Is $ f colon mathbb{R^2} to mathbb{R^2} , f(x,y) = ( x^2, y^2) $ uniformly continuous on $mathbb{R^2}$
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Is $ f colon mathbb{R^2} to mathbb{R^2} , f(x,y) = ( x^2, y^2) $ uniformly continuous on $mathbb{R^2} ?$
I think it is not, I tried proving it by contradiction but can't find the right $delta$
uniform-continuity
asked Nov 17 at 12:50
user15269
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up vote
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down vote
favorite
Is $ f colon mathbb{R^2} to mathbb{R^2} , f(x,y) = ( x^2, y^2) $ uniformly continuous on $mathbb{R^2} ?$
I think it is not, I tried proving it by contradiction but can't find the right $delta$
uniform-continuity
asked Nov 17 at 12:50
user15269
1608
It's not. Did you try the simpler example in one dimension of $f(x)=x^2$?
– Richard Rast
Nov 17 at 12:59
math.stackexchange.com/q/2993834/238946
– Daniel Camarena Perez
Nov 17 at 13:11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is $ f colon mathbb{R^2} to mathbb{R^2} , f(x,y) = ( x^2, y^2) $ uniformly continuous on $mathbb{R^2} ?$
I think it is not, I tried proving it by contradiction but can't find the right $delta$
uniform-continuity
asked Nov 17 at 12:50
user15269
1608
Is $ f colon mathbb{R^2} to mathbb{R^2} , f(x,y) = ( x^2, y^2) $ uniformly continuous on $mathbb{R^2} ?$
I think it is not, I tried proving it by contradiction but can't find the right $delta$
uniform-continuity
uniform-continuity
asked Nov 17 at 12:50
user15269
1608
asked Nov 17 at 12:50
user15269
1608
asked Nov 17 at 12:50
user15269
1608
asked Nov 17 at 12:50
user15269
1608
asked Nov 17 at 12:50
user15269
1608
1608
It's not. Did you try the simpler example in one dimension of $f(x)=x^2$?
– Richard Rast
Nov 17 at 12:59
math.stackexchange.com/q/2993834/238946
– Daniel Camarena Perez
Nov 17 at 13:11
add a comment |
It's not. Did you try the simpler example in one dimension of $f(x)=x^2$?
– Richard Rast
Nov 17 at 12:59
math.stackexchange.com/q/2993834/238946
– Daniel Camarena Perez
Nov 17 at 13:11
It's not. Did you try the simpler example in one dimension of $f(x)=x^2$?
– Richard Rast
Nov 17 at 12:59
It's not. Did you try the simpler example in one dimension of $f(x)=x^2$?
– Richard Rast
Nov 17 at 12:59
math.stackexchange.com/q/2993834/238946
– Daniel Camarena Perez
Nov 17 at 13:11
math.stackexchange.com/q/2993834/238946
– Daniel Camarena Perez
Nov 17 at 13:11
add a comment |
1 Answer
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First note that $g:x to x^2$ is not uniformly continuous on $mathbb{R}$.
Then note that $e: mathbb{R} to mathbb{R}^2$ given by $e(x) = (x,0)$ is uniformly continuous (even an isometry) and
$pi_1: mathbb{R}^2 to mathbb{R}$ given by $pi_1(x,y) = x$ is also uniformly continuous (a weak contraction even).
So if $f$ were uniformly continuous, so would $pi_1 circ f circ e$ be (compositions of uniformly continuous functions are still uniformly continuous), but this is just $g$ which is not, so $f$ also is not.
answered Nov 17 at 13:13
Henno Brandsma
102k344108
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
First note that $g:x to x^2$ is not uniformly continuous on $mathbb{R}$.
Then note that $e: mathbb{R} to mathbb{R}^2$ given by $e(x) = (x,0)$ is uniformly continuous (even an isometry) and
$pi_1: mathbb{R}^2 to mathbb{R}$ given by $pi_1(x,y) = x$ is also uniformly continuous (a weak contraction even).
So if $f$ were uniformly continuous, so would $pi_1 circ f circ e$ be (compositions of uniformly continuous functions are still uniformly continuous), but this is just $g$ which is not, so $f$ also is not.
answered Nov 17 at 13:13
Henno Brandsma
102k344108
add a comment |
up vote
0
down vote
First note that $g:x to x^2$ is not uniformly continuous on $mathbb{R}$.
Then note that $e: mathbb{R} to mathbb{R}^2$ given by $e(x) = (x,0)$ is uniformly continuous (even an isometry) and
$pi_1: mathbb{R}^2 to mathbb{R}$ given by $pi_1(x,y) = x$ is also uniformly continuous (a weak contraction even).
So if $f$ were uniformly continuous, so would $pi_1 circ f circ e$ be (compositions of uniformly continuous functions are still uniformly continuous), but this is just $g$ which is not, so $f$ also is not.
answered Nov 17 at 13:13
Henno Brandsma
102k344108
add a comment |
up vote
0
down vote
up vote
0
down vote
First note that $g:x to x^2$ is not uniformly continuous on $mathbb{R}$.
Then note that $e: mathbb{R} to mathbb{R}^2$ given by $e(x) = (x,0)$ is uniformly continuous (even an isometry) and
$pi_1: mathbb{R}^2 to mathbb{R}$ given by $pi_1(x,y) = x$ is also uniformly continuous (a weak contraction even).
So if $f$ were uniformly continuous, so would $pi_1 circ f circ e$ be (compositions of uniformly continuous functions are still uniformly continuous), but this is just $g$ which is not, so $f$ also is not.
answered Nov 17 at 13:13
Henno Brandsma
102k344108
First note that $g:x to x^2$ is not uniformly continuous on $mathbb{R}$.
Then note that $e: mathbb{R} to mathbb{R}^2$ given by $e(x) = (x,0)$ is uniformly continuous (even an isometry) and
$pi_1: mathbb{R}^2 to mathbb{R}$ given by $pi_1(x,y) = x$ is also uniformly continuous (a weak contraction even).
So if $f$ were uniformly continuous, so would $pi_1 circ f circ e$ be (compositions of uniformly continuous functions are still uniformly continuous), but this is just $g$ which is not, so $f$ also is not.
answered Nov 17 at 13:13
Henno Brandsma
102k344108
answered Nov 17 at 13:13
Henno Brandsma
102k344108
answered Nov 17 at 13:13
Henno Brandsma
102k344108
answered Nov 17 at 13:13
Henno Brandsma
102k344108
102k344108
add a comment |
add a comment |
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It's not. Did you try the simpler example in one dimension of $f(x)=x^2$?
– Richard Rast
Nov 17 at 12:59
math.stackexchange.com/q/2993834/238946
– Daniel Camarena Perez
Nov 17 at 13:11