Mixing
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If $f(x)=f(y)$ implies that $g(x)=g(y)$ is $g$ a function of $f$?
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Let $f: {mathbb R}^nto {mathbb R}^p$ and $g: {mathbb R}^n to {mathbb R}^q$ be two non-constant functions such that: $f(x)=f(y)$ implies that $g(x)=g(y)$, can we say that $g$ a function of $f$?
This is: $g(x)= h(f(x))$ for some function $h$.
functions
asked Jan 20 at 17:22
pirata piratapirata pirata
32
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add a comment |
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Let $f: {mathbb R}^nto {mathbb R}^p$ and $g: {mathbb R}^n to {mathbb R}^q$ be two non-constant functions such that: $f(x)=f(y)$ implies that $g(x)=g(y)$, can we say that $g$ a function of $f$?
This is: $g(x)= h(f(x))$ for some function $h$.
functions
asked Jan 20 at 17:22
pirata piratapirata pirata
32
$endgroup$
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How about a piece-wise function? Must $f$ and $g$ be continuous?
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– Aniruddh Venkatesan
Jan 20 at 17:24
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@AniruddhVenkatesan I am not necessarily assuming continuity. I have only proven that property in general. I just wonder if there is such implication.
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– pirata pirata
Jan 20 at 17:29
add a comment |
$begingroup$
Let $f: {mathbb R}^nto {mathbb R}^p$ and $g: {mathbb R}^n to {mathbb R}^q$ be two non-constant functions such that: $f(x)=f(y)$ implies that $g(x)=g(y)$, can we say that $g$ a function of $f$?
This is: $g(x)= h(f(x))$ for some function $h$.
functions
asked Jan 20 at 17:22
pirata piratapirata pirata
32
$endgroup$
Let $f: {mathbb R}^nto {mathbb R}^p$ and $g: {mathbb R}^n to {mathbb R}^q$ be two non-constant functions such that: $f(x)=f(y)$ implies that $g(x)=g(y)$, can we say that $g$ a function of $f$?
This is: $g(x)= h(f(x))$ for some function $h$.
functions
functions
asked Jan 20 at 17:22
pirata piratapirata pirata
32
asked Jan 20 at 17:22
pirata piratapirata pirata
32
edited Jan 20 at 17:37
pirata pirata
asked Jan 20 at 17:22
pirata piratapirata pirata
32
asked Jan 20 at 17:22
pirata piratapirata pirata
32
asked Jan 20 at 17:22
pirata piratapirata pirata
32
32
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How about a piece-wise function? Must $f$ and $g$ be continuous?
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– Aniruddh Venkatesan
Jan 20 at 17:24
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@AniruddhVenkatesan I am not necessarily assuming continuity. I have only proven that property in general. I just wonder if there is such implication.
$endgroup$
– pirata pirata
Jan 20 at 17:29
add a comment |
$begingroup$
How about a piece-wise function? Must $f$ and $g$ be continuous?
$endgroup$
– Aniruddh Venkatesan
Jan 20 at 17:24
$begingroup$
@AniruddhVenkatesan I am not necessarily assuming continuity. I have only proven that property in general. I just wonder if there is such implication.
$endgroup$
– pirata pirata
Jan 20 at 17:29
$begingroup$
How about a piece-wise function? Must $f$ and $g$ be continuous?
$endgroup$
– Aniruddh Venkatesan
Jan 20 at 17:24
$begingroup$
How about a piece-wise function? Must $f$ and $g$ be continuous?
$endgroup$
– Aniruddh Venkatesan
Jan 20 at 17:24
$begingroup$
@AniruddhVenkatesan I am not necessarily assuming continuity. I have only proven that property in general. I just wonder if there is such implication.
$endgroup$
– pirata pirata
Jan 20 at 17:29
$begingroup$
@AniruddhVenkatesan I am not necessarily assuming continuity. I have only proven that property in general. I just wonder if there is such implication.
$endgroup$
– pirata pirata
Jan 20 at 17:29
add a comment |
5 Answers
5
active
oldest
votes
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Define
$$h(x) = left{
begin{array}{lc}
g(z) & mbox{ if there exists some } z mbox{ such that } f(z)=x \
mbox{ anything} & mbox{otherwise}
end{array}
right.$$
Show that the given condition implies that $h$ is well defined, that is it does not depend on the choice of $z$.
Then, by definition
$$h(f(x))=g(x) forall x in mathbb R^d$$
answered Jan 20 at 17:43
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Your question is a bit unclear. But I'm gonna go on a limb here and assume that when you say "$g$ is a function of $f$" you mean as a composition, i.e. you're asking whether we can write that $g(x)=h(f(x))$ with some new function $h$. In that case, the answer is yes.
So, let's say we're given two functions $f:Ato B$ and $g:Ato C$, and we know that $f(x)=f(y)$ implies $g(x)=g(y)$ for any $x,yin A$. Then indeed we can construct a function $h:Bto C$ such that $g=hcirc f$, i.e. $g(x)=h(f(x))$ for all $xin A$.
Here's the definition of $h$. Take an arbitrary $zinoperatorname{Range}(f)$. This means that $z=f(x)$ for some $xin A$. Define $h(z)=g(x)$. Note that this definition may depend on the choice of $x$, but thanks to the given condition it doesn't! Indeed, if we have two different elements $x,yin A$ such that $f(x)=z$ and $f(y)=z$, then from $f(x)=f(y)$ we know that $g(x)=g(y)$, giving the same well-defined value for $h(z)$.
One remaining technicality is to define $h(z)$ for all remaining $zin Bsetminusoperatorname{Range}(f)$, and we can set those to be equal to anything we want.
Now, by the very definition this function satisfies $g(x)=h(f(x))$, as desired.
answered Jan 20 at 17:45
zipirovichzipirovich
11.3k11731
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add a comment |
$begingroup$
More general: Let $fcolon Xto Y, gcolon Xto Z$ be mappings such that for all $x,yin X$ the condition $f(x)=f(y)$ implies $g(x)=g(y)$. Then we may define $h:Yto Z$ by $h(c):=g(x)$ if $c=f(x)$ for some $xin X$ and arbitrary if $c notin f(X)$. By assumption $h$ is well defined, i.e., a function. And by construction $g(x)=h(g(x))$ for all $xin X$.
answered Jan 20 at 17:56
Jens SchwaigerJens Schwaiger
1,606138
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add a comment |
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I'm assuming that why you mean by "$g$ is a function of $f$" is "there exists a function $mathfrak{F}$ such that $g=mathfrak{F}(f)$".
$$bigl(forall x,y:f(x)=f(y)implies g(x)=g(y)bigr)impliesexists mathfrak{F}:g=mathfrak{F}(f)$$
Consider $f(x)=x$. Any function $g$ having the same domain as $f$ will satisfy the first half of the above assertion. No matter what metafunction $mathfrak{F}$ we examine, there will be functions $g$ satisfying the first half but not the second half of our assertion, so the assertion must be false.
So the answer to your question is no. Except possibly for some trivial and pathological universes of values for $x$ and $y$.
answered Jan 20 at 17:41
ShapeOfMatterShapeOfMatter
1794
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Isn't $mathfrak{F}=g$ working in your example?
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– N. S.
Jan 20 at 17:44
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No. That wouldn't even make sense; the domain (and range) of $mathfrak{F}$ is the space of functions like $f$ and $g$.
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– ShapeOfMatter
Jan 20 at 17:46
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That said, it's not clear that my interpretation of the question is consistent with everyone else's.
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– ShapeOfMatter
Jan 20 at 17:46
1
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If you consider $mathfrak{F}$ to be a function from some space of functions, to some space of functions, then the answer is trivially yes. You simply pick $mathfrak{F}$ to be the constant function $mathfrak{F}$. All you need is the domain of $mathfrak{F}$ to contain $f$ and the range to contain $g$... and since the question is bout the existence of such $mathfrak{F}$ you can pick the domain and range to be anything.
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– N. S.
Jan 20 at 17:49
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no. Suppose $f:Bbb{I}mapstoBbb{I}$. If $mathfrak{F}(f)=f$, then for most $f$ $g(x)=2f(x)$ will break our assertion.
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– ShapeOfMatter
Jan 20 at 17:54
|
show 3 more comments
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If $forall x : not exists y, s.t. f(x) = f(y)$, then the claim is true, but $f$, $g$, could be pretty anything
answered Jan 20 at 17:42
dEmigOddEmigOd
1,5411612
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(a) The claim is true regardless. (b) Your quantifiers don't make sense. It's impossible to have "$forall x : not exists y, text{ s.t. } f(x) = f(y)$", because $y=x$ always satisfies $f(x)=f(y)$. You probably meant "$forall x : not exists y, text{ s.t. } xne y text{ and } f(x) = f(y)$".
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– zipirovich
Jan 20 at 17:50
add a comment |
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5 Answers
5
active
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5 Answers
5
active
oldest
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$begingroup$
Define
$$h(x) = left{
begin{array}{lc}
g(z) & mbox{ if there exists some } z mbox{ such that } f(z)=x \
mbox{ anything} & mbox{otherwise}
end{array}
right.$$
Show that the given condition implies that $h$ is well defined, that is it does not depend on the choice of $z$.
Then, by definition
$$h(f(x))=g(x) forall x in mathbb R^d$$
answered Jan 20 at 17:43
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Define
$$h(x) = left{
begin{array}{lc}
g(z) & mbox{ if there exists some } z mbox{ such that } f(z)=x \
mbox{ anything} & mbox{otherwise}
end{array}
right.$$
Show that the given condition implies that $h$ is well defined, that is it does not depend on the choice of $z$.
Then, by definition
$$h(f(x))=g(x) forall x in mathbb R^d$$
answered Jan 20 at 17:43
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Define
$$h(x) = left{
begin{array}{lc}
g(z) & mbox{ if there exists some } z mbox{ such that } f(z)=x \
mbox{ anything} & mbox{otherwise}
end{array}
right.$$
Show that the given condition implies that $h$ is well defined, that is it does not depend on the choice of $z$.
Then, by definition
$$h(f(x))=g(x) forall x in mathbb R^d$$
answered Jan 20 at 17:43
N. S.N. S.
104k7114209
$endgroup$
Define
$$h(x) = left{
begin{array}{lc}
g(z) & mbox{ if there exists some } z mbox{ such that } f(z)=x \
mbox{ anything} & mbox{otherwise}
end{array}
right.$$
Show that the given condition implies that $h$ is well defined, that is it does not depend on the choice of $z$.
Then, by definition
$$h(f(x))=g(x) forall x in mathbb R^d$$
answered Jan 20 at 17:43
N. S.N. S.
104k7114209
answered Jan 20 at 17:43
N. S.N. S.
104k7114209
answered Jan 20 at 17:43
N. S.N. S.
104k7114209
answered Jan 20 at 17:43
N. S.N. S.
104k7114209
104k7114209
add a comment |
add a comment |
$begingroup$
Your question is a bit unclear. But I'm gonna go on a limb here and assume that when you say "$g$ is a function of $f$" you mean as a composition, i.e. you're asking whether we can write that $g(x)=h(f(x))$ with some new function $h$. In that case, the answer is yes.
So, let's say we're given two functions $f:Ato B$ and $g:Ato C$, and we know that $f(x)=f(y)$ implies $g(x)=g(y)$ for any $x,yin A$. Then indeed we can construct a function $h:Bto C$ such that $g=hcirc f$, i.e. $g(x)=h(f(x))$ for all $xin A$.
Here's the definition of $h$. Take an arbitrary $zinoperatorname{Range}(f)$. This means that $z=f(x)$ for some $xin A$. Define $h(z)=g(x)$. Note that this definition may depend on the choice of $x$, but thanks to the given condition it doesn't! Indeed, if we have two different elements $x,yin A$ such that $f(x)=z$ and $f(y)=z$, then from $f(x)=f(y)$ we know that $g(x)=g(y)$, giving the same well-defined value for $h(z)$.
One remaining technicality is to define $h(z)$ for all remaining $zin Bsetminusoperatorname{Range}(f)$, and we can set those to be equal to anything we want.
Now, by the very definition this function satisfies $g(x)=h(f(x))$, as desired.
answered Jan 20 at 17:45
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
Your question is a bit unclear. But I'm gonna go on a limb here and assume that when you say "$g$ is a function of $f$" you mean as a composition, i.e. you're asking whether we can write that $g(x)=h(f(x))$ with some new function $h$. In that case, the answer is yes.
So, let's say we're given two functions $f:Ato B$ and $g:Ato C$, and we know that $f(x)=f(y)$ implies $g(x)=g(y)$ for any $x,yin A$. Then indeed we can construct a function $h:Bto C$ such that $g=hcirc f$, i.e. $g(x)=h(f(x))$ for all $xin A$.
Here's the definition of $h$. Take an arbitrary $zinoperatorname{Range}(f)$. This means that $z=f(x)$ for some $xin A$. Define $h(z)=g(x)$. Note that this definition may depend on the choice of $x$, but thanks to the given condition it doesn't! Indeed, if we have two different elements $x,yin A$ such that $f(x)=z$ and $f(y)=z$, then from $f(x)=f(y)$ we know that $g(x)=g(y)$, giving the same well-defined value for $h(z)$.
One remaining technicality is to define $h(z)$ for all remaining $zin Bsetminusoperatorname{Range}(f)$, and we can set those to be equal to anything we want.
Now, by the very definition this function satisfies $g(x)=h(f(x))$, as desired.
answered Jan 20 at 17:45
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
Your question is a bit unclear. But I'm gonna go on a limb here and assume that when you say "$g$ is a function of $f$" you mean as a composition, i.e. you're asking whether we can write that $g(x)=h(f(x))$ with some new function $h$. In that case, the answer is yes.
So, let's say we're given two functions $f:Ato B$ and $g:Ato C$, and we know that $f(x)=f(y)$ implies $g(x)=g(y)$ for any $x,yin A$. Then indeed we can construct a function $h:Bto C$ such that $g=hcirc f$, i.e. $g(x)=h(f(x))$ for all $xin A$.
Here's the definition of $h$. Take an arbitrary $zinoperatorname{Range}(f)$. This means that $z=f(x)$ for some $xin A$. Define $h(z)=g(x)$. Note that this definition may depend on the choice of $x$, but thanks to the given condition it doesn't! Indeed, if we have two different elements $x,yin A$ such that $f(x)=z$ and $f(y)=z$, then from $f(x)=f(y)$ we know that $g(x)=g(y)$, giving the same well-defined value for $h(z)$.
One remaining technicality is to define $h(z)$ for all remaining $zin Bsetminusoperatorname{Range}(f)$, and we can set those to be equal to anything we want.
Now, by the very definition this function satisfies $g(x)=h(f(x))$, as desired.
answered Jan 20 at 17:45
zipirovichzipirovich
11.3k11731
$endgroup$
Your question is a bit unclear. But I'm gonna go on a limb here and assume that when you say "$g$ is a function of $f$" you mean as a composition, i.e. you're asking whether we can write that $g(x)=h(f(x))$ with some new function $h$. In that case, the answer is yes.
So, let's say we're given two functions $f:Ato B$ and $g:Ato C$, and we know that $f(x)=f(y)$ implies $g(x)=g(y)$ for any $x,yin A$. Then indeed we can construct a function $h:Bto C$ such that $g=hcirc f$, i.e. $g(x)=h(f(x))$ for all $xin A$.
Here's the definition of $h$. Take an arbitrary $zinoperatorname{Range}(f)$. This means that $z=f(x)$ for some $xin A$. Define $h(z)=g(x)$. Note that this definition may depend on the choice of $x$, but thanks to the given condition it doesn't! Indeed, if we have two different elements $x,yin A$ such that $f(x)=z$ and $f(y)=z$, then from $f(x)=f(y)$ we know that $g(x)=g(y)$, giving the same well-defined value for $h(z)$.
One remaining technicality is to define $h(z)$ for all remaining $zin Bsetminusoperatorname{Range}(f)$, and we can set those to be equal to anything we want.
Now, by the very definition this function satisfies $g(x)=h(f(x))$, as desired.
answered Jan 20 at 17:45
zipirovichzipirovich
11.3k11731
answered Jan 20 at 17:45
zipirovichzipirovich
11.3k11731
answered Jan 20 at 17:45
zipirovichzipirovich
11.3k11731
answered Jan 20 at 17:45
zipirovichzipirovich
11.3k11731
11.3k11731
add a comment |
add a comment |
$begingroup$
More general: Let $fcolon Xto Y, gcolon Xto Z$ be mappings such that for all $x,yin X$ the condition $f(x)=f(y)$ implies $g(x)=g(y)$. Then we may define $h:Yto Z$ by $h(c):=g(x)$ if $c=f(x)$ for some $xin X$ and arbitrary if $c notin f(X)$. By assumption $h$ is well defined, i.e., a function. And by construction $g(x)=h(g(x))$ for all $xin X$.
answered Jan 20 at 17:56
Jens SchwaigerJens Schwaiger
1,606138
$endgroup$
add a comment |
$begingroup$
More general: Let $fcolon Xto Y, gcolon Xto Z$ be mappings such that for all $x,yin X$ the condition $f(x)=f(y)$ implies $g(x)=g(y)$. Then we may define $h:Yto Z$ by $h(c):=g(x)$ if $c=f(x)$ for some $xin X$ and arbitrary if $c notin f(X)$. By assumption $h$ is well defined, i.e., a function. And by construction $g(x)=h(g(x))$ for all $xin X$.
answered Jan 20 at 17:56
Jens SchwaigerJens Schwaiger
1,606138
$endgroup$
add a comment |
$begingroup$
More general: Let $fcolon Xto Y, gcolon Xto Z$ be mappings such that for all $x,yin X$ the condition $f(x)=f(y)$ implies $g(x)=g(y)$. Then we may define $h:Yto Z$ by $h(c):=g(x)$ if $c=f(x)$ for some $xin X$ and arbitrary if $c notin f(X)$. By assumption $h$ is well defined, i.e., a function. And by construction $g(x)=h(g(x))$ for all $xin X$.
answered Jan 20 at 17:56
Jens SchwaigerJens Schwaiger
1,606138
$endgroup$
More general: Let $fcolon Xto Y, gcolon Xto Z$ be mappings such that for all $x,yin X$ the condition $f(x)=f(y)$ implies $g(x)=g(y)$. Then we may define $h:Yto Z$ by $h(c):=g(x)$ if $c=f(x)$ for some $xin X$ and arbitrary if $c notin f(X)$. By assumption $h$ is well defined, i.e., a function. And by construction $g(x)=h(g(x))$ for all $xin X$.
answered Jan 20 at 17:56
Jens SchwaigerJens Schwaiger
1,606138
answered Jan 20 at 17:56
Jens SchwaigerJens Schwaiger
1,606138
answered Jan 20 at 17:56
Jens SchwaigerJens Schwaiger
1,606138
answered Jan 20 at 17:56
Jens SchwaigerJens Schwaiger
1,606138
1,606138
add a comment |
add a comment |
$begingroup$
I'm assuming that why you mean by "$g$ is a function of $f$" is "there exists a function $mathfrak{F}$ such that $g=mathfrak{F}(f)$".
$$bigl(forall x,y:f(x)=f(y)implies g(x)=g(y)bigr)impliesexists mathfrak{F}:g=mathfrak{F}(f)$$
Consider $f(x)=x$. Any function $g$ having the same domain as $f$ will satisfy the first half of the above assertion. No matter what metafunction $mathfrak{F}$ we examine, there will be functions $g$ satisfying the first half but not the second half of our assertion, so the assertion must be false.
So the answer to your question is no. Except possibly for some trivial and pathological universes of values for $x$ and $y$.
answered Jan 20 at 17:41
ShapeOfMatterShapeOfMatter
1794
$endgroup$
$begingroup$
Isn't $mathfrak{F}=g$ working in your example?
$endgroup$
– N. S.
Jan 20 at 17:44
$begingroup$
No. That wouldn't even make sense; the domain (and range) of $mathfrak{F}$ is the space of functions like $f$ and $g$.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
$begingroup$
That said, it's not clear that my interpretation of the question is consistent with everyone else's.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
1
$begingroup$
If you consider $mathfrak{F}$ to be a function from some space of functions, to some space of functions, then the answer is trivially yes. You simply pick $mathfrak{F}$ to be the constant function $mathfrak{F}$. All you need is the domain of $mathfrak{F}$ to contain $f$ and the range to contain $g$... and since the question is bout the existence of such $mathfrak{F}$ you can pick the domain and range to be anything.
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– N. S.
Jan 20 at 17:49
$begingroup$
no. Suppose $f:Bbb{I}mapstoBbb{I}$. If $mathfrak{F}(f)=f$, then for most $f$ $g(x)=2f(x)$ will break our assertion.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:54
|
show 3 more comments
$begingroup$
I'm assuming that why you mean by "$g$ is a function of $f$" is "there exists a function $mathfrak{F}$ such that $g=mathfrak{F}(f)$".
$$bigl(forall x,y:f(x)=f(y)implies g(x)=g(y)bigr)impliesexists mathfrak{F}:g=mathfrak{F}(f)$$
Consider $f(x)=x$. Any function $g$ having the same domain as $f$ will satisfy the first half of the above assertion. No matter what metafunction $mathfrak{F}$ we examine, there will be functions $g$ satisfying the first half but not the second half of our assertion, so the assertion must be false.
So the answer to your question is no. Except possibly for some trivial and pathological universes of values for $x$ and $y$.
answered Jan 20 at 17:41
ShapeOfMatterShapeOfMatter
1794
$endgroup$
$begingroup$
Isn't $mathfrak{F}=g$ working in your example?
$endgroup$
– N. S.
Jan 20 at 17:44
$begingroup$
No. That wouldn't even make sense; the domain (and range) of $mathfrak{F}$ is the space of functions like $f$ and $g$.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
$begingroup$
That said, it's not clear that my interpretation of the question is consistent with everyone else's.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
1
$begingroup$
If you consider $mathfrak{F}$ to be a function from some space of functions, to some space of functions, then the answer is trivially yes. You simply pick $mathfrak{F}$ to be the constant function $mathfrak{F}$. All you need is the domain of $mathfrak{F}$ to contain $f$ and the range to contain $g$... and since the question is bout the existence of such $mathfrak{F}$ you can pick the domain and range to be anything.
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– N. S.
Jan 20 at 17:49
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no. Suppose $f:Bbb{I}mapstoBbb{I}$. If $mathfrak{F}(f)=f$, then for most $f$ $g(x)=2f(x)$ will break our assertion.
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– ShapeOfMatter
Jan 20 at 17:54
|
show 3 more comments
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I'm assuming that why you mean by "$g$ is a function of $f$" is "there exists a function $mathfrak{F}$ such that $g=mathfrak{F}(f)$".
$$bigl(forall x,y:f(x)=f(y)implies g(x)=g(y)bigr)impliesexists mathfrak{F}:g=mathfrak{F}(f)$$
Consider $f(x)=x$. Any function $g$ having the same domain as $f$ will satisfy the first half of the above assertion. No matter what metafunction $mathfrak{F}$ we examine, there will be functions $g$ satisfying the first half but not the second half of our assertion, so the assertion must be false.
So the answer to your question is no. Except possibly for some trivial and pathological universes of values for $x$ and $y$.
answered Jan 20 at 17:41
ShapeOfMatterShapeOfMatter
1794
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I'm assuming that why you mean by "$g$ is a function of $f$" is "there exists a function $mathfrak{F}$ such that $g=mathfrak{F}(f)$".
$$bigl(forall x,y:f(x)=f(y)implies g(x)=g(y)bigr)impliesexists mathfrak{F}:g=mathfrak{F}(f)$$
Consider $f(x)=x$. Any function $g$ having the same domain as $f$ will satisfy the first half of the above assertion. No matter what metafunction $mathfrak{F}$ we examine, there will be functions $g$ satisfying the first half but not the second half of our assertion, so the assertion must be false.
So the answer to your question is no. Except possibly for some trivial and pathological universes of values for $x$ and $y$.
answered Jan 20 at 17:41
ShapeOfMatterShapeOfMatter
1794
answered Jan 20 at 17:41
ShapeOfMatterShapeOfMatter
1794
answered Jan 20 at 17:41
ShapeOfMatterShapeOfMatter
1794
answered Jan 20 at 17:41
ShapeOfMatterShapeOfMatter
1794
1794
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Isn't $mathfrak{F}=g$ working in your example?
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– N. S.
Jan 20 at 17:44
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No. That wouldn't even make sense; the domain (and range) of $mathfrak{F}$ is the space of functions like $f$ and $g$.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
$begingroup$
That said, it's not clear that my interpretation of the question is consistent with everyone else's.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
1
$begingroup$
If you consider $mathfrak{F}$ to be a function from some space of functions, to some space of functions, then the answer is trivially yes. You simply pick $mathfrak{F}$ to be the constant function $mathfrak{F}$. All you need is the domain of $mathfrak{F}$ to contain $f$ and the range to contain $g$... and since the question is bout the existence of such $mathfrak{F}$ you can pick the domain and range to be anything.
$endgroup$
– N. S.
Jan 20 at 17:49
$begingroup$
no. Suppose $f:Bbb{I}mapstoBbb{I}$. If $mathfrak{F}(f)=f$, then for most $f$ $g(x)=2f(x)$ will break our assertion.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:54
|
show 3 more comments
$begingroup$
Isn't $mathfrak{F}=g$ working in your example?
$endgroup$
– N. S.
Jan 20 at 17:44
$begingroup$
No. That wouldn't even make sense; the domain (and range) of $mathfrak{F}$ is the space of functions like $f$ and $g$.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
$begingroup$
That said, it's not clear that my interpretation of the question is consistent with everyone else's.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
1
$begingroup$
If you consider $mathfrak{F}$ to be a function from some space of functions, to some space of functions, then the answer is trivially yes. You simply pick $mathfrak{F}$ to be the constant function $mathfrak{F}$. All you need is the domain of $mathfrak{F}$ to contain $f$ and the range to contain $g$... and since the question is bout the existence of such $mathfrak{F}$ you can pick the domain and range to be anything.
$endgroup$
– N. S.
Jan 20 at 17:49
$begingroup$
no. Suppose $f:Bbb{I}mapstoBbb{I}$. If $mathfrak{F}(f)=f$, then for most $f$ $g(x)=2f(x)$ will break our assertion.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:54
$begingroup$
Isn't $mathfrak{F}=g$ working in your example?
$endgroup$
– N. S.
Jan 20 at 17:44
$begingroup$
Isn't $mathfrak{F}=g$ working in your example?
$endgroup$
– N. S.
Jan 20 at 17:44
$begingroup$
No. That wouldn't even make sense; the domain (and range) of $mathfrak{F}$ is the space of functions like $f$ and $g$.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
$begingroup$
No. That wouldn't even make sense; the domain (and range) of $mathfrak{F}$ is the space of functions like $f$ and $g$.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
$begingroup$
That said, it's not clear that my interpretation of the question is consistent with everyone else's.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
$begingroup$
That said, it's not clear that my interpretation of the question is consistent with everyone else's.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:46
1
1
$begingroup$
If you consider $mathfrak{F}$ to be a function from some space of functions, to some space of functions, then the answer is trivially yes. You simply pick $mathfrak{F}$ to be the constant function $mathfrak{F}$. All you need is the domain of $mathfrak{F}$ to contain $f$ and the range to contain $g$... and since the question is bout the existence of such $mathfrak{F}$ you can pick the domain and range to be anything.
$endgroup$
– N. S.
Jan 20 at 17:49
$begingroup$
If you consider $mathfrak{F}$ to be a function from some space of functions, to some space of functions, then the answer is trivially yes. You simply pick $mathfrak{F}$ to be the constant function $mathfrak{F}$. All you need is the domain of $mathfrak{F}$ to contain $f$ and the range to contain $g$... and since the question is bout the existence of such $mathfrak{F}$ you can pick the domain and range to be anything.
$endgroup$
– N. S.
Jan 20 at 17:49
$begingroup$
no. Suppose $f:Bbb{I}mapstoBbb{I}$. If $mathfrak{F}(f)=f$, then for most $f$ $g(x)=2f(x)$ will break our assertion.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:54
$begingroup$
no. Suppose $f:Bbb{I}mapstoBbb{I}$. If $mathfrak{F}(f)=f$, then for most $f$ $g(x)=2f(x)$ will break our assertion.
$endgroup$
– ShapeOfMatter
Jan 20 at 17:54
|
show 3 more comments
$begingroup$
If $forall x : not exists y, s.t. f(x) = f(y)$, then the claim is true, but $f$, $g$, could be pretty anything
answered Jan 20 at 17:42
dEmigOddEmigOd
1,5411612
$endgroup$
$begingroup$
(a) The claim is true regardless. (b) Your quantifiers don't make sense. It's impossible to have "$forall x : not exists y, text{ s.t. } f(x) = f(y)$", because $y=x$ always satisfies $f(x)=f(y)$. You probably meant "$forall x : not exists y, text{ s.t. } xne y text{ and } f(x) = f(y)$".
$endgroup$
– zipirovich
Jan 20 at 17:50
add a comment |
$begingroup$
If $forall x : not exists y, s.t. f(x) = f(y)$, then the claim is true, but $f$, $g$, could be pretty anything
answered Jan 20 at 17:42
dEmigOddEmigOd
1,5411612
$endgroup$
$begingroup$
(a) The claim is true regardless. (b) Your quantifiers don't make sense. It's impossible to have "$forall x : not exists y, text{ s.t. } f(x) = f(y)$", because $y=x$ always satisfies $f(x)=f(y)$. You probably meant "$forall x : not exists y, text{ s.t. } xne y text{ and } f(x) = f(y)$".
$endgroup$
– zipirovich
Jan 20 at 17:50
add a comment |
$begingroup$
If $forall x : not exists y, s.t. f(x) = f(y)$, then the claim is true, but $f$, $g$, could be pretty anything
answered Jan 20 at 17:42
dEmigOddEmigOd
1,5411612
$endgroup$
If $forall x : not exists y, s.t. f(x) = f(y)$, then the claim is true, but $f$, $g$, could be pretty anything
answered Jan 20 at 17:42
dEmigOddEmigOd
1,5411612
answered Jan 20 at 17:42
dEmigOddEmigOd
1,5411612
answered Jan 20 at 17:42
dEmigOddEmigOd
1,5411612
answered Jan 20 at 17:42
dEmigOddEmigOd
1,5411612
1,5411612
$begingroup$
(a) The claim is true regardless. (b) Your quantifiers don't make sense. It's impossible to have "$forall x : not exists y, text{ s.t. } f(x) = f(y)$", because $y=x$ always satisfies $f(x)=f(y)$. You probably meant "$forall x : not exists y, text{ s.t. } xne y text{ and } f(x) = f(y)$".
$endgroup$
– zipirovich
Jan 20 at 17:50
add a comment |
$begingroup$
(a) The claim is true regardless. (b) Your quantifiers don't make sense. It's impossible to have "$forall x : not exists y, text{ s.t. } f(x) = f(y)$", because $y=x$ always satisfies $f(x)=f(y)$. You probably meant "$forall x : not exists y, text{ s.t. } xne y text{ and } f(x) = f(y)$".
$endgroup$
– zipirovich
Jan 20 at 17:50
$begingroup$
(a) The claim is true regardless. (b) Your quantifiers don't make sense. It's impossible to have "$forall x : not exists y, text{ s.t. } f(x) = f(y)$", because $y=x$ always satisfies $f(x)=f(y)$. You probably meant "$forall x : not exists y, text{ s.t. } xne y text{ and } f(x) = f(y)$".
$endgroup$
– zipirovich
Jan 20 at 17:50
$begingroup$
(a) The claim is true regardless. (b) Your quantifiers don't make sense. It's impossible to have "$forall x : not exists y, text{ s.t. } f(x) = f(y)$", because $y=x$ always satisfies $f(x)=f(y)$. You probably meant "$forall x : not exists y, text{ s.t. } xne y text{ and } f(x) = f(y)$".
$endgroup$
– zipirovich
Jan 20 at 17:50
add a comment |
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$begingroup$
How about a piece-wise function? Must $f$ and $g$ be continuous?
$endgroup$
– Aniruddh Venkatesan
Jan 20 at 17:24
$begingroup$
@AniruddhVenkatesan I am not necessarily assuming continuity. I have only proven that property in general. I just wonder if there is such implication.
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– pirata pirata
Jan 20 at 17:29