Mixing
Continuity of Delta in Delta-Epsilon Argument
$begingroup$
This is a strange question but I am interested to know in this problem.
Let $Omega subsetmathbb{R}^{N}$ be a bounded smooth domain. Assume $f : overline{Omega}tomathbb{R}$ is continuous. Then, we have the following definition :
$$forall x in overline{Omega}, forall varepsilon > 0, exists delta(x,varepsilon) > 0 ni |x-y|<delta(x,varepsilon)implies|f(x)-f(y)|<varepsilon$$
Now, my question is, since $overline{Omega}$ is a compact set, can we always find $delta(x,varepsilon)>0$ so that it becomes a continuous function with respect to $x$ and $varepsilon$?
For simple function like $y=mx+c$ in one-dimensional case for finite interval $[a,b]$ the question is trivial but I am not sure about $N$-dimensional case with arbitrary continuous function.
real-analysis functions continuity epsilon-delta
asked Jan 22 at 1:46
Evan William ChandraEvan William Chandra
628313
$endgroup$
add a comment |
$begingroup$
This is a strange question but I am interested to know in this problem.
Let $Omega subsetmathbb{R}^{N}$ be a bounded smooth domain. Assume $f : overline{Omega}tomathbb{R}$ is continuous. Then, we have the following definition :
$$forall x in overline{Omega}, forall varepsilon > 0, exists delta(x,varepsilon) > 0 ni |x-y|<delta(x,varepsilon)implies|f(x)-f(y)|<varepsilon$$
Now, my question is, since $overline{Omega}$ is a compact set, can we always find $delta(x,varepsilon)>0$ so that it becomes a continuous function with respect to $x$ and $varepsilon$?
For simple function like $y=mx+c$ in one-dimensional case for finite interval $[a,b]$ the question is trivial but I am not sure about $N$-dimensional case with arbitrary continuous function.
real-analysis functions continuity epsilon-delta
asked Jan 22 at 1:46
Evan William ChandraEvan William Chandra
628313
$endgroup$
1
$begingroup$
$Omega subset mathbb{R}^{mathbb{N}}$ or $Omega subset mathbb{R}^N$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:50
$begingroup$
How does $Omega$ compact follow from bounded + smooth? Or should it be $bar{Omega}$ compact?
$endgroup$
– coffeemath
Jan 22 at 1:51
$begingroup$
I don't think $delta(x,epsilon)$ is a well-defined function from $Omega times(0,infty) to mathbb{R}$. If $delta(x,epsilon)$ works for a given $x$ and $epsilon$ then so does $delta/2$. So, how should one define the mapping $(x,epsilon)mapsto delta(x,epsilon)$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:53
1
$begingroup$
Thank you so much for the clarification, I have fixed the assumptions. But can we define $delta(x,varepsilon)$ so that it becomes continuous? I understand that $delta(x,varepsilon)$ is not unique
$endgroup$
– Evan William Chandra
Jan 22 at 1:55
$begingroup$
Single-valued functions of any number of variables must satisfy the uniqueness of output per input property. So, you cannot even define $delta(x,epsilon)$ as a function let alone ask for continuity. But if you had multi-valued functions in mind from the start then it's beyond my realm, sorry.
$endgroup$
– LoveTooNap29
Jan 22 at 2:11
add a comment |
$begingroup$
This is a strange question but I am interested to know in this problem.
Let $Omega subsetmathbb{R}^{N}$ be a bounded smooth domain. Assume $f : overline{Omega}tomathbb{R}$ is continuous. Then, we have the following definition :
$$forall x in overline{Omega}, forall varepsilon > 0, exists delta(x,varepsilon) > 0 ni |x-y|<delta(x,varepsilon)implies|f(x)-f(y)|<varepsilon$$
Now, my question is, since $overline{Omega}$ is a compact set, can we always find $delta(x,varepsilon)>0$ so that it becomes a continuous function with respect to $x$ and $varepsilon$?
For simple function like $y=mx+c$ in one-dimensional case for finite interval $[a,b]$ the question is trivial but I am not sure about $N$-dimensional case with arbitrary continuous function.
real-analysis functions continuity epsilon-delta
asked Jan 22 at 1:46
Evan William ChandraEvan William Chandra
628313
$endgroup$
This is a strange question but I am interested to know in this problem.
Let $Omega subsetmathbb{R}^{N}$ be a bounded smooth domain. Assume $f : overline{Omega}tomathbb{R}$ is continuous. Then, we have the following definition :
$$forall x in overline{Omega}, forall varepsilon > 0, exists delta(x,varepsilon) > 0 ni |x-y|<delta(x,varepsilon)implies|f(x)-f(y)|<varepsilon$$
Now, my question is, since $overline{Omega}$ is a compact set, can we always find $delta(x,varepsilon)>0$ so that it becomes a continuous function with respect to $x$ and $varepsilon$?
For simple function like $y=mx+c$ in one-dimensional case for finite interval $[a,b]$ the question is trivial but I am not sure about $N$-dimensional case with arbitrary continuous function.
real-analysis functions continuity epsilon-delta
real-analysis functions continuity epsilon-delta
asked Jan 22 at 1:46
Evan William ChandraEvan William Chandra
628313
asked Jan 22 at 1:46
Evan William ChandraEvan William Chandra
628313
edited Jan 22 at 1:54
Evan William Chandra
asked Jan 22 at 1:46
Evan William ChandraEvan William Chandra
628313
asked Jan 22 at 1:46
Evan William ChandraEvan William Chandra
628313
asked Jan 22 at 1:46
Evan William ChandraEvan William Chandra
628313
628313
1
$begingroup$
$Omega subset mathbb{R}^{mathbb{N}}$ or $Omega subset mathbb{R}^N$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:50
$begingroup$
How does $Omega$ compact follow from bounded + smooth? Or should it be $bar{Omega}$ compact?
$endgroup$
– coffeemath
Jan 22 at 1:51
$begingroup$
I don't think $delta(x,epsilon)$ is a well-defined function from $Omega times(0,infty) to mathbb{R}$. If $delta(x,epsilon)$ works for a given $x$ and $epsilon$ then so does $delta/2$. So, how should one define the mapping $(x,epsilon)mapsto delta(x,epsilon)$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:53
1
$begingroup$
Thank you so much for the clarification, I have fixed the assumptions. But can we define $delta(x,varepsilon)$ so that it becomes continuous? I understand that $delta(x,varepsilon)$ is not unique
$endgroup$
– Evan William Chandra
Jan 22 at 1:55
$begingroup$
Single-valued functions of any number of variables must satisfy the uniqueness of output per input property. So, you cannot even define $delta(x,epsilon)$ as a function let alone ask for continuity. But if you had multi-valued functions in mind from the start then it's beyond my realm, sorry.
$endgroup$
– LoveTooNap29
Jan 22 at 2:11
add a comment |
1
$begingroup$
$Omega subset mathbb{R}^{mathbb{N}}$ or $Omega subset mathbb{R}^N$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:50
$begingroup$
How does $Omega$ compact follow from bounded + smooth? Or should it be $bar{Omega}$ compact?
$endgroup$
– coffeemath
Jan 22 at 1:51
$begingroup$
I don't think $delta(x,epsilon)$ is a well-defined function from $Omega times(0,infty) to mathbb{R}$. If $delta(x,epsilon)$ works for a given $x$ and $epsilon$ then so does $delta/2$. So, how should one define the mapping $(x,epsilon)mapsto delta(x,epsilon)$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:53
1
$begingroup$
Thank you so much for the clarification, I have fixed the assumptions. But can we define $delta(x,varepsilon)$ so that it becomes continuous? I understand that $delta(x,varepsilon)$ is not unique
$endgroup$
– Evan William Chandra
Jan 22 at 1:55
$begingroup$
Single-valued functions of any number of variables must satisfy the uniqueness of output per input property. So, you cannot even define $delta(x,epsilon)$ as a function let alone ask for continuity. But if you had multi-valued functions in mind from the start then it's beyond my realm, sorry.
$endgroup$
– LoveTooNap29
Jan 22 at 2:11
1
1
$begingroup$
$Omega subset mathbb{R}^{mathbb{N}}$ or $Omega subset mathbb{R}^N$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:50
$begingroup$
$Omega subset mathbb{R}^{mathbb{N}}$ or $Omega subset mathbb{R}^N$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:50
$begingroup$
How does $Omega$ compact follow from bounded + smooth? Or should it be $bar{Omega}$ compact?
$endgroup$
– coffeemath
Jan 22 at 1:51
$begingroup$
How does $Omega$ compact follow from bounded + smooth? Or should it be $bar{Omega}$ compact?
$endgroup$
– coffeemath
Jan 22 at 1:51
$begingroup$
I don't think $delta(x,epsilon)$ is a well-defined function from $Omega times(0,infty) to mathbb{R}$. If $delta(x,epsilon)$ works for a given $x$ and $epsilon$ then so does $delta/2$. So, how should one define the mapping $(x,epsilon)mapsto delta(x,epsilon)$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:53
$begingroup$
I don't think $delta(x,epsilon)$ is a well-defined function from $Omega times(0,infty) to mathbb{R}$. If $delta(x,epsilon)$ works for a given $x$ and $epsilon$ then so does $delta/2$. So, how should one define the mapping $(x,epsilon)mapsto delta(x,epsilon)$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:53
1
1
$begingroup$
Thank you so much for the clarification, I have fixed the assumptions. But can we define $delta(x,varepsilon)$ so that it becomes continuous? I understand that $delta(x,varepsilon)$ is not unique
$endgroup$
– Evan William Chandra
Jan 22 at 1:55
$begingroup$
Thank you so much for the clarification, I have fixed the assumptions. But can we define $delta(x,varepsilon)$ so that it becomes continuous? I understand that $delta(x,varepsilon)$ is not unique
$endgroup$
– Evan William Chandra
Jan 22 at 1:55
$begingroup$
Single-valued functions of any number of variables must satisfy the uniqueness of output per input property. So, you cannot even define $delta(x,epsilon)$ as a function let alone ask for continuity. But if you had multi-valued functions in mind from the start then it's beyond my realm, sorry.
$endgroup$
– LoveTooNap29
Jan 22 at 2:11
$begingroup$
Single-valued functions of any number of variables must satisfy the uniqueness of output per input property. So, you cannot even define $delta(x,epsilon)$ as a function let alone ask for continuity. But if you had multi-valued functions in mind from the start then it's beyond my realm, sorry.
$endgroup$
– LoveTooNap29
Jan 22 at 2:11
add a comment |
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$begingroup$
$Omega subset mathbb{R}^{mathbb{N}}$ or $Omega subset mathbb{R}^N$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:50
$begingroup$
How does $Omega$ compact follow from bounded + smooth? Or should it be $bar{Omega}$ compact?
$endgroup$
– coffeemath
Jan 22 at 1:51
$begingroup$
I don't think $delta(x,epsilon)$ is a well-defined function from $Omega times(0,infty) to mathbb{R}$. If $delta(x,epsilon)$ works for a given $x$ and $epsilon$ then so does $delta/2$. So, how should one define the mapping $(x,epsilon)mapsto delta(x,epsilon)$?
$endgroup$
– LoveTooNap29
Jan 22 at 1:53
1
$begingroup$
Thank you so much for the clarification, I have fixed the assumptions. But can we define $delta(x,varepsilon)$ so that it becomes continuous? I understand that $delta(x,varepsilon)$ is not unique
$endgroup$
– Evan William Chandra
Jan 22 at 1:55
$begingroup$
Single-valued functions of any number of variables must satisfy the uniqueness of output per input property. So, you cannot even define $delta(x,epsilon)$ as a function let alone ask for continuity. But if you had multi-valued functions in mind from the start then it's beyond my realm, sorry.
$endgroup$
– LoveTooNap29
Jan 22 at 2:11