Lipschitz constant higher dimension
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Suppose $f:mathbb{R^2} rightarrow mathbb{R}$ is Lipschitz with Lipschitz constant $K>0$, i.e. $|f(x)-f(y)|le K||x-y||$.
Define $g:mathbb{Rtimes mathbb{R}^2} rightarrow mathbb{R^3}$ by $g(t, x)=(x, f(x))$. I want to prove $g$ is Lipschitz in the second variable with Lipschitz constant $K+1$ i.e. $||g(t, x)-g(t, y)||le (1+K) ||x-y||$, $forall x, y in mathbb{R^2}$
I want to know if my proof is correct:
We have: $||g(t, x)-g(t, y)||=||(x, f(x))-(y, f(y))||=||(x-y, f(x)-f(y))||le||(x-y, 0)||+$ $ +||(0, 0, f(x)-f(y))||le(1+K)||x-y||$ as $f$ is Lipschitz.
Is my proof correct? Thank you.
analysis
asked Dec 16 '18 at 12:59
hermithermit
353
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add a comment |
$begingroup$
Suppose $f:mathbb{R^2} rightarrow mathbb{R}$ is Lipschitz with Lipschitz constant $K>0$, i.e. $|f(x)-f(y)|le K||x-y||$.
Define $g:mathbb{Rtimes mathbb{R}^2} rightarrow mathbb{R^3}$ by $g(t, x)=(x, f(x))$. I want to prove $g$ is Lipschitz in the second variable with Lipschitz constant $K+1$ i.e. $||g(t, x)-g(t, y)||le (1+K) ||x-y||$, $forall x, y in mathbb{R^2}$
I want to know if my proof is correct:
We have: $||g(t, x)-g(t, y)||=||(x, f(x))-(y, f(y))||=||(x-y, f(x)-f(y))||le||(x-y, 0)||+$ $ +||(0, 0, f(x)-f(y))||le(1+K)||x-y||$ as $f$ is Lipschitz.
Is my proof correct? Thank you.
analysis
asked Dec 16 '18 at 12:59
hermithermit
353
$endgroup$
add a comment |
$begingroup$
Suppose $f:mathbb{R^2} rightarrow mathbb{R}$ is Lipschitz with Lipschitz constant $K>0$, i.e. $|f(x)-f(y)|le K||x-y||$.
Define $g:mathbb{Rtimes mathbb{R}^2} rightarrow mathbb{R^3}$ by $g(t, x)=(x, f(x))$. I want to prove $g$ is Lipschitz in the second variable with Lipschitz constant $K+1$ i.e. $||g(t, x)-g(t, y)||le (1+K) ||x-y||$, $forall x, y in mathbb{R^2}$
I want to know if my proof is correct:
We have: $||g(t, x)-g(t, y)||=||(x, f(x))-(y, f(y))||=||(x-y, f(x)-f(y))||le||(x-y, 0)||+$ $ +||(0, 0, f(x)-f(y))||le(1+K)||x-y||$ as $f$ is Lipschitz.
Is my proof correct? Thank you.
analysis
asked Dec 16 '18 at 12:59
hermithermit
353
$endgroup$
Suppose $f:mathbb{R^2} rightarrow mathbb{R}$ is Lipschitz with Lipschitz constant $K>0$, i.e. $|f(x)-f(y)|le K||x-y||$.
Define $g:mathbb{Rtimes mathbb{R}^2} rightarrow mathbb{R^3}$ by $g(t, x)=(x, f(x))$. I want to prove $g$ is Lipschitz in the second variable with Lipschitz constant $K+1$ i.e. $||g(t, x)-g(t, y)||le (1+K) ||x-y||$, $forall x, y in mathbb{R^2}$
I want to know if my proof is correct:
We have: $||g(t, x)-g(t, y)||=||(x, f(x))-(y, f(y))||=||(x-y, f(x)-f(y))||le||(x-y, 0)||+$ $ +||(0, 0, f(x)-f(y))||le(1+K)||x-y||$ as $f$ is Lipschitz.
Is my proof correct? Thank you.
analysis
analysis
asked Dec 16 '18 at 12:59
hermithermit
353
asked Dec 16 '18 at 12:59
hermithermit
353
asked Dec 16 '18 at 12:59
hermithermit
353
asked Dec 16 '18 at 12:59
hermithermit
353
asked Dec 16 '18 at 12:59
hermithermit
353
353
add a comment |
add a comment |
1 Answer
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Your proof is correct, up to some typos and details to be added:
$||(0, 0, f(x)-f(y))||$ should be $||(0, f(x)-f(y))||$;- maybe it would be better to add the step $||(0, f(x)-f(y))||=leftlvert f(x)-f(y)rightrvert$.
answered Dec 16 '18 at 13:36
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
$begingroup$
Thank you. I wrote $||(0, 0, f(x)-f(y))||$ because $x, y in mathbb{R^2}$
$endgroup$
– hermit
Dec 16 '18 at 13:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your proof is correct, up to some typos and details to be added:
$||(0, 0, f(x)-f(y))||$ should be $||(0, f(x)-f(y))||$;- maybe it would be better to add the step $||(0, f(x)-f(y))||=leftlvert f(x)-f(y)rightrvert$.
answered Dec 16 '18 at 13:36
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
$begingroup$
Thank you. I wrote $||(0, 0, f(x)-f(y))||$ because $x, y in mathbb{R^2}$
$endgroup$
– hermit
Dec 16 '18 at 13:40
add a comment |
$begingroup$
Your proof is correct, up to some typos and details to be added:
$||(0, 0, f(x)-f(y))||$ should be $||(0, f(x)-f(y))||$;- maybe it would be better to add the step $||(0, f(x)-f(y))||=leftlvert f(x)-f(y)rightrvert$.
answered Dec 16 '18 at 13:36
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
$begingroup$
Thank you. I wrote $||(0, 0, f(x)-f(y))||$ because $x, y in mathbb{R^2}$
$endgroup$
– hermit
Dec 16 '18 at 13:40
add a comment |
$begingroup$
Your proof is correct, up to some typos and details to be added:
$||(0, 0, f(x)-f(y))||$ should be $||(0, f(x)-f(y))||$;- maybe it would be better to add the step $||(0, f(x)-f(y))||=leftlvert f(x)-f(y)rightrvert$.
answered Dec 16 '18 at 13:36
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
Your proof is correct, up to some typos and details to be added:
$||(0, 0, f(x)-f(y))||$ should be $||(0, f(x)-f(y))||$;- maybe it would be better to add the step $||(0, f(x)-f(y))||=leftlvert f(x)-f(y)rightrvert$.
answered Dec 16 '18 at 13:36
Davide GiraudoDavide Giraudo
125k16150261
answered Dec 16 '18 at 13:36
Davide GiraudoDavide Giraudo
125k16150261
answered Dec 16 '18 at 13:36
Davide GiraudoDavide Giraudo
125k16150261
answered Dec 16 '18 at 13:36
Davide GiraudoDavide Giraudo
125k16150261
125k16150261
$begingroup$
Thank you. I wrote $||(0, 0, f(x)-f(y))||$ because $x, y in mathbb{R^2}$
$endgroup$
– hermit
Dec 16 '18 at 13:40
add a comment |
$begingroup$
Thank you. I wrote $||(0, 0, f(x)-f(y))||$ because $x, y in mathbb{R^2}$
$endgroup$
– hermit
Dec 16 '18 at 13:40
$begingroup$
Thank you. I wrote $||(0, 0, f(x)-f(y))||$ because $x, y in mathbb{R^2}$
$endgroup$
– hermit
Dec 16 '18 at 13:40
$begingroup$
Thank you. I wrote $||(0, 0, f(x)-f(y))||$ because $x, y in mathbb{R^2}$
$endgroup$
– hermit
Dec 16 '18 at 13:40
add a comment |
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