Mixing
Show that collection of Lipschitz functions with Lipschitz norm is a Banach space
$begingroup$
Question: Let $mathcal{L}$ be the normed space of all Lipschitz functions on a Banach space $X$ that are equal to $0$ at the origin, under the norm
$$|f| = supleft{ frac{|f(x)-f(y)|}{|x-y|}:x,yin X right}.$$
Show that $mathcal{L}$ is a Banach space.
My attempt:
Let $(f_n)_{n=1}^infty$ be a Cauchy sequence in $(mathcal{L},|cdot|).$
Define $f:Xto mathbb{R}$ by
$$ f(x) = lim_{ntoinfty} f_n(x).
$$
We claim that $f$ is well-defined, that is, the limit exists.
Fix $xin Xsetminus {0}$ and $epsilon>0.$
Since $(f_n)_{n=1}^infty$ is Cauchy, there exists $Nin mathbb{N}$ such that for any $m,ngeq N,$
$$ |f_m-f_n| <frac{epsilon}{|x|}.
$$
Note that
$$ |f_m-f_n| = sup_{x,yin X} frac{|(f_m-f_n)(x) - (f_m-f_n)(y) |}{|x-y|} geq frac{|f_m(x) - f_n(x)|}{|x|}.
$$
So we have $|f_m(x) - f_n(x)| < epsilon.$
Therefore, $(f_n(x))_{n=1}^infty$ is Cauchy in $mathbb{R}.$
Since $mathbb{R}$ is complete, so $lim_{ntoinfty} f_n(x)$ exists, that is, $f$ is well-defined.
We claim that $fin mathcal{L}.$
By construction, $f(0) = 0.$
Since $(f_n)_{n=1}^infty$ is Cauchy in $(mathcal{L},|cdot|),$ therefore it is bounded, that is, there exists $B>0$ such that $|f_n|leq B$ for all $ngeq 1.$
Since
$$f = f_N + (f-f_N),$$
so for any $x,yin X,$
begin{align*}
|f(x)-f(y)| & leq |f_N(x)-f_N(y)| + |(f-f_N)(x) - (f-f_N)(y)| \
& leq |f_N|| x-y| + |f-f_N| |x-y| \
& leq |f_N|| x-y| + |x-y| \
& = (|f_N|+1)|x-y|.
end{align*}
Therefore, $|f|$ is finite and hence $fin mathcal{L}.$
Lastly, we wish to prove that $(f_n)$ converges to $f$ in $(mathcal{L},|cdot|).$
Indeed, for every $epsilon>0,$ by Cauchyness of $(f_n),$ there exists $Ninmathbb{N}$ such that for any $m,ngeq N,$
$$ |f_n-f_m|<frac{epsilon}{2}.
$$
Then
$$ |f-f_n| = |lim_{mtoinfty} f_m - f_n| = lim_{mtoinfty} |f_m-f_n| < epsilon.
$$
We conclude that $mathcal{L}$ is complete.
Is my proof above correct?
real-analysis functional-analysis proof-verification banach-spaces
asked Feb 3 at 5:33
IdonknowIdonknow
2,610950119
$endgroup$
add a comment |
$begingroup$
Question: Let $mathcal{L}$ be the normed space of all Lipschitz functions on a Banach space $X$ that are equal to $0$ at the origin, under the norm
$$|f| = supleft{ frac{|f(x)-f(y)|}{|x-y|}:x,yin X right}.$$
Show that $mathcal{L}$ is a Banach space.
My attempt:
Let $(f_n)_{n=1}^infty$ be a Cauchy sequence in $(mathcal{L},|cdot|).$
Define $f:Xto mathbb{R}$ by
$$ f(x) = lim_{ntoinfty} f_n(x).
$$
We claim that $f$ is well-defined, that is, the limit exists.
Fix $xin Xsetminus {0}$ and $epsilon>0.$
Since $(f_n)_{n=1}^infty$ is Cauchy, there exists $Nin mathbb{N}$ such that for any $m,ngeq N,$
$$ |f_m-f_n| <frac{epsilon}{|x|}.
$$
Note that
$$ |f_m-f_n| = sup_{x,yin X} frac{|(f_m-f_n)(x) - (f_m-f_n)(y) |}{|x-y|} geq frac{|f_m(x) - f_n(x)|}{|x|}.
$$
So we have $|f_m(x) - f_n(x)| < epsilon.$
Therefore, $(f_n(x))_{n=1}^infty$ is Cauchy in $mathbb{R}.$
Since $mathbb{R}$ is complete, so $lim_{ntoinfty} f_n(x)$ exists, that is, $f$ is well-defined.
We claim that $fin mathcal{L}.$
By construction, $f(0) = 0.$
Since $(f_n)_{n=1}^infty$ is Cauchy in $(mathcal{L},|cdot|),$ therefore it is bounded, that is, there exists $B>0$ such that $|f_n|leq B$ for all $ngeq 1.$
Since
$$f = f_N + (f-f_N),$$
so for any $x,yin X,$
begin{align*}
|f(x)-f(y)| & leq |f_N(x)-f_N(y)| + |(f-f_N)(x) - (f-f_N)(y)| \
& leq |f_N|| x-y| + |f-f_N| |x-y| \
& leq |f_N|| x-y| + |x-y| \
& = (|f_N|+1)|x-y|.
end{align*}
Therefore, $|f|$ is finite and hence $fin mathcal{L}.$
Lastly, we wish to prove that $(f_n)$ converges to $f$ in $(mathcal{L},|cdot|).$
Indeed, for every $epsilon>0,$ by Cauchyness of $(f_n),$ there exists $Ninmathbb{N}$ such that for any $m,ngeq N,$
$$ |f_n-f_m|<frac{epsilon}{2}.
$$
Then
$$ |f-f_n| = |lim_{mtoinfty} f_m - f_n| = lim_{mtoinfty} |f_m-f_n| < epsilon.
$$
We conclude that $mathcal{L}$ is complete.
Is my proof above correct?
real-analysis functional-analysis proof-verification banach-spaces
asked Feb 3 at 5:33
IdonknowIdonknow
2,610950119
$endgroup$
$begingroup$
Why define $f(0) = 0$ instead of $f(0) = lim_{n to infty} f_n(0)$? This means, if your Cauchy sequence is constantly, say, the constant function $1$, then $f(x)$ will be the function that is constantly $1$ except at $0$, where it's $0$. Such a function is not continuous, let alone Lipschitz.
$endgroup$
– Theo Bendit
Feb 3 at 5:50
$begingroup$
@TheoBendit Yes, you are right. Edited my proof.
$endgroup$
– Idonknow
Feb 3 at 5:52
$begingroup$
Wait, I'm an idiot, I just noticed the requirement that $f(0) = 0$ for membership in the space. The constant function $1$ does not belong to the space in the first place!
$endgroup$
– Theo Bendit
Feb 3 at 5:54
$begingroup$
@TheoBendit Yes. But I think your suggestion still works and it is neater than my suggestion.
$endgroup$
– Idonknow
Feb 3 at 5:59
add a comment |
$begingroup$
Question: Let $mathcal{L}$ be the normed space of all Lipschitz functions on a Banach space $X$ that are equal to $0$ at the origin, under the norm
$$|f| = supleft{ frac{|f(x)-f(y)|}{|x-y|}:x,yin X right}.$$
Show that $mathcal{L}$ is a Banach space.
My attempt:
Let $(f_n)_{n=1}^infty$ be a Cauchy sequence in $(mathcal{L},|cdot|).$
Define $f:Xto mathbb{R}$ by
$$ f(x) = lim_{ntoinfty} f_n(x).
$$
We claim that $f$ is well-defined, that is, the limit exists.
Fix $xin Xsetminus {0}$ and $epsilon>0.$
Since $(f_n)_{n=1}^infty$ is Cauchy, there exists $Nin mathbb{N}$ such that for any $m,ngeq N,$
$$ |f_m-f_n| <frac{epsilon}{|x|}.
$$
Note that
$$ |f_m-f_n| = sup_{x,yin X} frac{|(f_m-f_n)(x) - (f_m-f_n)(y) |}{|x-y|} geq frac{|f_m(x) - f_n(x)|}{|x|}.
$$
So we have $|f_m(x) - f_n(x)| < epsilon.$
Therefore, $(f_n(x))_{n=1}^infty$ is Cauchy in $mathbb{R}.$
Since $mathbb{R}$ is complete, so $lim_{ntoinfty} f_n(x)$ exists, that is, $f$ is well-defined.
We claim that $fin mathcal{L}.$
By construction, $f(0) = 0.$
Since $(f_n)_{n=1}^infty$ is Cauchy in $(mathcal{L},|cdot|),$ therefore it is bounded, that is, there exists $B>0$ such that $|f_n|leq B$ for all $ngeq 1.$
Since
$$f = f_N + (f-f_N),$$
so for any $x,yin X,$
begin{align*}
|f(x)-f(y)| & leq |f_N(x)-f_N(y)| + |(f-f_N)(x) - (f-f_N)(y)| \
& leq |f_N|| x-y| + |f-f_N| |x-y| \
& leq |f_N|| x-y| + |x-y| \
& = (|f_N|+1)|x-y|.
end{align*}
Therefore, $|f|$ is finite and hence $fin mathcal{L}.$
Lastly, we wish to prove that $(f_n)$ converges to $f$ in $(mathcal{L},|cdot|).$
Indeed, for every $epsilon>0,$ by Cauchyness of $(f_n),$ there exists $Ninmathbb{N}$ such that for any $m,ngeq N,$
$$ |f_n-f_m|<frac{epsilon}{2}.
$$
Then
$$ |f-f_n| = |lim_{mtoinfty} f_m - f_n| = lim_{mtoinfty} |f_m-f_n| < epsilon.
$$
We conclude that $mathcal{L}$ is complete.
Is my proof above correct?
real-analysis functional-analysis proof-verification banach-spaces
asked Feb 3 at 5:33
IdonknowIdonknow
2,610950119
$endgroup$
Question: Let $mathcal{L}$ be the normed space of all Lipschitz functions on a Banach space $X$ that are equal to $0$ at the origin, under the norm
$$|f| = supleft{ frac{|f(x)-f(y)|}{|x-y|}:x,yin X right}.$$
Show that $mathcal{L}$ is a Banach space.
My attempt:
Let $(f_n)_{n=1}^infty$ be a Cauchy sequence in $(mathcal{L},|cdot|).$
Define $f:Xto mathbb{R}$ by
$$ f(x) = lim_{ntoinfty} f_n(x).
$$
We claim that $f$ is well-defined, that is, the limit exists.
Fix $xin Xsetminus {0}$ and $epsilon>0.$
Since $(f_n)_{n=1}^infty$ is Cauchy, there exists $Nin mathbb{N}$ such that for any $m,ngeq N,$
$$ |f_m-f_n| <frac{epsilon}{|x|}.
$$
Note that
$$ |f_m-f_n| = sup_{x,yin X} frac{|(f_m-f_n)(x) - (f_m-f_n)(y) |}{|x-y|} geq frac{|f_m(x) - f_n(x)|}{|x|}.
$$
So we have $|f_m(x) - f_n(x)| < epsilon.$
Therefore, $(f_n(x))_{n=1}^infty$ is Cauchy in $mathbb{R}.$
Since $mathbb{R}$ is complete, so $lim_{ntoinfty} f_n(x)$ exists, that is, $f$ is well-defined.
We claim that $fin mathcal{L}.$
By construction, $f(0) = 0.$
Since $(f_n)_{n=1}^infty$ is Cauchy in $(mathcal{L},|cdot|),$ therefore it is bounded, that is, there exists $B>0$ such that $|f_n|leq B$ for all $ngeq 1.$
Since
$$f = f_N + (f-f_N),$$
so for any $x,yin X,$
begin{align*}
|f(x)-f(y)| & leq |f_N(x)-f_N(y)| + |(f-f_N)(x) - (f-f_N)(y)| \
& leq |f_N|| x-y| + |f-f_N| |x-y| \
& leq |f_N|| x-y| + |x-y| \
& = (|f_N|+1)|x-y|.
end{align*}
Therefore, $|f|$ is finite and hence $fin mathcal{L}.$
Lastly, we wish to prove that $(f_n)$ converges to $f$ in $(mathcal{L},|cdot|).$
Indeed, for every $epsilon>0,$ by Cauchyness of $(f_n),$ there exists $Ninmathbb{N}$ such that for any $m,ngeq N,$
$$ |f_n-f_m|<frac{epsilon}{2}.
$$
Then
$$ |f-f_n| = |lim_{mtoinfty} f_m - f_n| = lim_{mtoinfty} |f_m-f_n| < epsilon.
$$
We conclude that $mathcal{L}$ is complete.
Is my proof above correct?
real-analysis functional-analysis proof-verification banach-spaces
real-analysis functional-analysis proof-verification banach-spaces
asked Feb 3 at 5:33
IdonknowIdonknow
2,610950119
asked Feb 3 at 5:33
IdonknowIdonknow
2,610950119
edited Feb 3 at 5:52
Idonknow
asked Feb 3 at 5:33
IdonknowIdonknow
2,610950119
asked Feb 3 at 5:33
IdonknowIdonknow
2,610950119
asked Feb 3 at 5:33
IdonknowIdonknow
2,610950119
2,610950119
$begingroup$
Why define $f(0) = 0$ instead of $f(0) = lim_{n to infty} f_n(0)$? This means, if your Cauchy sequence is constantly, say, the constant function $1$, then $f(x)$ will be the function that is constantly $1$ except at $0$, where it's $0$. Such a function is not continuous, let alone Lipschitz.
$endgroup$
– Theo Bendit
Feb 3 at 5:50
$begingroup$
@TheoBendit Yes, you are right. Edited my proof.
$endgroup$
– Idonknow
Feb 3 at 5:52
$begingroup$
Wait, I'm an idiot, I just noticed the requirement that $f(0) = 0$ for membership in the space. The constant function $1$ does not belong to the space in the first place!
$endgroup$
– Theo Bendit
Feb 3 at 5:54
$begingroup$
@TheoBendit Yes. But I think your suggestion still works and it is neater than my suggestion.
$endgroup$
– Idonknow
Feb 3 at 5:59
add a comment |
$begingroup$
Why define $f(0) = 0$ instead of $f(0) = lim_{n to infty} f_n(0)$? This means, if your Cauchy sequence is constantly, say, the constant function $1$, then $f(x)$ will be the function that is constantly $1$ except at $0$, where it's $0$. Such a function is not continuous, let alone Lipschitz.
$endgroup$
– Theo Bendit
Feb 3 at 5:50
$begingroup$
@TheoBendit Yes, you are right. Edited my proof.
$endgroup$
– Idonknow
Feb 3 at 5:52
$begingroup$
Wait, I'm an idiot, I just noticed the requirement that $f(0) = 0$ for membership in the space. The constant function $1$ does not belong to the space in the first place!
$endgroup$
– Theo Bendit
Feb 3 at 5:54
$begingroup$
@TheoBendit Yes. But I think your suggestion still works and it is neater than my suggestion.
$endgroup$
– Idonknow
Feb 3 at 5:59
$begingroup$
Why define $f(0) = 0$ instead of $f(0) = lim_{n to infty} f_n(0)$? This means, if your Cauchy sequence is constantly, say, the constant function $1$, then $f(x)$ will be the function that is constantly $1$ except at $0$, where it's $0$. Such a function is not continuous, let alone Lipschitz.
$endgroup$
– Theo Bendit
Feb 3 at 5:50
$begingroup$
Why define $f(0) = 0$ instead of $f(0) = lim_{n to infty} f_n(0)$? This means, if your Cauchy sequence is constantly, say, the constant function $1$, then $f(x)$ will be the function that is constantly $1$ except at $0$, where it's $0$. Such a function is not continuous, let alone Lipschitz.
$endgroup$
– Theo Bendit
Feb 3 at 5:50
$begingroup$
@TheoBendit Yes, you are right. Edited my proof.
$endgroup$
– Idonknow
Feb 3 at 5:52
$begingroup$
@TheoBendit Yes, you are right. Edited my proof.
$endgroup$
– Idonknow
Feb 3 at 5:52
$begingroup$
Wait, I'm an idiot, I just noticed the requirement that $f(0) = 0$ for membership in the space. The constant function $1$ does not belong to the space in the first place!
$endgroup$
– Theo Bendit
Feb 3 at 5:54
$begingroup$
Wait, I'm an idiot, I just noticed the requirement that $f(0) = 0$ for membership in the space. The constant function $1$ does not belong to the space in the first place!
$endgroup$
– Theo Bendit
Feb 3 at 5:54
$begingroup$
@TheoBendit Yes. But I think your suggestion still works and it is neater than my suggestion.
$endgroup$
– Idonknow
Feb 3 at 5:59
$begingroup$
@TheoBendit Yes. But I think your suggestion still works and it is neater than my suggestion.
$endgroup$
– Idonknow
Feb 3 at 5:59
add a comment |
1 Answer
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oldest
votes
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Your proof has all of the right ideas but suffers from a couple of defects since you end up implicitly doing things like assuming $f in mathcal{L}$ before you've proved it and confusing the pointwise limit with the limit in $mathcal{L}$. To fix this, you just need to write everything down in terms of the definition of the norm and make sure you're working with pointwise limits. I include details below.
Firstly, when trying to prove that $f in mathcal{L}$, on the second line you write down $|f-f_N|$ and then bound this by $1$. Doing this already requires you to assume $f-f_N in mathcal{L}$ and hence $f in mathcal{L}$ and also something like $f_n to f$ in $mathcal{L}$.
Instead, write
begin{align*}
|f(x) - f(y)| &= |lim_{n to infty} (f_n(x) - f_n(y))|
\&= lim_{n to infty} |f_n(x) - f_n(y)|
\& leq lim_{n to infty} |f_n| |x-y|
\& leq C|x-y|
end{align*}
where in the last line we used that $|f_n|$ is a bounded sequence. Note that in the second line, I only use that $f_n to f$ pointwise and not in $mathcal{L}$.
This is important because when trying to prove that $f_n to f$, you take a pointwise limit and take it outside of the $mathcal{L}$-norm. This doesn't work since you can only do that once you know that the limit converges in $mathcal{L}$ and not just pointwise. Instead, work again with a pointwise limit by writing
begin{align*}
|f(x) - f(y) - f_n(x) + f_n(y)| &= lim_{m to infty} |f_m(x) - f_m(y) - f_n(x) + f_n(y)|\
& leq lim_{m to infty} |f_m - f_n| |x-y|
end{align*}
Now since $(f_n)$ is Cauchy in $mathcal{L}$, there is $N$ independent of $x,y$ such that for $n,m geq N$ $|f_n - f_m| leq varepsilon$. So we get
$$|f(x) - f(y) - f_n(x) + f_n(y)| leq varepsilon |x-y|$$
which implies for $n geq N$, $|f-f_n| leq varepsilon$ so $f_n to f$ in $mathcal{L}$.
answered Feb 3 at 9:51
Rhys SteeleRhys Steele
7,9301931
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add a comment |
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$begingroup$
Your proof has all of the right ideas but suffers from a couple of defects since you end up implicitly doing things like assuming $f in mathcal{L}$ before you've proved it and confusing the pointwise limit with the limit in $mathcal{L}$. To fix this, you just need to write everything down in terms of the definition of the norm and make sure you're working with pointwise limits. I include details below.
Firstly, when trying to prove that $f in mathcal{L}$, on the second line you write down $|f-f_N|$ and then bound this by $1$. Doing this already requires you to assume $f-f_N in mathcal{L}$ and hence $f in mathcal{L}$ and also something like $f_n to f$ in $mathcal{L}$.
Instead, write
begin{align*}
|f(x) - f(y)| &= |lim_{n to infty} (f_n(x) - f_n(y))|
\&= lim_{n to infty} |f_n(x) - f_n(y)|
\& leq lim_{n to infty} |f_n| |x-y|
\& leq C|x-y|
end{align*}
where in the last line we used that $|f_n|$ is a bounded sequence. Note that in the second line, I only use that $f_n to f$ pointwise and not in $mathcal{L}$.
This is important because when trying to prove that $f_n to f$, you take a pointwise limit and take it outside of the $mathcal{L}$-norm. This doesn't work since you can only do that once you know that the limit converges in $mathcal{L}$ and not just pointwise. Instead, work again with a pointwise limit by writing
begin{align*}
|f(x) - f(y) - f_n(x) + f_n(y)| &= lim_{m to infty} |f_m(x) - f_m(y) - f_n(x) + f_n(y)|\
& leq lim_{m to infty} |f_m - f_n| |x-y|
end{align*}
Now since $(f_n)$ is Cauchy in $mathcal{L}$, there is $N$ independent of $x,y$ such that for $n,m geq N$ $|f_n - f_m| leq varepsilon$. So we get
$$|f(x) - f(y) - f_n(x) + f_n(y)| leq varepsilon |x-y|$$
which implies for $n geq N$, $|f-f_n| leq varepsilon$ so $f_n to f$ in $mathcal{L}$.
answered Feb 3 at 9:51
Rhys SteeleRhys Steele
7,9301931
$endgroup$
add a comment |
$begingroup$
Your proof has all of the right ideas but suffers from a couple of defects since you end up implicitly doing things like assuming $f in mathcal{L}$ before you've proved it and confusing the pointwise limit with the limit in $mathcal{L}$. To fix this, you just need to write everything down in terms of the definition of the norm and make sure you're working with pointwise limits. I include details below.
Firstly, when trying to prove that $f in mathcal{L}$, on the second line you write down $|f-f_N|$ and then bound this by $1$. Doing this already requires you to assume $f-f_N in mathcal{L}$ and hence $f in mathcal{L}$ and also something like $f_n to f$ in $mathcal{L}$.
Instead, write
begin{align*}
|f(x) - f(y)| &= |lim_{n to infty} (f_n(x) - f_n(y))|
\&= lim_{n to infty} |f_n(x) - f_n(y)|
\& leq lim_{n to infty} |f_n| |x-y|
\& leq C|x-y|
end{align*}
where in the last line we used that $|f_n|$ is a bounded sequence. Note that in the second line, I only use that $f_n to f$ pointwise and not in $mathcal{L}$.
This is important because when trying to prove that $f_n to f$, you take a pointwise limit and take it outside of the $mathcal{L}$-norm. This doesn't work since you can only do that once you know that the limit converges in $mathcal{L}$ and not just pointwise. Instead, work again with a pointwise limit by writing
begin{align*}
|f(x) - f(y) - f_n(x) + f_n(y)| &= lim_{m to infty} |f_m(x) - f_m(y) - f_n(x) + f_n(y)|\
& leq lim_{m to infty} |f_m - f_n| |x-y|
end{align*}
Now since $(f_n)$ is Cauchy in $mathcal{L}$, there is $N$ independent of $x,y$ such that for $n,m geq N$ $|f_n - f_m| leq varepsilon$. So we get
$$|f(x) - f(y) - f_n(x) + f_n(y)| leq varepsilon |x-y|$$
which implies for $n geq N$, $|f-f_n| leq varepsilon$ so $f_n to f$ in $mathcal{L}$.
answered Feb 3 at 9:51
Rhys SteeleRhys Steele
7,9301931
$endgroup$
add a comment |
$begingroup$
Your proof has all of the right ideas but suffers from a couple of defects since you end up implicitly doing things like assuming $f in mathcal{L}$ before you've proved it and confusing the pointwise limit with the limit in $mathcal{L}$. To fix this, you just need to write everything down in terms of the definition of the norm and make sure you're working with pointwise limits. I include details below.
Firstly, when trying to prove that $f in mathcal{L}$, on the second line you write down $|f-f_N|$ and then bound this by $1$. Doing this already requires you to assume $f-f_N in mathcal{L}$ and hence $f in mathcal{L}$ and also something like $f_n to f$ in $mathcal{L}$.
Instead, write
begin{align*}
|f(x) - f(y)| &= |lim_{n to infty} (f_n(x) - f_n(y))|
\&= lim_{n to infty} |f_n(x) - f_n(y)|
\& leq lim_{n to infty} |f_n| |x-y|
\& leq C|x-y|
end{align*}
where in the last line we used that $|f_n|$ is a bounded sequence. Note that in the second line, I only use that $f_n to f$ pointwise and not in $mathcal{L}$.
This is important because when trying to prove that $f_n to f$, you take a pointwise limit and take it outside of the $mathcal{L}$-norm. This doesn't work since you can only do that once you know that the limit converges in $mathcal{L}$ and not just pointwise. Instead, work again with a pointwise limit by writing
begin{align*}
|f(x) - f(y) - f_n(x) + f_n(y)| &= lim_{m to infty} |f_m(x) - f_m(y) - f_n(x) + f_n(y)|\
& leq lim_{m to infty} |f_m - f_n| |x-y|
end{align*}
Now since $(f_n)$ is Cauchy in $mathcal{L}$, there is $N$ independent of $x,y$ such that for $n,m geq N$ $|f_n - f_m| leq varepsilon$. So we get
$$|f(x) - f(y) - f_n(x) + f_n(y)| leq varepsilon |x-y|$$
which implies for $n geq N$, $|f-f_n| leq varepsilon$ so $f_n to f$ in $mathcal{L}$.
answered Feb 3 at 9:51
Rhys SteeleRhys Steele
7,9301931
$endgroup$
Your proof has all of the right ideas but suffers from a couple of defects since you end up implicitly doing things like assuming $f in mathcal{L}$ before you've proved it and confusing the pointwise limit with the limit in $mathcal{L}$. To fix this, you just need to write everything down in terms of the definition of the norm and make sure you're working with pointwise limits. I include details below.
Firstly, when trying to prove that $f in mathcal{L}$, on the second line you write down $|f-f_N|$ and then bound this by $1$. Doing this already requires you to assume $f-f_N in mathcal{L}$ and hence $f in mathcal{L}$ and also something like $f_n to f$ in $mathcal{L}$.
Instead, write
begin{align*}
|f(x) - f(y)| &= |lim_{n to infty} (f_n(x) - f_n(y))|
\&= lim_{n to infty} |f_n(x) - f_n(y)|
\& leq lim_{n to infty} |f_n| |x-y|
\& leq C|x-y|
end{align*}
where in the last line we used that $|f_n|$ is a bounded sequence. Note that in the second line, I only use that $f_n to f$ pointwise and not in $mathcal{L}$.
This is important because when trying to prove that $f_n to f$, you take a pointwise limit and take it outside of the $mathcal{L}$-norm. This doesn't work since you can only do that once you know that the limit converges in $mathcal{L}$ and not just pointwise. Instead, work again with a pointwise limit by writing
begin{align*}
|f(x) - f(y) - f_n(x) + f_n(y)| &= lim_{m to infty} |f_m(x) - f_m(y) - f_n(x) + f_n(y)|\
& leq lim_{m to infty} |f_m - f_n| |x-y|
end{align*}
Now since $(f_n)$ is Cauchy in $mathcal{L}$, there is $N$ independent of $x,y$ such that for $n,m geq N$ $|f_n - f_m| leq varepsilon$. So we get
$$|f(x) - f(y) - f_n(x) + f_n(y)| leq varepsilon |x-y|$$
which implies for $n geq N$, $|f-f_n| leq varepsilon$ so $f_n to f$ in $mathcal{L}$.
answered Feb 3 at 9:51
Rhys SteeleRhys Steele
7,9301931
answered Feb 3 at 9:51
Rhys SteeleRhys Steele
7,9301931
answered Feb 3 at 9:51
Rhys SteeleRhys Steele
7,9301931
answered Feb 3 at 9:51
Rhys SteeleRhys Steele
7,9301931
7,9301931
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$begingroup$
Why define $f(0) = 0$ instead of $f(0) = lim_{n to infty} f_n(0)$? This means, if your Cauchy sequence is constantly, say, the constant function $1$, then $f(x)$ will be the function that is constantly $1$ except at $0$, where it's $0$. Such a function is not continuous, let alone Lipschitz.
$endgroup$
– Theo Bendit
Feb 3 at 5:50
$begingroup$
@TheoBendit Yes, you are right. Edited my proof.
$endgroup$
– Idonknow
Feb 3 at 5:52
$begingroup$
Wait, I'm an idiot, I just noticed the requirement that $f(0) = 0$ for membership in the space. The constant function $1$ does not belong to the space in the first place!
$endgroup$
– Theo Bendit
Feb 3 at 5:54
$begingroup$
@TheoBendit Yes. But I think your suggestion still works and it is neater than my suggestion.
$endgroup$
– Idonknow
Feb 3 at 5:59