Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$.
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Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$, which is a non-Lipschitz function.
uniform-convergence lipschitz-functions sequence-of-function
asked Dec 2 '18 at 1:48
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Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$, which is a non-Lipschitz function.
uniform-convergence lipschitz-functions sequence-of-function
asked Dec 2 '18 at 1:48
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Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$, which is a non-Lipschitz function.
uniform-convergence lipschitz-functions sequence-of-function
asked Dec 2 '18 at 1:48
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Find a sequence of Lipschitz continuous functions on $[0,1]$ whose uniform limit is $sqrt{x}$, which is a non-Lipschitz function.
uniform-convergence lipschitz-functions sequence-of-function
uniform-convergence lipschitz-functions sequence-of-function
asked Dec 2 '18 at 1:48
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asked Dec 2 '18 at 1:48
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asked Dec 2 '18 at 1:48
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asked Dec 2 '18 at 1:48
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asked Dec 2 '18 at 1:48
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1 Answer
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$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 '18 at 1:52
AnupamAnupam
2,4731824
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$f_n$ are not Lipschitz continuous functions, take $n=1$.
$endgroup$
– Rebuild
Dec 2 '18 at 2:04
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$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
$endgroup$
– Anupam
Dec 2 '18 at 2:08
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Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
$endgroup$
– Rebuild
Dec 2 '18 at 2:11
$begingroup$
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
$endgroup$
– Rebuild
Dec 2 '18 at 2:18
add a comment |
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1 Answer
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1 Answer
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0
$begingroup$
$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 '18 at 1:52
AnupamAnupam
2,4731824
$endgroup$
$begingroup$
$f_n$ are not Lipschitz continuous functions, take $n=1$.
$endgroup$
– Rebuild
Dec 2 '18 at 2:04
$begingroup$
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
$endgroup$
– Anupam
Dec 2 '18 at 2:08
$begingroup$
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
$endgroup$
– Rebuild
Dec 2 '18 at 2:11
$begingroup$
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
$endgroup$
– Rebuild
Dec 2 '18 at 2:18
add a comment |
0
$begingroup$
$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 '18 at 1:52
AnupamAnupam
2,4731824
$endgroup$
$begingroup$
$f_n$ are not Lipschitz continuous functions, take $n=1$.
$endgroup$
– Rebuild
Dec 2 '18 at 2:04
$begingroup$
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
$endgroup$
– Anupam
Dec 2 '18 at 2:08
$begingroup$
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
$endgroup$
– Rebuild
Dec 2 '18 at 2:11
$begingroup$
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
$endgroup$
– Rebuild
Dec 2 '18 at 2:18
add a comment |
0
0
0
$begingroup$
$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 '18 at 1:52
AnupamAnupam
2,4731824
$endgroup$
$f_n:[0,1]to mathbb{R}$ defined by $f_n(x)=sqrt{x+frac{1}{n}}$ for all $xin [0,1]$ and for all $nin mathbb{N}$.
answered Dec 2 '18 at 1:52
AnupamAnupam
2,4731824
answered Dec 2 '18 at 1:52
AnupamAnupam
2,4731824
answered Dec 2 '18 at 1:52
AnupamAnupam
2,4731824
answered Dec 2 '18 at 1:52
AnupamAnupam
2,4731824
2,4731824
$begingroup$
$f_n$ are not Lipschitz continuous functions, take $n=1$.
$endgroup$
– Rebuild
Dec 2 '18 at 2:04
$begingroup$
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
$endgroup$
– Anupam
Dec 2 '18 at 2:08
$begingroup$
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
$endgroup$
– Rebuild
Dec 2 '18 at 2:11
$begingroup$
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
$endgroup$
– Rebuild
Dec 2 '18 at 2:18
add a comment |
$begingroup$
$f_n$ are not Lipschitz continuous functions, take $n=1$.
$endgroup$
– Rebuild
Dec 2 '18 at 2:04
$begingroup$
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
$endgroup$
– Anupam
Dec 2 '18 at 2:08
$begingroup$
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
$endgroup$
– Rebuild
Dec 2 '18 at 2:11
$begingroup$
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
$endgroup$
– Rebuild
Dec 2 '18 at 2:18
$begingroup$
$f_n$ are not Lipschitz continuous functions, take $n=1$.
$endgroup$
– Rebuild
Dec 2 '18 at 2:04
$begingroup$
$f_n$ are not Lipschitz continuous functions, take $n=1$.
$endgroup$
– Rebuild
Dec 2 '18 at 2:04
$begingroup$
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
$endgroup$
– Anupam
Dec 2 '18 at 2:08
$begingroup$
$f_1(x)=sqrt{x+1}$ is differentiable and $|f_1^{prime}(x)|=|frac{1}{2sqrt{x+1}}|leq frac{1}{2}$. Thus $f_1$ is Lipschitz continuous.
$endgroup$
– Anupam
Dec 2 '18 at 2:08
$begingroup$
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
$endgroup$
– Rebuild
Dec 2 '18 at 2:11
$begingroup$
Oh, I see, but when $n to infty$, does this mean the bound $M to infty$, so that the function is not Lipschitz?
$endgroup$
– Rebuild
Dec 2 '18 at 2:11
$begingroup$
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
$endgroup$
– Rebuild
Dec 2 '18 at 2:18
$begingroup$
I got it that fix $n$, we can have every $f_n$ is Lipschitz. Thanks
$endgroup$
– Rebuild
Dec 2 '18 at 2:18
add a comment |
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