Show that if f : A → Y is Lipschitz then it is uniformly continuous on A
There's a step that I don't understand from my notes:
Proof.
$forallvarepsilon>0 hspace{3pt}existsdelta>0: forall x_1,x_2in Ahspace{3pt}d_X(x_1,x_2)<deltahspace{3pt}d_Y(f(x_1),f(x_2))<varepsilon$
Choose $delta=frac{epsilon}{L}$
This is the step i don't understand, why am I allowed to choose $delta$ when delta itself is not arbitrary and it depends by the choose of $varepsilon$
$d_Y(f(x_1),f(x_2))leq Ld_x(x_1,x_2)=frac{Lcdotvarepsilon}{L}=varepsilon$
real-analysis
asked Dec 4 '18 at 21:16
ArchimedessArchimedess
114
add a comment |
There's a step that I don't understand from my notes:
Proof.
$forallvarepsilon>0 hspace{3pt}existsdelta>0: forall x_1,x_2in Ahspace{3pt}d_X(x_1,x_2)<deltahspace{3pt}d_Y(f(x_1),f(x_2))<varepsilon$
Choose $delta=frac{epsilon}{L}$
This is the step i don't understand, why am I allowed to choose $delta$ when delta itself is not arbitrary and it depends by the choose of $varepsilon$
$d_Y(f(x_1),f(x_2))leq Ld_x(x_1,x_2)=frac{Lcdotvarepsilon}{L}=varepsilon$
real-analysis
asked Dec 4 '18 at 21:16
ArchimedessArchimedess
114
1
$delta$ is not supposed to be arbitrary, and it may well depend on $epsilon$. The point is that it should not depend on $x_1$ and $x_2$.
– MisterRiemann
Dec 4 '18 at 21:18
Does your definition allow $L=0$?
– Math_QED
Dec 4 '18 at 21:21
No, it says $exists L>0$ such that...
– Archimedess
Dec 4 '18 at 21:22
add a comment |
There's a step that I don't understand from my notes:
Proof.
$forallvarepsilon>0 hspace{3pt}existsdelta>0: forall x_1,x_2in Ahspace{3pt}d_X(x_1,x_2)<deltahspace{3pt}d_Y(f(x_1),f(x_2))<varepsilon$
Choose $delta=frac{epsilon}{L}$
This is the step i don't understand, why am I allowed to choose $delta$ when delta itself is not arbitrary and it depends by the choose of $varepsilon$
$d_Y(f(x_1),f(x_2))leq Ld_x(x_1,x_2)=frac{Lcdotvarepsilon}{L}=varepsilon$
real-analysis
asked Dec 4 '18 at 21:16
ArchimedessArchimedess
114
There's a step that I don't understand from my notes:
Proof.
$forallvarepsilon>0 hspace{3pt}existsdelta>0: forall x_1,x_2in Ahspace{3pt}d_X(x_1,x_2)<deltahspace{3pt}d_Y(f(x_1),f(x_2))<varepsilon$
Choose $delta=frac{epsilon}{L}$
This is the step i don't understand, why am I allowed to choose $delta$ when delta itself is not arbitrary and it depends by the choose of $varepsilon$
$d_Y(f(x_1),f(x_2))leq Ld_x(x_1,x_2)=frac{Lcdotvarepsilon}{L}=varepsilon$
real-analysis
real-analysis
asked Dec 4 '18 at 21:16
ArchimedessArchimedess
114
asked Dec 4 '18 at 21:16
ArchimedessArchimedess
114
asked Dec 4 '18 at 21:16
ArchimedessArchimedess
114
asked Dec 4 '18 at 21:16
ArchimedessArchimedess
114
asked Dec 4 '18 at 21:16
ArchimedessArchimedess
114
114
1
$delta$ is not supposed to be arbitrary, and it may well depend on $epsilon$. The point is that it should not depend on $x_1$ and $x_2$.
– MisterRiemann
Dec 4 '18 at 21:18
Does your definition allow $L=0$?
– Math_QED
Dec 4 '18 at 21:21
No, it says $exists L>0$ such that...
– Archimedess
Dec 4 '18 at 21:22
add a comment |
1
$delta$ is not supposed to be arbitrary, and it may well depend on $epsilon$. The point is that it should not depend on $x_1$ and $x_2$.
– MisterRiemann
Dec 4 '18 at 21:18
Does your definition allow $L=0$?
– Math_QED
Dec 4 '18 at 21:21
No, it says $exists L>0$ such that...
– Archimedess
Dec 4 '18 at 21:22
1
1
$delta$ is not supposed to be arbitrary, and it may well depend on $epsilon$. The point is that it should not depend on $x_1$ and $x_2$.
– MisterRiemann
Dec 4 '18 at 21:18
$delta$ is not supposed to be arbitrary, and it may well depend on $epsilon$. The point is that it should not depend on $x_1$ and $x_2$.
– MisterRiemann
Dec 4 '18 at 21:18
Does your definition allow $L=0$?
– Math_QED
Dec 4 '18 at 21:21
Does your definition allow $L=0$?
– Math_QED
Dec 4 '18 at 21:21
No, it says $exists L>0$ such that...
– Archimedess
Dec 4 '18 at 21:22
No, it says $exists L>0$ such that...
– Archimedess
Dec 4 '18 at 21:22
add a comment |
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1
$delta$ is not supposed to be arbitrary, and it may well depend on $epsilon$. The point is that it should not depend on $x_1$ and $x_2$.
– MisterRiemann
Dec 4 '18 at 21:18
Does your definition allow $L=0$?
– Math_QED
Dec 4 '18 at 21:21
No, it says $exists L>0$ such that...
– Archimedess
Dec 4 '18 at 21:22