Mixing
Non-constant function with 0 Lipschitz semi-norm
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Suppose we have a bounded metric space $(X,d)$. We say a function $f:Xto mathbb{R}$ is Lipschitz if
$|f|=sup_{substack{xneq y\x,yin X}}frac{left|f(x)-f(y)right|}{d(x,y)}<infty$.
This is a semi-norm. For example any constant function $f$ has $|f|=0$. Is there a non-constant function $f$ with $|f|=0$? My guess is no, but I can't seem to convince myself. Apologies if this is a trivial question. Thanks in advance
metric-spaces norm examples-counterexamples normed-spaces lipschitz-functions
asked Jan 22 at 14:59
Mr MartingaleMr Martingale
25417
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add a comment |
$begingroup$
Suppose we have a bounded metric space $(X,d)$. We say a function $f:Xto mathbb{R}$ is Lipschitz if
$|f|=sup_{substack{xneq y\x,yin X}}frac{left|f(x)-f(y)right|}{d(x,y)}<infty$.
This is a semi-norm. For example any constant function $f$ has $|f|=0$. Is there a non-constant function $f$ with $|f|=0$? My guess is no, but I can't seem to convince myself. Apologies if this is a trivial question. Thanks in advance
metric-spaces norm examples-counterexamples normed-spaces lipschitz-functions
asked Jan 22 at 14:59
Mr MartingaleMr Martingale
25417
$endgroup$
add a comment |
$begingroup$
Suppose we have a bounded metric space $(X,d)$. We say a function $f:Xto mathbb{R}$ is Lipschitz if
$|f|=sup_{substack{xneq y\x,yin X}}frac{left|f(x)-f(y)right|}{d(x,y)}<infty$.
This is a semi-norm. For example any constant function $f$ has $|f|=0$. Is there a non-constant function $f$ with $|f|=0$? My guess is no, but I can't seem to convince myself. Apologies if this is a trivial question. Thanks in advance
metric-spaces norm examples-counterexamples normed-spaces lipschitz-functions
asked Jan 22 at 14:59
Mr MartingaleMr Martingale
25417
$endgroup$
Suppose we have a bounded metric space $(X,d)$. We say a function $f:Xto mathbb{R}$ is Lipschitz if
$|f|=sup_{substack{xneq y\x,yin X}}frac{left|f(x)-f(y)right|}{d(x,y)}<infty$.
This is a semi-norm. For example any constant function $f$ has $|f|=0$. Is there a non-constant function $f$ with $|f|=0$? My guess is no, but I can't seem to convince myself. Apologies if this is a trivial question. Thanks in advance
metric-spaces norm examples-counterexamples normed-spaces lipschitz-functions
metric-spaces norm examples-counterexamples normed-spaces lipschitz-functions
asked Jan 22 at 14:59
Mr MartingaleMr Martingale
25417
asked Jan 22 at 14:59
Mr MartingaleMr Martingale
25417
asked Jan 22 at 14:59
Mr MartingaleMr Martingale
25417
asked Jan 22 at 14:59
Mr MartingaleMr Martingale
25417
asked Jan 22 at 14:59
Mr MartingaleMr Martingale
25417
25417
add a comment |
add a comment |
1 Answer
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If $|f|=0$, then for any $xneq y$, $|f(x)-f(y)| leq |f| d(x,y)=0$ so $f(x)=f(y)$. Thus $f$ is constant.
answered Jan 22 at 15:01
MindlackMindlack
4,920211
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This made me smile. I am convinced!
$endgroup$
– Mr Martingale
Jan 22 at 15:04
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $|f|=0$, then for any $xneq y$, $|f(x)-f(y)| leq |f| d(x,y)=0$ so $f(x)=f(y)$. Thus $f$ is constant.
answered Jan 22 at 15:01
MindlackMindlack
4,920211
$endgroup$
$begingroup$
This made me smile. I am convinced!
$endgroup$
– Mr Martingale
Jan 22 at 15:04
add a comment |
$begingroup$
If $|f|=0$, then for any $xneq y$, $|f(x)-f(y)| leq |f| d(x,y)=0$ so $f(x)=f(y)$. Thus $f$ is constant.
answered Jan 22 at 15:01
MindlackMindlack
4,920211
$endgroup$
$begingroup$
This made me smile. I am convinced!
$endgroup$
– Mr Martingale
Jan 22 at 15:04
add a comment |
$begingroup$
If $|f|=0$, then for any $xneq y$, $|f(x)-f(y)| leq |f| d(x,y)=0$ so $f(x)=f(y)$. Thus $f$ is constant.
answered Jan 22 at 15:01
MindlackMindlack
4,920211
$endgroup$
If $|f|=0$, then for any $xneq y$, $|f(x)-f(y)| leq |f| d(x,y)=0$ so $f(x)=f(y)$. Thus $f$ is constant.
answered Jan 22 at 15:01
MindlackMindlack
4,920211
answered Jan 22 at 15:01
MindlackMindlack
4,920211
answered Jan 22 at 15:01
MindlackMindlack
4,920211
answered Jan 22 at 15:01
MindlackMindlack
4,920211
4,920211
$begingroup$
This made me smile. I am convinced!
$endgroup$
– Mr Martingale
Jan 22 at 15:04
add a comment |
$begingroup$
This made me smile. I am convinced!
$endgroup$
– Mr Martingale
Jan 22 at 15:04
$begingroup$
This made me smile. I am convinced!
$endgroup$
– Mr Martingale
Jan 22 at 15:04
$begingroup$
This made me smile. I am convinced!
$endgroup$
– Mr Martingale
Jan 22 at 15:04
add a comment |
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