Mixing
Prove discontinuity of $U(f)= f^3$
$begingroup$
Let $U:C([0,1])to C([0,1])$ defined as $U(f):=f^3$. I want to show it's not continuous in the supremum norm (i.e. $L^infty$ norm) nor in the $L^1$ norm.
I can show the discontinuity with a counter example, but I couldn't find one. Moreover I know the space $(C^0([a,b]), L^1)$ is not complete, maybe this can help.
I tought to find a sequence of functions that converges in $C([0,1])$ but its cubes don't. Is this the right way?
functional-analysis continuity
asked Jan 30 at 15:14
james wattjames watt
373110
$endgroup$
add a comment |
$begingroup$
Let $U:C([0,1])to C([0,1])$ defined as $U(f):=f^3$. I want to show it's not continuous in the supremum norm (i.e. $L^infty$ norm) nor in the $L^1$ norm.
I can show the discontinuity with a counter example, but I couldn't find one. Moreover I know the space $(C^0([a,b]), L^1)$ is not complete, maybe this can help.
I tought to find a sequence of functions that converges in $C([0,1])$ but its cubes don't. Is this the right way?
functional-analysis continuity
asked Jan 30 at 15:14
james wattjames watt
373110
$endgroup$
add a comment |
$begingroup$
Let $U:C([0,1])to C([0,1])$ defined as $U(f):=f^3$. I want to show it's not continuous in the supremum norm (i.e. $L^infty$ norm) nor in the $L^1$ norm.
I can show the discontinuity with a counter example, but I couldn't find one. Moreover I know the space $(C^0([a,b]), L^1)$ is not complete, maybe this can help.
I tought to find a sequence of functions that converges in $C([0,1])$ but its cubes don't. Is this the right way?
functional-analysis continuity
asked Jan 30 at 15:14
james wattjames watt
373110
$endgroup$
Let $U:C([0,1])to C([0,1])$ defined as $U(f):=f^3$. I want to show it's not continuous in the supremum norm (i.e. $L^infty$ norm) nor in the $L^1$ norm.
I can show the discontinuity with a counter example, but I couldn't find one. Moreover I know the space $(C^0([a,b]), L^1)$ is not complete, maybe this can help.
I tought to find a sequence of functions that converges in $C([0,1])$ but its cubes don't. Is this the right way?
functional-analysis continuity
functional-analysis continuity
asked Jan 30 at 15:14
james wattjames watt
373110
asked Jan 30 at 15:14
james wattjames watt
373110
asked Jan 30 at 15:14
james wattjames watt
373110
asked Jan 30 at 15:14
james wattjames watt
373110
asked Jan 30 at 15:14
james wattjames watt
373110
373110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is continuous with respect to the supremum norm.
If $f_nrightarrow f$ in the sup norm, let $M$ be such that $|f_n|_infty,|f|_infty<M$. For any two functions $f,g$ we have $f^3-g^3 = (f-g)(f^2+fg+g^2)$ and $|fg|_inftyleq |f|_inftycdot|g|_infty$. From this we see that $$|f^3-f_n^3|_infty leq |f-f_n|_infty|f^2+fcdot f_n+f_n^2|_infty$$
But $|f^2+fcdot f_n+f_n^2|leq 3M^2$ so we have
$$|f^3-f_n^3|_infty leq 3M^2|f-f_n|_inftyrightarrow 0$$
It is not continuous with respect to the $L^1$ norm
First note that if we can find $M$ such that $|f_n|_infty,|f|_infty<M$ then by using the inequality $|fg|_{L^1}leq |f|_{L^1}cdot |g|_infty$ and arguing as before we will get a convergence.
So in order to find a counter-example we need that no such $M$ exists. Let me first construct a counter-example with non-continuous functions and then explain how one can correct this:
Let $f_n(x)$ be the function which takes the value $n$ on the inverval $[0,frac{1}{n^2}]$ and zero otherwise then
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = ncdot frac{1}{n^2}=frac{1}{n}rightarrow 0$$ therefore $f_nrightarrow 0$ in $L^1$ (and is not bounded of course).
However $(f_n(x))^3$ takes the value $n^3$ again on $[0,frac{1}{n^2}]$. Same calculation as before gives
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = n^3cdot frac{1}{n^2}=nrightarrow infty$$
which diverge!
In order to deal with the discontinuity one can replace $f_n$ on a small interval $[frac{1}{n^2},frac{1}{n^2}+varepsilon]$ by a "line" which connectes $f_n(frac{1}{n^2})$ with $0$ (try to draw it) and then argue as before.
answered Jan 30 at 15:21
YankoYanko
8,3142830
$endgroup$
$begingroup$
Why $|f^2+fcdot f_n + f_n^2| leq 3$ ?
$endgroup$
– lisyarus
Jan 30 at 15:25
$begingroup$
When you say $le 3$, you mean $le 3 sup_{ninmathbb N}|f_n|_infty^2$ (which is finite since the sequence $(f_n)$ is convergent and hence bounded with respect to $|cdot|_infty$).
$endgroup$
– Mars Plastic
Jan 30 at 15:25
$begingroup$
@lisyarus because $f,f_n$ are into $[0,1]$.
$endgroup$
– Yanko
Jan 30 at 15:25
1
$begingroup$
@Yanko No, they are not. They are functions from $[0,1]$ into $mathbb C$ (or $mathbb R$). At least this is what is usually denoted by $C([0,1])$.
$endgroup$
– Mars Plastic
Jan 30 at 15:27
1
$begingroup$
@Yanko Why are they into [0, 1]? $C([0, 1])$ means continuous functions with the domain being [0,1], not the range.
$endgroup$
– lisyarus
Jan 30 at 15:27
|
show 6 more comments
$begingroup$
Considering the $L^1$-norm, one might think of functions like $f_n=1_{[0,n^{-2}]}cdot n$. They converge to zero with respect to the $L^1$-norm, but the cubes diverge. However, these functions are obviously not continuous. This is not a major problem though, you just have to "smoothen" the jump. Do you know how to do that?
answered Jan 30 at 15:41
Mars PlasticMars Plastic
1,477122
$endgroup$
$begingroup$
(+1) Looks like you beat me to it.
$endgroup$
– Yanko
Jan 30 at 15:47
$begingroup$
Haha, no competition intended. ;-)
$endgroup$
– Mars Plastic
Jan 30 at 15:50
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It is continuous with respect to the supremum norm.
If $f_nrightarrow f$ in the sup norm, let $M$ be such that $|f_n|_infty,|f|_infty<M$. For any two functions $f,g$ we have $f^3-g^3 = (f-g)(f^2+fg+g^2)$ and $|fg|_inftyleq |f|_inftycdot|g|_infty$. From this we see that $$|f^3-f_n^3|_infty leq |f-f_n|_infty|f^2+fcdot f_n+f_n^2|_infty$$
But $|f^2+fcdot f_n+f_n^2|leq 3M^2$ so we have
$$|f^3-f_n^3|_infty leq 3M^2|f-f_n|_inftyrightarrow 0$$
It is not continuous with respect to the $L^1$ norm
First note that if we can find $M$ such that $|f_n|_infty,|f|_infty<M$ then by using the inequality $|fg|_{L^1}leq |f|_{L^1}cdot |g|_infty$ and arguing as before we will get a convergence.
So in order to find a counter-example we need that no such $M$ exists. Let me first construct a counter-example with non-continuous functions and then explain how one can correct this:
Let $f_n(x)$ be the function which takes the value $n$ on the inverval $[0,frac{1}{n^2}]$ and zero otherwise then
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = ncdot frac{1}{n^2}=frac{1}{n}rightarrow 0$$ therefore $f_nrightarrow 0$ in $L^1$ (and is not bounded of course).
However $(f_n(x))^3$ takes the value $n^3$ again on $[0,frac{1}{n^2}]$. Same calculation as before gives
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = n^3cdot frac{1}{n^2}=nrightarrow infty$$
which diverge!
In order to deal with the discontinuity one can replace $f_n$ on a small interval $[frac{1}{n^2},frac{1}{n^2}+varepsilon]$ by a "line" which connectes $f_n(frac{1}{n^2})$ with $0$ (try to draw it) and then argue as before.
answered Jan 30 at 15:21
YankoYanko
8,3142830
$endgroup$
$begingroup$
Why $|f^2+fcdot f_n + f_n^2| leq 3$ ?
$endgroup$
– lisyarus
Jan 30 at 15:25
$begingroup$
When you say $le 3$, you mean $le 3 sup_{ninmathbb N}|f_n|_infty^2$ (which is finite since the sequence $(f_n)$ is convergent and hence bounded with respect to $|cdot|_infty$).
$endgroup$
– Mars Plastic
Jan 30 at 15:25
$begingroup$
@lisyarus because $f,f_n$ are into $[0,1]$.
$endgroup$
– Yanko
Jan 30 at 15:25
1
$begingroup$
@Yanko No, they are not. They are functions from $[0,1]$ into $mathbb C$ (or $mathbb R$). At least this is what is usually denoted by $C([0,1])$.
$endgroup$
– Mars Plastic
Jan 30 at 15:27
1
$begingroup$
@Yanko Why are they into [0, 1]? $C([0, 1])$ means continuous functions with the domain being [0,1], not the range.
$endgroup$
– lisyarus
Jan 30 at 15:27
|
show 6 more comments
$begingroup$
It is continuous with respect to the supremum norm.
If $f_nrightarrow f$ in the sup norm, let $M$ be such that $|f_n|_infty,|f|_infty<M$. For any two functions $f,g$ we have $f^3-g^3 = (f-g)(f^2+fg+g^2)$ and $|fg|_inftyleq |f|_inftycdot|g|_infty$. From this we see that $$|f^3-f_n^3|_infty leq |f-f_n|_infty|f^2+fcdot f_n+f_n^2|_infty$$
But $|f^2+fcdot f_n+f_n^2|leq 3M^2$ so we have
$$|f^3-f_n^3|_infty leq 3M^2|f-f_n|_inftyrightarrow 0$$
It is not continuous with respect to the $L^1$ norm
First note that if we can find $M$ such that $|f_n|_infty,|f|_infty<M$ then by using the inequality $|fg|_{L^1}leq |f|_{L^1}cdot |g|_infty$ and arguing as before we will get a convergence.
So in order to find a counter-example we need that no such $M$ exists. Let me first construct a counter-example with non-continuous functions and then explain how one can correct this:
Let $f_n(x)$ be the function which takes the value $n$ on the inverval $[0,frac{1}{n^2}]$ and zero otherwise then
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = ncdot frac{1}{n^2}=frac{1}{n}rightarrow 0$$ therefore $f_nrightarrow 0$ in $L^1$ (and is not bounded of course).
However $(f_n(x))^3$ takes the value $n^3$ again on $[0,frac{1}{n^2}]$. Same calculation as before gives
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = n^3cdot frac{1}{n^2}=nrightarrow infty$$
which diverge!
In order to deal with the discontinuity one can replace $f_n$ on a small interval $[frac{1}{n^2},frac{1}{n^2}+varepsilon]$ by a "line" which connectes $f_n(frac{1}{n^2})$ with $0$ (try to draw it) and then argue as before.
answered Jan 30 at 15:21
YankoYanko
8,3142830
$endgroup$
$begingroup$
Why $|f^2+fcdot f_n + f_n^2| leq 3$ ?
$endgroup$
– lisyarus
Jan 30 at 15:25
$begingroup$
When you say $le 3$, you mean $le 3 sup_{ninmathbb N}|f_n|_infty^2$ (which is finite since the sequence $(f_n)$ is convergent and hence bounded with respect to $|cdot|_infty$).
$endgroup$
– Mars Plastic
Jan 30 at 15:25
$begingroup$
@lisyarus because $f,f_n$ are into $[0,1]$.
$endgroup$
– Yanko
Jan 30 at 15:25
1
$begingroup$
@Yanko No, they are not. They are functions from $[0,1]$ into $mathbb C$ (or $mathbb R$). At least this is what is usually denoted by $C([0,1])$.
$endgroup$
– Mars Plastic
Jan 30 at 15:27
1
$begingroup$
@Yanko Why are they into [0, 1]? $C([0, 1])$ means continuous functions with the domain being [0,1], not the range.
$endgroup$
– lisyarus
Jan 30 at 15:27
|
show 6 more comments
$begingroup$
It is continuous with respect to the supremum norm.
If $f_nrightarrow f$ in the sup norm, let $M$ be such that $|f_n|_infty,|f|_infty<M$. For any two functions $f,g$ we have $f^3-g^3 = (f-g)(f^2+fg+g^2)$ and $|fg|_inftyleq |f|_inftycdot|g|_infty$. From this we see that $$|f^3-f_n^3|_infty leq |f-f_n|_infty|f^2+fcdot f_n+f_n^2|_infty$$
But $|f^2+fcdot f_n+f_n^2|leq 3M^2$ so we have
$$|f^3-f_n^3|_infty leq 3M^2|f-f_n|_inftyrightarrow 0$$
It is not continuous with respect to the $L^1$ norm
First note that if we can find $M$ such that $|f_n|_infty,|f|_infty<M$ then by using the inequality $|fg|_{L^1}leq |f|_{L^1}cdot |g|_infty$ and arguing as before we will get a convergence.
So in order to find a counter-example we need that no such $M$ exists. Let me first construct a counter-example with non-continuous functions and then explain how one can correct this:
Let $f_n(x)$ be the function which takes the value $n$ on the inverval $[0,frac{1}{n^2}]$ and zero otherwise then
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = ncdot frac{1}{n^2}=frac{1}{n}rightarrow 0$$ therefore $f_nrightarrow 0$ in $L^1$ (and is not bounded of course).
However $(f_n(x))^3$ takes the value $n^3$ again on $[0,frac{1}{n^2}]$. Same calculation as before gives
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = n^3cdot frac{1}{n^2}=nrightarrow infty$$
which diverge!
In order to deal with the discontinuity one can replace $f_n$ on a small interval $[frac{1}{n^2},frac{1}{n^2}+varepsilon]$ by a "line" which connectes $f_n(frac{1}{n^2})$ with $0$ (try to draw it) and then argue as before.
answered Jan 30 at 15:21
YankoYanko
8,3142830
$endgroup$
It is continuous with respect to the supremum norm.
If $f_nrightarrow f$ in the sup norm, let $M$ be such that $|f_n|_infty,|f|_infty<M$. For any two functions $f,g$ we have $f^3-g^3 = (f-g)(f^2+fg+g^2)$ and $|fg|_inftyleq |f|_inftycdot|g|_infty$. From this we see that $$|f^3-f_n^3|_infty leq |f-f_n|_infty|f^2+fcdot f_n+f_n^2|_infty$$
But $|f^2+fcdot f_n+f_n^2|leq 3M^2$ so we have
$$|f^3-f_n^3|_infty leq 3M^2|f-f_n|_inftyrightarrow 0$$
It is not continuous with respect to the $L^1$ norm
First note that if we can find $M$ such that $|f_n|_infty,|f|_infty<M$ then by using the inequality $|fg|_{L^1}leq |f|_{L^1}cdot |g|_infty$ and arguing as before we will get a convergence.
So in order to find a counter-example we need that no such $M$ exists. Let me first construct a counter-example with non-continuous functions and then explain how one can correct this:
Let $f_n(x)$ be the function which takes the value $n$ on the inverval $[0,frac{1}{n^2}]$ and zero otherwise then
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = ncdot frac{1}{n^2}=frac{1}{n}rightarrow 0$$ therefore $f_nrightarrow 0$ in $L^1$ (and is not bounded of course).
However $(f_n(x))^3$ takes the value $n^3$ again on $[0,frac{1}{n^2}]$. Same calculation as before gives
$$int_0^1 |f_n(x)|dx = int_0^{frac{1}{n^2}}n = n^3cdot frac{1}{n^2}=nrightarrow infty$$
which diverge!
In order to deal with the discontinuity one can replace $f_n$ on a small interval $[frac{1}{n^2},frac{1}{n^2}+varepsilon]$ by a "line" which connectes $f_n(frac{1}{n^2})$ with $0$ (try to draw it) and then argue as before.
answered Jan 30 at 15:21
YankoYanko
8,3142830
edited Jan 30 at 15:46
answered Jan 30 at 15:21
YankoYanko
8,3142830
answered Jan 30 at 15:21
YankoYanko
8,3142830
answered Jan 30 at 15:21
YankoYanko
8,3142830
8,3142830
$begingroup$
Why $|f^2+fcdot f_n + f_n^2| leq 3$ ?
$endgroup$
– lisyarus
Jan 30 at 15:25
$begingroup$
When you say $le 3$, you mean $le 3 sup_{ninmathbb N}|f_n|_infty^2$ (which is finite since the sequence $(f_n)$ is convergent and hence bounded with respect to $|cdot|_infty$).
$endgroup$
– Mars Plastic
Jan 30 at 15:25
$begingroup$
@lisyarus because $f,f_n$ are into $[0,1]$.
$endgroup$
– Yanko
Jan 30 at 15:25
1
$begingroup$
@Yanko No, they are not. They are functions from $[0,1]$ into $mathbb C$ (or $mathbb R$). At least this is what is usually denoted by $C([0,1])$.
$endgroup$
– Mars Plastic
Jan 30 at 15:27
1
$begingroup$
@Yanko Why are they into [0, 1]? $C([0, 1])$ means continuous functions with the domain being [0,1], not the range.
$endgroup$
– lisyarus
Jan 30 at 15:27
|
show 6 more comments
$begingroup$
Why $|f^2+fcdot f_n + f_n^2| leq 3$ ?
$endgroup$
– lisyarus
Jan 30 at 15:25
$begingroup$
When you say $le 3$, you mean $le 3 sup_{ninmathbb N}|f_n|_infty^2$ (which is finite since the sequence $(f_n)$ is convergent and hence bounded with respect to $|cdot|_infty$).
$endgroup$
– Mars Plastic
Jan 30 at 15:25
$begingroup$
@lisyarus because $f,f_n$ are into $[0,1]$.
$endgroup$
– Yanko
Jan 30 at 15:25
1
$begingroup$
@Yanko No, they are not. They are functions from $[0,1]$ into $mathbb C$ (or $mathbb R$). At least this is what is usually denoted by $C([0,1])$.
$endgroup$
– Mars Plastic
Jan 30 at 15:27
1
$begingroup$
@Yanko Why are they into [0, 1]? $C([0, 1])$ means continuous functions with the domain being [0,1], not the range.
$endgroup$
– lisyarus
Jan 30 at 15:27
$begingroup$
Why $|f^2+fcdot f_n + f_n^2| leq 3$ ?
$endgroup$
– lisyarus
Jan 30 at 15:25
$begingroup$
Why $|f^2+fcdot f_n + f_n^2| leq 3$ ?
$endgroup$
– lisyarus
Jan 30 at 15:25
$begingroup$
When you say $le 3$, you mean $le 3 sup_{ninmathbb N}|f_n|_infty^2$ (which is finite since the sequence $(f_n)$ is convergent and hence bounded with respect to $|cdot|_infty$).
$endgroup$
– Mars Plastic
Jan 30 at 15:25
$begingroup$
When you say $le 3$, you mean $le 3 sup_{ninmathbb N}|f_n|_infty^2$ (which is finite since the sequence $(f_n)$ is convergent and hence bounded with respect to $|cdot|_infty$).
$endgroup$
– Mars Plastic
Jan 30 at 15:25
$begingroup$
@lisyarus because $f,f_n$ are into $[0,1]$.
$endgroup$
– Yanko
Jan 30 at 15:25
$begingroup$
@lisyarus because $f,f_n$ are into $[0,1]$.
$endgroup$
– Yanko
Jan 30 at 15:25
1
1
$begingroup$
@Yanko No, they are not. They are functions from $[0,1]$ into $mathbb C$ (or $mathbb R$). At least this is what is usually denoted by $C([0,1])$.
$endgroup$
– Mars Plastic
Jan 30 at 15:27
$begingroup$
@Yanko No, they are not. They are functions from $[0,1]$ into $mathbb C$ (or $mathbb R$). At least this is what is usually denoted by $C([0,1])$.
$endgroup$
– Mars Plastic
Jan 30 at 15:27
1
1
$begingroup$
@Yanko Why are they into [0, 1]? $C([0, 1])$ means continuous functions with the domain being [0,1], not the range.
$endgroup$
– lisyarus
Jan 30 at 15:27
$begingroup$
@Yanko Why are they into [0, 1]? $C([0, 1])$ means continuous functions with the domain being [0,1], not the range.
$endgroup$
– lisyarus
Jan 30 at 15:27
|
show 6 more comments
$begingroup$
Considering the $L^1$-norm, one might think of functions like $f_n=1_{[0,n^{-2}]}cdot n$. They converge to zero with respect to the $L^1$-norm, but the cubes diverge. However, these functions are obviously not continuous. This is not a major problem though, you just have to "smoothen" the jump. Do you know how to do that?
answered Jan 30 at 15:41
Mars PlasticMars Plastic
1,477122
$endgroup$
$begingroup$
(+1) Looks like you beat me to it.
$endgroup$
– Yanko
Jan 30 at 15:47
$begingroup$
Haha, no competition intended. ;-)
$endgroup$
– Mars Plastic
Jan 30 at 15:50
add a comment |
$begingroup$
Considering the $L^1$-norm, one might think of functions like $f_n=1_{[0,n^{-2}]}cdot n$. They converge to zero with respect to the $L^1$-norm, but the cubes diverge. However, these functions are obviously not continuous. This is not a major problem though, you just have to "smoothen" the jump. Do you know how to do that?
answered Jan 30 at 15:41
Mars PlasticMars Plastic
1,477122
$endgroup$
$begingroup$
(+1) Looks like you beat me to it.
$endgroup$
– Yanko
Jan 30 at 15:47
$begingroup$
Haha, no competition intended. ;-)
$endgroup$
– Mars Plastic
Jan 30 at 15:50
add a comment |
$begingroup$
Considering the $L^1$-norm, one might think of functions like $f_n=1_{[0,n^{-2}]}cdot n$. They converge to zero with respect to the $L^1$-norm, but the cubes diverge. However, these functions are obviously not continuous. This is not a major problem though, you just have to "smoothen" the jump. Do you know how to do that?
answered Jan 30 at 15:41
Mars PlasticMars Plastic
1,477122
$endgroup$
Considering the $L^1$-norm, one might think of functions like $f_n=1_{[0,n^{-2}]}cdot n$. They converge to zero with respect to the $L^1$-norm, but the cubes diverge. However, these functions are obviously not continuous. This is not a major problem though, you just have to "smoothen" the jump. Do you know how to do that?
answered Jan 30 at 15:41
Mars PlasticMars Plastic
1,477122
answered Jan 30 at 15:41
Mars PlasticMars Plastic
1,477122
answered Jan 30 at 15:41
Mars PlasticMars Plastic
1,477122
answered Jan 30 at 15:41
Mars PlasticMars Plastic
1,477122
1,477122
$begingroup$
(+1) Looks like you beat me to it.
$endgroup$
– Yanko
Jan 30 at 15:47
$begingroup$
Haha, no competition intended. ;-)
$endgroup$
– Mars Plastic
Jan 30 at 15:50
add a comment |
$begingroup$
(+1) Looks like you beat me to it.
$endgroup$
– Yanko
Jan 30 at 15:47
$begingroup$
Haha, no competition intended. ;-)
$endgroup$
– Mars Plastic
Jan 30 at 15:50
$begingroup$
(+1) Looks like you beat me to it.
$endgroup$
– Yanko
Jan 30 at 15:47
$begingroup$
(+1) Looks like you beat me to it.
$endgroup$
– Yanko
Jan 30 at 15:47
$begingroup$
Haha, no competition intended. ;-)
$endgroup$
– Mars Plastic
Jan 30 at 15:50
$begingroup$
Haha, no competition intended. ;-)
$endgroup$
– Mars Plastic
Jan 30 at 15:50
add a comment |
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