. Tanner
2,1471117
edited Jan 20 at 18:49
J. W. Tanner
2,1471117
edited Jan 20 at 18:49
J. W. Tanner
2,1471117
2,1471117
asked Jan 20 at 18:37
Lucas CorrêaLucas Corrêa
1,5711321
asked Jan 20 at 18:37
Lucas CorrêaLucas Corrêa
1,5711321
asked Jan 20 at 18:37
Lucas CorrêaLucas Corrêa
1,5711321
1,5711321
1
$begingroup$
$f(x) = x/|x|$ would be a counter-example for $n=1$. I assume that the connectedness of $mathbb{R}^nsetminus{0} $ for $n ge 2$ plays a role.
$endgroup$
– Martin R
Jan 20 at 18:44
2
$begingroup$
The problem is to show that the limit exist... To show abstracly that the limit exists, use sequential caracterisation and show the sequence is a cauchy sequence with your mean value inequality. If n=1 the mean value inequality fails. Do you see why ?
$endgroup$
– DLeMeur
Jan 20 at 18:44
1
$begingroup$
@DLeMeur, I think fails because $mathbb{R}setminus{0}$ is disconnected, as Martin said. We can define $f(x) = -1$ for $x<0$ and $f(x)=1$ for $x>0$. Then we can find $x,y$ such that $|f(x) - f(y)| = 2$ but $f'(x) = 0$ for all $x$.
$endgroup$
– Lucas Corrêa
Jan 20 at 18:52
1
$begingroup$
Yes. But you can t conclude directly the convergence of $g $.You must also show that L does not depend of $(x_n)$.
$endgroup$
– DLeMeur
Jan 20 at 19:06
1
$begingroup$
To have the MVI apply to $|f(x)-f(y)|,$ you need to know $[x,y]$ does not go through $0.$
$endgroup$
– zhw.
Jan 20 at 19:33
|
show 6 more comments
1
$begingroup$
$f(x) = x/|x|$ would be a counter-example for $n=1$. I assume that the connectedness of $mathbb{R}^nsetminus{0} $ for $n ge 2$ plays a role.
$endgroup$
– Martin R
Jan 20 at 18:44
2
$begingroup$
The problem is to show that the limit exist... To show abstracly that the limit exists, use sequential caracterisation and show the sequence is a cauchy sequence with your mean value inequality. If n=1 the mean value inequality fails. Do you see why ?
$endgroup$
– DLeMeur
Jan 20 at 18:44
1
$begingroup$
@DLeMeur, I think fails because $mathbb{R}setminus{0}$ is disconnected, as Martin said. We can define $f(x) = -1$ for $x<0$ and $f(x)=1$ for $x>0$. Then we can find $x,y$ such that $|f(x) - f(y)| = 2$ but $f'(x) = 0$ for all $x$.
$endgroup$
– Lucas Corrêa
Jan 20 at 18:52
1
$begingroup$
Yes. But you can t conclude directly the convergence of $g $.You must also show that L does not depend of $(x_n)$.
$endgroup$
– DLeMeur
Jan 20 at 19:06
1
$begingroup$
To have the MVI apply to $|f(x)-f(y)|,$ you need to know $[x,y]$ does not go through $0.$
$endgroup$
– zhw.
Jan 20 at 19:33
1
1
$begingroup$
$f(x) = x/|x|$ would be a counter-example for $n=1$. I assume that the connectedness of $mathbb{R}^nsetminus{0} $ for $n ge 2$ plays a role.
$endgroup$
– Martin R
Jan 20 at 18:44
$begingroup$
$f(x) = x/|x|$ would be a counter-example for $n=1$. I assume that the connectedness of $mathbb{R}^nsetminus{0} $ for $n ge 2$ plays a role.
$endgroup$
– Martin R
Jan 20 at 18:44
2
2
$begingroup$
The problem is to show that the limit exist... To show abstracly that the limit exists, use sequential caracterisation and show the sequence is a cauchy sequence with your mean value inequality. If n=1 the mean value inequality fails. Do you see why ?
$endgroup$
– DLeMeur
Jan 20 at 18:44
$begingroup$
The problem is to show that the limit exist... To show abstracly that the limit exists, use sequential caracterisation and show the sequence is a cauchy sequence with your mean value inequality. If n=1 the mean value inequality fails. Do you see why ?
$endgroup$
– DLeMeur
Jan 20 at 18:44
1
1
$begingroup$
@DLeMeur, I think fails because $mathbb{R}setminus{0}$ is disconnected, as Martin said. We can define $f(x) = -1$ for $x<0$ and $f(x)=1$ for $x>0$. Then we can find $x,y$ such that $|f(x) - f(y)| = 2$ but $f'(x) = 0$ for all $x$.
$endgroup$
– Lucas Corrêa
Jan 20 at 18:52
$begingroup$
@DLeMeur, I think fails because $mathbb{R}setminus{0}$ is disconnected, as Martin said. We can define $f(x) = -1$ for $x<0$ and $f(x)=1$ for $x>0$. Then we can find $x,y$ such that $|f(x) - f(y)| = 2$ but $f'(x) = 0$ for all $x$.
$endgroup$
– Lucas Corrêa
Jan 20 at 18:52
1
1
$begingroup$
Yes. But you can t conclude directly the convergence of $g $.You must also show that L does not depend of $(x_n)$.
$endgroup$
– DLeMeur
Jan 20 at 19:06
$begingroup$
Yes. But you can t conclude directly the convergence of $g $.You must also show that L does not depend of $(x_n)$.
$endgroup$
– DLeMeur
Jan 20 at 19:06
1
1
$begingroup$
To have the MVI apply to $|f(x)-f(y)|,$ you need to know $[x,y]$ does not go through $0.$
$endgroup$
– zhw.
Jan 20 at 19:33
$begingroup$
To have the MVI apply to $|f(x)-f(y)|,$ you need to know $[x,y]$ does not go through $0.$
$endgroup$
– zhw.
Jan 20 at 19:33
|
show 6 more comments
1 Answer
1
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oldest
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$begingroup$
Hint: Suppose $x,yne 0.$ Let $L$ be the line through $x$ and $0.$ If $y$ is not on $L,$ then $[x,y]$ does not pass through $0.$ Hence $|f(y)-f(x)|le M|y-x|$ by the MVI. If $y$ is on $L,$ Choose $y'ne 0$ close to $y$ but not on $L ,...$
answered Jan 20 at 19:05
zhw.zhw.
73.4k43175
$endgroup$
add a comment |
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$begingroup$
Hint: Suppose $x,yne 0.$ Let $L$ be the line through $x$ and $0.$ If $y$ is not on $L,$ then $[x,y]$ does not pass through $0.$ Hence $|f(y)-f(x)|le M|y-x|$ by the MVI. If $y$ is on $L,$ Choose $y'ne 0$ close to $y$ but not on $L ,...$
answered Jan 20 at 19:05
zhw.zhw.
73.4k43175
$endgroup$
add a comment |
$begingroup$
Hint: Suppose $x,yne 0.$ Let $L$ be the line through $x$ and $0.$ If $y$ is not on $L,$ then $[x,y]$ does not pass through $0.$ Hence $|f(y)-f(x)|le M|y-x|$ by the MVI. If $y$ is on $L,$ Choose $y'ne 0$ close to $y$ but not on $L ,...$
answered Jan 20 at 19:05
zhw.zhw.
73.4k43175
$endgroup$
add a comment |
$begingroup$
Hint: Suppose $x,yne 0.$ Let $L$ be the line through $x$ and $0.$ If $y$ is not on $L,$ then $[x,y]$ does not pass through $0.$ Hence $|f(y)-f(x)|le M|y-x|$ by the MVI. If $y$ is on $L,$ Choose $y'ne 0$ close to $y$ but not on $L ,...$
answered Jan 20 at 19:05
zhw.zhw.
73.4k43175
$endgroup$
Hint: Suppose $x,yne 0.$ Let $L$ be the line through $x$ and $0.$ If $y$ is not on $L,$ then $[x,y]$ does not pass through $0.$ Hence $|f(y)-f(x)|le M|y-x|$ by the MVI. If $y$ is on $L,$ Choose $y'ne 0$ close to $y$ but not on $L ,...$
answered Jan 20 at 19:05
zhw.zhw.
73.4k43175
edited Jan 20 at 19:14
answered Jan 20 at 19:05
zhw.zhw.
73.4k43175
answered Jan 20 at 19:05
zhw.zhw.
73.4k43175
answered Jan 20 at 19:05
zhw.zhw.
73.4k43175
73.4k43175
add a comment |
add a comment |
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$begingroup$
$f(x) = x/|x|$ would be a counter-example for $n=1$. I assume that the connectedness of $mathbb{R}^nsetminus{0} $ for $n ge 2$ plays a role.
$endgroup$
– Martin R
Jan 20 at 18:44
2
$begingroup$
The problem is to show that the limit exist... To show abstracly that the limit exists, use sequential caracterisation and show the sequence is a cauchy sequence with your mean value inequality. If n=1 the mean value inequality fails. Do you see why ?
$endgroup$
– DLeMeur
Jan 20 at 18:44
1
$begingroup$
@DLeMeur, I think fails because $mathbb{R}setminus{0}$ is disconnected, as Martin said. We can define $f(x) = -1$ for $x<0$ and $f(x)=1$ for $x>0$. Then we can find $x,y$ such that $|f(x) - f(y)| = 2$ but $f'(x) = 0$ for all $x$.
$endgroup$
– Lucas Corrêa
Jan 20 at 18:52
1
$begingroup$
Yes. But you can t conclude directly the convergence of $g $.You must also show that L does not depend of $(x_n)$.
$endgroup$
– DLeMeur
Jan 20 at 19:06
1
$begingroup$
To have the MVI apply to $|f(x)-f(y)|,$ you need to know $[x,y]$ does not go through $0.$
$endgroup$
– zhw.
Jan 20 at 19:33