an example of a non-linear isometry between normed spaces which is not one-to-one
0
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Can anybody give an example of a non-linear isometry between normed spaces which is not one-to-one?
Or, does there exist such a mapping?
linear-algebra functional-analysis
edited Jan 15 at 17:10
Bernard
125k743118
asked Jan 15 at 16:14
serenusserenus
24316
$endgroup$
add a comment |
0
$begingroup$
Can anybody give an example of a non-linear isometry between normed spaces which is not one-to-one?
Or, does there exist such a mapping?
linear-algebra functional-analysis
edited Jan 15 at 17:10
Bernard
125k743118
asked Jan 15 at 16:14
serenusserenus
24316
$endgroup$
$begingroup$
If $f(x)=f(y)$ then $0=d(f(x),f(y))=d(x,y)$.
$endgroup$
– Robert Z
Jan 15 at 16:19
$begingroup$
So, the answer is negative !
$endgroup$
– serenus
Jan 15 at 16:27
$begingroup$
I mean, any isometry is necessarily one-to-one.
$endgroup$
– serenus
Jan 15 at 20:50
$begingroup$
Yes, see also math.stackexchange.com/questions/568415/…
$endgroup$
– Robert Z
Jan 16 at 7:46
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:03
add a comment |
0
0
0
$begingroup$
Can anybody give an example of a non-linear isometry between normed spaces which is not one-to-one?
Or, does there exist such a mapping?
linear-algebra functional-analysis
edited Jan 15 at 17:10
Bernard
125k743118
asked Jan 15 at 16:14
serenusserenus
24316
$endgroup$
Can anybody give an example of a non-linear isometry between normed spaces which is not one-to-one?
Or, does there exist such a mapping?
linear-algebra functional-analysis
linear-algebra functional-analysis
edited Jan 15 at 17:10
Bernard
125k743118
asked Jan 15 at 16:14
serenusserenus
24316
edited Jan 15 at 17:10
Bernard
125k743118
asked Jan 15 at 16:14
serenusserenus
24316
edited Jan 15 at 17:10
Bernard
125k743118
edited Jan 15 at 17:10
Bernard
125k743118
edited Jan 15 at 17:10
Bernard
125k743118
125k743118
asked Jan 15 at 16:14
serenusserenus
24316
asked Jan 15 at 16:14
serenusserenus
24316
asked Jan 15 at 16:14
serenusserenus
24316
24316
$begingroup$
If $f(x)=f(y)$ then $0=d(f(x),f(y))=d(x,y)$.
$endgroup$
– Robert Z
Jan 15 at 16:19
$begingroup$
So, the answer is negative !
$endgroup$
– serenus
Jan 15 at 16:27
$begingroup$
I mean, any isometry is necessarily one-to-one.
$endgroup$
– serenus
Jan 15 at 20:50
$begingroup$
Yes, see also math.stackexchange.com/questions/568415/…
$endgroup$
– Robert Z
Jan 16 at 7:46
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:03
add a comment |
$begingroup$
If $f(x)=f(y)$ then $0=d(f(x),f(y))=d(x,y)$.
$endgroup$
– Robert Z
Jan 15 at 16:19
$begingroup$
So, the answer is negative !
$endgroup$
– serenus
Jan 15 at 16:27
$begingroup$
I mean, any isometry is necessarily one-to-one.
$endgroup$
– serenus
Jan 15 at 20:50
$begingroup$
Yes, see also math.stackexchange.com/questions/568415/…
$endgroup$
– Robert Z
Jan 16 at 7:46
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:03
$begingroup$
If $f(x)=f(y)$ then $0=d(f(x),f(y))=d(x,y)$.
$endgroup$
– Robert Z
Jan 15 at 16:19
$begingroup$
If $f(x)=f(y)$ then $0=d(f(x),f(y))=d(x,y)$.
$endgroup$
– Robert Z
Jan 15 at 16:19
$begingroup$
So, the answer is negative !
$endgroup$
– serenus
Jan 15 at 16:27
$begingroup$
So, the answer is negative !
$endgroup$
– serenus
Jan 15 at 16:27
$begingroup$
I mean, any isometry is necessarily one-to-one.
$endgroup$
– serenus
Jan 15 at 20:50
$begingroup$
I mean, any isometry is necessarily one-to-one.
$endgroup$
– serenus
Jan 15 at 20:50
$begingroup$
Yes, see also math.stackexchange.com/questions/568415/…
$endgroup$
– Robert Z
Jan 16 at 7:46
$begingroup$
Yes, see also math.stackexchange.com/questions/568415/…
$endgroup$
– Robert Z
Jan 16 at 7:46
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:03
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:03
add a comment |
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$begingroup$
If $f(x)=f(y)$ then $0=d(f(x),f(y))=d(x,y)$.
$endgroup$
– Robert Z
Jan 15 at 16:19
$begingroup$
So, the answer is negative !
$endgroup$
– serenus
Jan 15 at 16:27
$begingroup$
I mean, any isometry is necessarily one-to-one.
$endgroup$
– serenus
Jan 15 at 20:50
$begingroup$
Yes, see also math.stackexchange.com/questions/568415/…
$endgroup$
– Robert Z
Jan 16 at 7:46
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:03