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Prove that, the function $f$ is injective: $f big(f(x)f(y)big) + f(x +y) = f(xy).$
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I need to learn, by which method, I can prove that the function $f$ is injective. I would like to ask you to explain this problem using more detailed, more understandable, clearer and simpler English words. There's only one way I can understand. For example;
Let $f: mathbb{R} tomathbb{R}$ and $f(x)=x-10$.
We have, $f(y)=y-10$, then $f(x)=f(y)$, we get $f(x)=f(y) Rightarrow x-10=y-10 Rightarrow x=y$. So, $f$ is injective.
And here is my problem:
Why can't we apply this method to this problem?
$f: mathbb{R} tomathbb{R}$ and $f(0)≠0$, such that, for all real numbers $x$ and $y$, $$f big(f(x)f(y)big) + f(x +y) = f(xy).$$
Prove that, $f$ is injective.
Thank you very much for teaching.
algebra-precalculus proof-writing contest-math functional-equations alternative-proof
asked Nov 12 at 21:23
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up vote
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favorite
I need to learn, by which method, I can prove that the function $f$ is injective. I would like to ask you to explain this problem using more detailed, more understandable, clearer and simpler English words. There's only one way I can understand. For example;
Let $f: mathbb{R} tomathbb{R}$ and $f(x)=x-10$.
We have, $f(y)=y-10$, then $f(x)=f(y)$, we get $f(x)=f(y) Rightarrow x-10=y-10 Rightarrow x=y$. So, $f$ is injective.
And here is my problem:
Why can't we apply this method to this problem?
$f: mathbb{R} tomathbb{R}$ and $f(0)≠0$, such that, for all real numbers $x$ and $y$, $$f big(f(x)f(y)big) + f(x +y) = f(xy).$$
Prove that, $f$ is injective.
Thank you very much for teaching.
algebra-precalculus proof-writing contest-math functional-equations alternative-proof
asked Nov 12 at 21:23
Student
5881418
This question has an open bounty worth +150
reputation from Student ending in 2 days.
This question has not received enough attention.
1
If anyone would like to post as a comment any function that satisfies this, that'd be very helpful.
– DreamConspiracy
Nov 12 at 22:58
1
Because you have nowhere explicitly expresed $x$ (or $y$). They are only implicitly given in $f$.
– greedoid
Nov 13 at 8:34
1
@EricWofsey The OP posted this functional equation here before (see math.stackexchange.com/questions/2968359/…). There is a correct and complete answer that includes a proof of injectivity. A similar proof was given here by JavaTeachMe2018, but the answerer deleted the answer. The OP never specified what is not clear about the injectivity proofs in these answers (by Zvi in the link I gave, and by JavaTeachMe2018 here).
– Batominovski
Nov 14 at 21:57
1
There is also another old link: math.stackexchange.com/questions/2362996. Riemann gave basically the same proof. If the OP wants a different injectivity proof, that may be difficult. From my search, the similar solutions given by these three users are the only known ways, with some small twists, to solve the problem.
– Batominovski
Nov 14 at 22:03
4
I don't see any way to prove injectivity except to catalog all the functions that satisfy the requirement and then to prove each one injective. Your example of $f(x)=x-1$ does work and is injective by the same proof as you gave for $x-10$ at the start. The solution $f(x)=0$ is not injective, which shows that not all solutions are injective. One therefore cannot prove injectivity for all solutions.
– Ross Millikan
Nov 15 at 0:10
|
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need to learn, by which method, I can prove that the function $f$ is injective. I would like to ask you to explain this problem using more detailed, more understandable, clearer and simpler English words. There's only one way I can understand. For example;
Let $f: mathbb{R} tomathbb{R}$ and $f(x)=x-10$.
We have, $f(y)=y-10$, then $f(x)=f(y)$, we get $f(x)=f(y) Rightarrow x-10=y-10 Rightarrow x=y$. So, $f$ is injective.
And here is my problem:
Why can't we apply this method to this problem?
$f: mathbb{R} tomathbb{R}$ and $f(0)≠0$, such that, for all real numbers $x$ and $y$, $$f big(f(x)f(y)big) + f(x +y) = f(xy).$$
Prove that, $f$ is injective.
Thank you very much for teaching.
algebra-precalculus proof-writing contest-math functional-equations alternative-proof
asked Nov 12 at 21:23
Student
5881418
I need to learn, by which method, I can prove that the function $f$ is injective. I would like to ask you to explain this problem using more detailed, more understandable, clearer and simpler English words. There's only one way I can understand. For example;
Let $f: mathbb{R} tomathbb{R}$ and $f(x)=x-10$.
We have, $f(y)=y-10$, then $f(x)=f(y)$, we get $f(x)=f(y) Rightarrow x-10=y-10 Rightarrow x=y$. So, $f$ is injective.
And here is my problem:
Why can't we apply this method to this problem?
$f: mathbb{R} tomathbb{R}$ and $f(0)≠0$, such that, for all real numbers $x$ and $y$, $$f big(f(x)f(y)big) + f(x +y) = f(xy).$$
Prove that, $f$ is injective.
Thank you very much for teaching.
algebra-precalculus proof-writing contest-math functional-equations alternative-proof
algebra-precalculus proof-writing contest-math functional-equations alternative-proof
asked Nov 12 at 21:23
Student
5881418
asked Nov 12 at 21:23
Student
5881418
edited Nov 14 at 21:48
asked Nov 12 at 21:23
Student
5881418
asked Nov 12 at 21:23
Student
5881418
asked Nov 12 at 21:23
Student
5881418
5881418
This question has an open bounty worth +150
reputation from Student ending in 2 days.
This question has not received enough attention.
This question has an open bounty worth +150
reputation from Student ending in 2 days.
This question has not received enough attention.
1
If anyone would like to post as a comment any function that satisfies this, that'd be very helpful.
– DreamConspiracy
Nov 12 at 22:58
1
Because you have nowhere explicitly expresed $x$ (or $y$). They are only implicitly given in $f$.
– greedoid
Nov 13 at 8:34
1
@EricWofsey The OP posted this functional equation here before (see math.stackexchange.com/questions/2968359/…). There is a correct and complete answer that includes a proof of injectivity. A similar proof was given here by JavaTeachMe2018, but the answerer deleted the answer. The OP never specified what is not clear about the injectivity proofs in these answers (by Zvi in the link I gave, and by JavaTeachMe2018 here).
– Batominovski
Nov 14 at 21:57
1
There is also another old link: math.stackexchange.com/questions/2362996. Riemann gave basically the same proof. If the OP wants a different injectivity proof, that may be difficult. From my search, the similar solutions given by these three users are the only known ways, with some small twists, to solve the problem.
– Batominovski
Nov 14 at 22:03
4
I don't see any way to prove injectivity except to catalog all the functions that satisfy the requirement and then to prove each one injective. Your example of $f(x)=x-1$ does work and is injective by the same proof as you gave for $x-10$ at the start. The solution $f(x)=0$ is not injective, which shows that not all solutions are injective. One therefore cannot prove injectivity for all solutions.
– Ross Millikan
Nov 15 at 0:10
|
show 4 more comments
1
If anyone would like to post as a comment any function that satisfies this, that'd be very helpful.
– DreamConspiracy
Nov 12 at 22:58
1
Because you have nowhere explicitly expresed $x$ (or $y$). They are only implicitly given in $f$.
– greedoid
Nov 13 at 8:34
1
@EricWofsey The OP posted this functional equation here before (see math.stackexchange.com/questions/2968359/…). There is a correct and complete answer that includes a proof of injectivity. A similar proof was given here by JavaTeachMe2018, but the answerer deleted the answer. The OP never specified what is not clear about the injectivity proofs in these answers (by Zvi in the link I gave, and by JavaTeachMe2018 here).
– Batominovski
Nov 14 at 21:57
1
There is also another old link: math.stackexchange.com/questions/2362996. Riemann gave basically the same proof. If the OP wants a different injectivity proof, that may be difficult. From my search, the similar solutions given by these three users are the only known ways, with some small twists, to solve the problem.
– Batominovski
Nov 14 at 22:03
4
I don't see any way to prove injectivity except to catalog all the functions that satisfy the requirement and then to prove each one injective. Your example of $f(x)=x-1$ does work and is injective by the same proof as you gave for $x-10$ at the start. The solution $f(x)=0$ is not injective, which shows that not all solutions are injective. One therefore cannot prove injectivity for all solutions.
– Ross Millikan
Nov 15 at 0:10
1
1
If anyone would like to post as a comment any function that satisfies this, that'd be very helpful.
– DreamConspiracy
Nov 12 at 22:58
If anyone would like to post as a comment any function that satisfies this, that'd be very helpful.
– DreamConspiracy
Nov 12 at 22:58
1
1
Because you have nowhere explicitly expresed $x$ (or $y$). They are only implicitly given in $f$.
– greedoid
Nov 13 at 8:34
Because you have nowhere explicitly expresed $x$ (or $y$). They are only implicitly given in $f$.
– greedoid
Nov 13 at 8:34
1
1
@EricWofsey The OP posted this functional equation here before (see math.stackexchange.com/questions/2968359/…). There is a correct and complete answer that includes a proof of injectivity. A similar proof was given here by JavaTeachMe2018, but the answerer deleted the answer. The OP never specified what is not clear about the injectivity proofs in these answers (by Zvi in the link I gave, and by JavaTeachMe2018 here).
– Batominovski
Nov 14 at 21:57
@EricWofsey The OP posted this functional equation here before (see math.stackexchange.com/questions/2968359/…). There is a correct and complete answer that includes a proof of injectivity. A similar proof was given here by JavaTeachMe2018, but the answerer deleted the answer. The OP never specified what is not clear about the injectivity proofs in these answers (by Zvi in the link I gave, and by JavaTeachMe2018 here).
– Batominovski
Nov 14 at 21:57
1
1
There is also another old link: math.stackexchange.com/questions/2362996. Riemann gave basically the same proof. If the OP wants a different injectivity proof, that may be difficult. From my search, the similar solutions given by these three users are the only known ways, with some small twists, to solve the problem.
– Batominovski
Nov 14 at 22:03
There is also another old link: math.stackexchange.com/questions/2362996. Riemann gave basically the same proof. If the OP wants a different injectivity proof, that may be difficult. From my search, the similar solutions given by these three users are the only known ways, with some small twists, to solve the problem.
– Batominovski
Nov 14 at 22:03
4
4
I don't see any way to prove injectivity except to catalog all the functions that satisfy the requirement and then to prove each one injective. Your example of $f(x)=x-1$ does work and is injective by the same proof as you gave for $x-10$ at the start. The solution $f(x)=0$ is not injective, which shows that not all solutions are injective. One therefore cannot prove injectivity for all solutions.
– Ross Millikan
Nov 15 at 0:10
I don't see any way to prove injectivity except to catalog all the functions that satisfy the requirement and then to prove each one injective. Your example of $f(x)=x-1$ does work and is injective by the same proof as you gave for $x-10$ at the start. The solution $f(x)=0$ is not injective, which shows that not all solutions are injective. One therefore cannot prove injectivity for all solutions.
– Ross Millikan
Nov 15 at 0:10
|
show 4 more comments
2 Answers
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4
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You cannot apply the simple test straightforwardly, mainly because we do not have explicit information about $f$.
You may be able to arrive at a contradiction by assuming $f(a)=f(b)$ for some $ane b$ and playing around cleverly with the functional equation.
For example, if $f(a)=f(b)$, we find
$$f(a^2)-f(2a)=f(ab)-f(a+b)=f(b^2)-f(2b)=f(f(a)f(b)) $$
This gives you further points of $f$ that are somwehat related. This may help, but it is not straightforward to see how this could shortcut the way to injectivity (as in, as was said in the comments, being able to see that $f$ is injective much easier than determining the complete set of solutions of the functional equation).
Another approach: By letting $x=y=0$, we find
$$f(f(0)^2)=0 $$
and from $y_0:=f(0)ne 0$ can conclude that $f$ has a non-zero root, i.e., there exists $x_0(=f(0)^2)$ such that $f(x_0)=0$.
Then $$f(f(x_0)f(y)) +f(x_0+y)=f(x_0y)$$
for all $y$, i.e.,
$$f(x_0y)=f(x_0+y)+y_0ne f(x_0+y). $$
Here, we arrive at a contradiction if we exhibit $y$ with $x_0y=x_0+y$. This is possibly (namely, $y=frac{x_0}{x_0-1}$) as long as $x_0ne 1$. So we conclude
$$ x_0=1$$
before getting any closer to the goal of showing that $f$ is injective (i.e., we seem to only exclude that $|a-b|=1$).
We can use this (letting $x=y=1$) to show
$$ f(2)=-f(0)$$
and other tricky things (in fact, ultimately that $f(x)=pm(x-1)$), but I still do not see any way to show injectivity much simpler than solving the functional equation completely.
answered Nov 15 at 7:36
Hagen von Eitzen
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1
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We can prove the solution is injective and from this fact, solve the equation completely, getting the solution $f(x) = 1-x,forall xinmathbf{R}$ or $f(x)= x-1, forall x$. I'll start from the scratch.
Let $gamma := f(0)$. Then, from the FE, we have $f(gamma^2) =0$. Our first claim is that
$$f(y) = 0 Rightarrow y = 1.$$
Assume to the contrary that $yneq 1$, but $f(y)=0$. Then by setting $x=frac{y}{y-1}$, we should have $$0 = f(f(x)f(y)) = f(0) =gamma$$, contradicting $gamma neq 0$.
From the above argument we get that $gamma^2 = 1$ and for all $y neq 1$, it holds that $$f(y)f(frac{y}{y-1})=1 cdots(*)$$
So far we have $gamma =pm1$, and we will see if $gamma =1$, then $f(x) = 1-x$ and otherwise, $f(x) = x-1$.
Assume that $gamma = -1$. From the functional equation, we have
$$f(x+1) = f(x) + 1,quadforall xin mathbf{R},$$
$$f(x+n) = f(x) + n,quadforall xin mathbf{R}, ;nin mathbf{Z},$$
$$f(n) = n-1, quad forall nin mathbf{Z}.$$
Using this fact, after a change of variable $y-1 mapsto y$, we get
$$left(f(y)+1right)left(f(frac{1}{y})+1right)=1,quad forall yneq 0 cdots(**)$$
Notice this implies $f^{-1}(-1) = {0}$. Our next claim is that
$$f(alpha) =f(beta) Rightarrow f(qalpha) = f(qbeta), f(alpha+q) = f(beta+q),quadforall qin mathbf{Q}.$$
We may assume $alpha neq 0, beta neq 0$. To prove this, note that $f(alpha) =f(beta)$ implies $f(nalpha)=f(nbeta)$ for all $ninmathbf{Z}$ by the FE. Then, by $(**)$, for non-zero $n$, we have $f(frac{1}{nalpha})=f(frac{1}{nbeta}).$ Hence, it holds $f(frac{m}{nalpha})=f(frac{m}{nbeta})$ and again by $(**)$, $$f(frac{n}{m}alpha)=f(frac{n}{m}alpha), quadforall ninmathbf{Z}, min mathbf{Z} setminus {0}.$$ This prove the fisrt half. For the second half, note that by the FE,
$$f(alpha x) -f(beta x) = f(alpha +x )-f(beta +x) quadforall xin mathbf{R}.$$
So, in particular, $f(alpha +q ) -f(beta +q) =0,forall qin mathbf{Q}$.
Our (almost) final claim is that $f$ is injective. Assume to the contrary that $f(alpha) = f(beta) $ for $alphaneqbeta$. We proceed by finding the solution $(x,y,q)in mathbf{R}timesmathbf{R}timesmathbf{Q}$ satisfying:
$$xy = alpha +q,; x+y = beta +1 +q.$$ Once this is done, then by the FE,
$$f(f(x)f(y)) = f(xy) -f(x+y) = f(alpha +q)-f(beta +1 +q) = f(alpha +q)-f(beta +q)-1 = -1, $$ getting $f(x)f(y) = 0$. Without loss of generality, $f(x) = 0$ and hence $x=1$. So we end up with $y -q=alpha = beta ,$ contradiction!
We notice that the real solution $(x,y)$ exists if and only if its quadratic discriminant
$$(x-y)^2 = (x+y)^2 -4xy = (beta+q+1)^2 -4(alpha+q)= q^2 +2(beta-1)q +(beta+1)^2 -4alpha geq 0.$$ Of course we can choose $q_0inmathbf{Q}$ such that
$$q_0^2 +2(beta-1)q_0 +(beta+1)^2 -4alpha > 0.$$
Hence, the injectivity of $f$ is established. Finally, by letting $y=0$ in the FE, it holds
$$f(-f(x)) =-f(x) -1,$$ which is saying that $f(t) = t-1$ if $(-t)$ belongs to the range of $f$. Let $f(u) = u'$. By the above FE, $-(1+u')$ also belongs to the range of $f$. Thus, $f(1+u') = u'$. By injectivity, we get $u= 1+u'$, hence $f(u) = u-1$ for all $uin mathbf{R}$, as desired.
Note: exactly the same argument shows that $f(u) = 1-u$ in the case $gamma =1.$
answered 2 days ago
Song
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
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up vote
4
down vote
You cannot apply the simple test straightforwardly, mainly because we do not have explicit information about $f$.
You may be able to arrive at a contradiction by assuming $f(a)=f(b)$ for some $ane b$ and playing around cleverly with the functional equation.
For example, if $f(a)=f(b)$, we find
$$f(a^2)-f(2a)=f(ab)-f(a+b)=f(b^2)-f(2b)=f(f(a)f(b)) $$
This gives you further points of $f$ that are somwehat related. This may help, but it is not straightforward to see how this could shortcut the way to injectivity (as in, as was said in the comments, being able to see that $f$ is injective much easier than determining the complete set of solutions of the functional equation).
Another approach: By letting $x=y=0$, we find
$$f(f(0)^2)=0 $$
and from $y_0:=f(0)ne 0$ can conclude that $f$ has a non-zero root, i.e., there exists $x_0(=f(0)^2)$ such that $f(x_0)=0$.
Then $$f(f(x_0)f(y)) +f(x_0+y)=f(x_0y)$$
for all $y$, i.e.,
$$f(x_0y)=f(x_0+y)+y_0ne f(x_0+y). $$
Here, we arrive at a contradiction if we exhibit $y$ with $x_0y=x_0+y$. This is possibly (namely, $y=frac{x_0}{x_0-1}$) as long as $x_0ne 1$. So we conclude
$$ x_0=1$$
before getting any closer to the goal of showing that $f$ is injective (i.e., we seem to only exclude that $|a-b|=1$).
We can use this (letting $x=y=1$) to show
$$ f(2)=-f(0)$$
and other tricky things (in fact, ultimately that $f(x)=pm(x-1)$), but I still do not see any way to show injectivity much simpler than solving the functional equation completely.
answered Nov 15 at 7:36
Hagen von Eitzen
273k21266493
add a comment |
up vote
4
down vote
You cannot apply the simple test straightforwardly, mainly because we do not have explicit information about $f$.
You may be able to arrive at a contradiction by assuming $f(a)=f(b)$ for some $ane b$ and playing around cleverly with the functional equation.
For example, if $f(a)=f(b)$, we find
$$f(a^2)-f(2a)=f(ab)-f(a+b)=f(b^2)-f(2b)=f(f(a)f(b)) $$
This gives you further points of $f$ that are somwehat related. This may help, but it is not straightforward to see how this could shortcut the way to injectivity (as in, as was said in the comments, being able to see that $f$ is injective much easier than determining the complete set of solutions of the functional equation).
Another approach: By letting $x=y=0$, we find
$$f(f(0)^2)=0 $$
and from $y_0:=f(0)ne 0$ can conclude that $f$ has a non-zero root, i.e., there exists $x_0(=f(0)^2)$ such that $f(x_0)=0$.
Then $$f(f(x_0)f(y)) +f(x_0+y)=f(x_0y)$$
for all $y$, i.e.,
$$f(x_0y)=f(x_0+y)+y_0ne f(x_0+y). $$
Here, we arrive at a contradiction if we exhibit $y$ with $x_0y=x_0+y$. This is possibly (namely, $y=frac{x_0}{x_0-1}$) as long as $x_0ne 1$. So we conclude
$$ x_0=1$$
before getting any closer to the goal of showing that $f$ is injective (i.e., we seem to only exclude that $|a-b|=1$).
We can use this (letting $x=y=1$) to show
$$ f(2)=-f(0)$$
and other tricky things (in fact, ultimately that $f(x)=pm(x-1)$), but I still do not see any way to show injectivity much simpler than solving the functional equation completely.
answered Nov 15 at 7:36
Hagen von Eitzen
273k21266493
add a comment |
up vote
4
down vote
up vote
4
down vote
You cannot apply the simple test straightforwardly, mainly because we do not have explicit information about $f$.
You may be able to arrive at a contradiction by assuming $f(a)=f(b)$ for some $ane b$ and playing around cleverly with the functional equation.
For example, if $f(a)=f(b)$, we find
$$f(a^2)-f(2a)=f(ab)-f(a+b)=f(b^2)-f(2b)=f(f(a)f(b)) $$
This gives you further points of $f$ that are somwehat related. This may help, but it is not straightforward to see how this could shortcut the way to injectivity (as in, as was said in the comments, being able to see that $f$ is injective much easier than determining the complete set of solutions of the functional equation).
Another approach: By letting $x=y=0$, we find
$$f(f(0)^2)=0 $$
and from $y_0:=f(0)ne 0$ can conclude that $f$ has a non-zero root, i.e., there exists $x_0(=f(0)^2)$ such that $f(x_0)=0$.
Then $$f(f(x_0)f(y)) +f(x_0+y)=f(x_0y)$$
for all $y$, i.e.,
$$f(x_0y)=f(x_0+y)+y_0ne f(x_0+y). $$
Here, we arrive at a contradiction if we exhibit $y$ with $x_0y=x_0+y$. This is possibly (namely, $y=frac{x_0}{x_0-1}$) as long as $x_0ne 1$. So we conclude
$$ x_0=1$$
before getting any closer to the goal of showing that $f$ is injective (i.e., we seem to only exclude that $|a-b|=1$).
We can use this (letting $x=y=1$) to show
$$ f(2)=-f(0)$$
and other tricky things (in fact, ultimately that $f(x)=pm(x-1)$), but I still do not see any way to show injectivity much simpler than solving the functional equation completely.
answered Nov 15 at 7:36
Hagen von Eitzen
273k21266493
You cannot apply the simple test straightforwardly, mainly because we do not have explicit information about $f$.
You may be able to arrive at a contradiction by assuming $f(a)=f(b)$ for some $ane b$ and playing around cleverly with the functional equation.
For example, if $f(a)=f(b)$, we find
$$f(a^2)-f(2a)=f(ab)-f(a+b)=f(b^2)-f(2b)=f(f(a)f(b)) $$
This gives you further points of $f$ that are somwehat related. This may help, but it is not straightforward to see how this could shortcut the way to injectivity (as in, as was said in the comments, being able to see that $f$ is injective much easier than determining the complete set of solutions of the functional equation).
Another approach: By letting $x=y=0$, we find
$$f(f(0)^2)=0 $$
and from $y_0:=f(0)ne 0$ can conclude that $f$ has a non-zero root, i.e., there exists $x_0(=f(0)^2)$ such that $f(x_0)=0$.
Then $$f(f(x_0)f(y)) +f(x_0+y)=f(x_0y)$$
for all $y$, i.e.,
$$f(x_0y)=f(x_0+y)+y_0ne f(x_0+y). $$
Here, we arrive at a contradiction if we exhibit $y$ with $x_0y=x_0+y$. This is possibly (namely, $y=frac{x_0}{x_0-1}$) as long as $x_0ne 1$. So we conclude
$$ x_0=1$$
before getting any closer to the goal of showing that $f$ is injective (i.e., we seem to only exclude that $|a-b|=1$).
We can use this (letting $x=y=1$) to show
$$ f(2)=-f(0)$$
and other tricky things (in fact, ultimately that $f(x)=pm(x-1)$), but I still do not see any way to show injectivity much simpler than solving the functional equation completely.
answered Nov 15 at 7:36
Hagen von Eitzen
273k21266493
answered Nov 15 at 7:36
Hagen von Eitzen
273k21266493
answered Nov 15 at 7:36
Hagen von Eitzen
273k21266493
answered Nov 15 at 7:36
Hagen von Eitzen
273k21266493
273k21266493
add a comment |
add a comment |
up vote
1
down vote
We can prove the solution is injective and from this fact, solve the equation completely, getting the solution $f(x) = 1-x,forall xinmathbf{R}$ or $f(x)= x-1, forall x$. I'll start from the scratch.
Let $gamma := f(0)$. Then, from the FE, we have $f(gamma^2) =0$. Our first claim is that
$$f(y) = 0 Rightarrow y = 1.$$
Assume to the contrary that $yneq 1$, but $f(y)=0$. Then by setting $x=frac{y}{y-1}$, we should have $$0 = f(f(x)f(y)) = f(0) =gamma$$, contradicting $gamma neq 0$.
From the above argument we get that $gamma^2 = 1$ and for all $y neq 1$, it holds that $$f(y)f(frac{y}{y-1})=1 cdots(*)$$
So far we have $gamma =pm1$, and we will see if $gamma =1$, then $f(x) = 1-x$ and otherwise, $f(x) = x-1$.
Assume that $gamma = -1$. From the functional equation, we have
$$f(x+1) = f(x) + 1,quadforall xin mathbf{R},$$
$$f(x+n) = f(x) + n,quadforall xin mathbf{R}, ;nin mathbf{Z},$$
$$f(n) = n-1, quad forall nin mathbf{Z}.$$
Using this fact, after a change of variable $y-1 mapsto y$, we get
$$left(f(y)+1right)left(f(frac{1}{y})+1right)=1,quad forall yneq 0 cdots(**)$$
Notice this implies $f^{-1}(-1) = {0}$. Our next claim is that
$$f(alpha) =f(beta) Rightarrow f(qalpha) = f(qbeta), f(alpha+q) = f(beta+q),quadforall qin mathbf{Q}.$$
We may assume $alpha neq 0, beta neq 0$. To prove this, note that $f(alpha) =f(beta)$ implies $f(nalpha)=f(nbeta)$ for all $ninmathbf{Z}$ by the FE. Then, by $(**)$, for non-zero $n$, we have $f(frac{1}{nalpha})=f(frac{1}{nbeta}).$ Hence, it holds $f(frac{m}{nalpha})=f(frac{m}{nbeta})$ and again by $(**)$, $$f(frac{n}{m}alpha)=f(frac{n}{m}alpha), quadforall ninmathbf{Z}, min mathbf{Z} setminus {0}.$$ This prove the fisrt half. For the second half, note that by the FE,
$$f(alpha x) -f(beta x) = f(alpha +x )-f(beta +x) quadforall xin mathbf{R}.$$
So, in particular, $f(alpha +q ) -f(beta +q) =0,forall qin mathbf{Q}$.
Our (almost) final claim is that $f$ is injective. Assume to the contrary that $f(alpha) = f(beta) $ for $alphaneqbeta$. We proceed by finding the solution $(x,y,q)in mathbf{R}timesmathbf{R}timesmathbf{Q}$ satisfying:
$$xy = alpha +q,; x+y = beta +1 +q.$$ Once this is done, then by the FE,
$$f(f(x)f(y)) = f(xy) -f(x+y) = f(alpha +q)-f(beta +1 +q) = f(alpha +q)-f(beta +q)-1 = -1, $$ getting $f(x)f(y) = 0$. Without loss of generality, $f(x) = 0$ and hence $x=1$. So we end up with $y -q=alpha = beta ,$ contradiction!
We notice that the real solution $(x,y)$ exists if and only if its quadratic discriminant
$$(x-y)^2 = (x+y)^2 -4xy = (beta+q+1)^2 -4(alpha+q)= q^2 +2(beta-1)q +(beta+1)^2 -4alpha geq 0.$$ Of course we can choose $q_0inmathbf{Q}$ such that
$$q_0^2 +2(beta-1)q_0 +(beta+1)^2 -4alpha > 0.$$
Hence, the injectivity of $f$ is established. Finally, by letting $y=0$ in the FE, it holds
$$f(-f(x)) =-f(x) -1,$$ which is saying that $f(t) = t-1$ if $(-t)$ belongs to the range of $f$. Let $f(u) = u'$. By the above FE, $-(1+u')$ also belongs to the range of $f$. Thus, $f(1+u') = u'$. By injectivity, we get $u= 1+u'$, hence $f(u) = u-1$ for all $uin mathbf{R}$, as desired.
Note: exactly the same argument shows that $f(u) = 1-u$ in the case $gamma =1.$
answered 2 days ago
Song
5239
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1
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We can prove the solution is injective and from this fact, solve the equation completely, getting the solution $f(x) = 1-x,forall xinmathbf{R}$ or $f(x)= x-1, forall x$. I'll start from the scratch.
Let $gamma := f(0)$. Then, from the FE, we have $f(gamma^2) =0$. Our first claim is that
$$f(y) = 0 Rightarrow y = 1.$$
Assume to the contrary that $yneq 1$, but $f(y)=0$. Then by setting $x=frac{y}{y-1}$, we should have $$0 = f(f(x)f(y)) = f(0) =gamma$$, contradicting $gamma neq 0$.
From the above argument we get that $gamma^2 = 1$ and for all $y neq 1$, it holds that $$f(y)f(frac{y}{y-1})=1 cdots(*)$$
So far we have $gamma =pm1$, and we will see if $gamma =1$, then $f(x) = 1-x$ and otherwise, $f(x) = x-1$.
Assume that $gamma = -1$. From the functional equation, we have
$$f(x+1) = f(x) + 1,quadforall xin mathbf{R},$$
$$f(x+n) = f(x) + n,quadforall xin mathbf{R}, ;nin mathbf{Z},$$
$$f(n) = n-1, quad forall nin mathbf{Z}.$$
Using this fact, after a change of variable $y-1 mapsto y$, we get
$$left(f(y)+1right)left(f(frac{1}{y})+1right)=1,quad forall yneq 0 cdots(**)$$
Notice this implies $f^{-1}(-1) = {0}$. Our next claim is that
$$f(alpha) =f(beta) Rightarrow f(qalpha) = f(qbeta), f(alpha+q) = f(beta+q),quadforall qin mathbf{Q}.$$
We may assume $alpha neq 0, beta neq 0$. To prove this, note that $f(alpha) =f(beta)$ implies $f(nalpha)=f(nbeta)$ for all $ninmathbf{Z}$ by the FE. Then, by $(**)$, for non-zero $n$, we have $f(frac{1}{nalpha})=f(frac{1}{nbeta}).$ Hence, it holds $f(frac{m}{nalpha})=f(frac{m}{nbeta})$ and again by $(**)$, $$f(frac{n}{m}alpha)=f(frac{n}{m}alpha), quadforall ninmathbf{Z}, min mathbf{Z} setminus {0}.$$ This prove the fisrt half. For the second half, note that by the FE,
$$f(alpha x) -f(beta x) = f(alpha +x )-f(beta +x) quadforall xin mathbf{R}.$$
So, in particular, $f(alpha +q ) -f(beta +q) =0,forall qin mathbf{Q}$.
Our (almost) final claim is that $f$ is injective. Assume to the contrary that $f(alpha) = f(beta) $ for $alphaneqbeta$. We proceed by finding the solution $(x,y,q)in mathbf{R}timesmathbf{R}timesmathbf{Q}$ satisfying:
$$xy = alpha +q,; x+y = beta +1 +q.$$ Once this is done, then by the FE,
$$f(f(x)f(y)) = f(xy) -f(x+y) = f(alpha +q)-f(beta +1 +q) = f(alpha +q)-f(beta +q)-1 = -1, $$ getting $f(x)f(y) = 0$. Without loss of generality, $f(x) = 0$ and hence $x=1$. So we end up with $y -q=alpha = beta ,$ contradiction!
We notice that the real solution $(x,y)$ exists if and only if its quadratic discriminant
$$(x-y)^2 = (x+y)^2 -4xy = (beta+q+1)^2 -4(alpha+q)= q^2 +2(beta-1)q +(beta+1)^2 -4alpha geq 0.$$ Of course we can choose $q_0inmathbf{Q}$ such that
$$q_0^2 +2(beta-1)q_0 +(beta+1)^2 -4alpha > 0.$$
Hence, the injectivity of $f$ is established. Finally, by letting $y=0$ in the FE, it holds
$$f(-f(x)) =-f(x) -1,$$ which is saying that $f(t) = t-1$ if $(-t)$ belongs to the range of $f$. Let $f(u) = u'$. By the above FE, $-(1+u')$ also belongs to the range of $f$. Thus, $f(1+u') = u'$. By injectivity, we get $u= 1+u'$, hence $f(u) = u-1$ for all $uin mathbf{R}$, as desired.
Note: exactly the same argument shows that $f(u) = 1-u$ in the case $gamma =1.$
answered 2 days ago
Song
5239
add a comment |
up vote
1
down vote
up vote
1
down vote
We can prove the solution is injective and from this fact, solve the equation completely, getting the solution $f(x) = 1-x,forall xinmathbf{R}$ or $f(x)= x-1, forall x$. I'll start from the scratch.
Let $gamma := f(0)$. Then, from the FE, we have $f(gamma^2) =0$. Our first claim is that
$$f(y) = 0 Rightarrow y = 1.$$
Assume to the contrary that $yneq 1$, but $f(y)=0$. Then by setting $x=frac{y}{y-1}$, we should have $$0 = f(f(x)f(y)) = f(0) =gamma$$, contradicting $gamma neq 0$.
From the above argument we get that $gamma^2 = 1$ and for all $y neq 1$, it holds that $$f(y)f(frac{y}{y-1})=1 cdots(*)$$
So far we have $gamma =pm1$, and we will see if $gamma =1$, then $f(x) = 1-x$ and otherwise, $f(x) = x-1$.
Assume that $gamma = -1$. From the functional equation, we have
$$f(x+1) = f(x) + 1,quadforall xin mathbf{R},$$
$$f(x+n) = f(x) + n,quadforall xin mathbf{R}, ;nin mathbf{Z},$$
$$f(n) = n-1, quad forall nin mathbf{Z}.$$
Using this fact, after a change of variable $y-1 mapsto y$, we get
$$left(f(y)+1right)left(f(frac{1}{y})+1right)=1,quad forall yneq 0 cdots(**)$$
Notice this implies $f^{-1}(-1) = {0}$. Our next claim is that
$$f(alpha) =f(beta) Rightarrow f(qalpha) = f(qbeta), f(alpha+q) = f(beta+q),quadforall qin mathbf{Q}.$$
We may assume $alpha neq 0, beta neq 0$. To prove this, note that $f(alpha) =f(beta)$ implies $f(nalpha)=f(nbeta)$ for all $ninmathbf{Z}$ by the FE. Then, by $(**)$, for non-zero $n$, we have $f(frac{1}{nalpha})=f(frac{1}{nbeta}).$ Hence, it holds $f(frac{m}{nalpha})=f(frac{m}{nbeta})$ and again by $(**)$, $$f(frac{n}{m}alpha)=f(frac{n}{m}alpha), quadforall ninmathbf{Z}, min mathbf{Z} setminus {0}.$$ This prove the fisrt half. For the second half, note that by the FE,
$$f(alpha x) -f(beta x) = f(alpha +x )-f(beta +x) quadforall xin mathbf{R}.$$
So, in particular, $f(alpha +q ) -f(beta +q) =0,forall qin mathbf{Q}$.
Our (almost) final claim is that $f$ is injective. Assume to the contrary that $f(alpha) = f(beta) $ for $alphaneqbeta$. We proceed by finding the solution $(x,y,q)in mathbf{R}timesmathbf{R}timesmathbf{Q}$ satisfying:
$$xy = alpha +q,; x+y = beta +1 +q.$$ Once this is done, then by the FE,
$$f(f(x)f(y)) = f(xy) -f(x+y) = f(alpha +q)-f(beta +1 +q) = f(alpha +q)-f(beta +q)-1 = -1, $$ getting $f(x)f(y) = 0$. Without loss of generality, $f(x) = 0$ and hence $x=1$. So we end up with $y -q=alpha = beta ,$ contradiction!
We notice that the real solution $(x,y)$ exists if and only if its quadratic discriminant
$$(x-y)^2 = (x+y)^2 -4xy = (beta+q+1)^2 -4(alpha+q)= q^2 +2(beta-1)q +(beta+1)^2 -4alpha geq 0.$$ Of course we can choose $q_0inmathbf{Q}$ such that
$$q_0^2 +2(beta-1)q_0 +(beta+1)^2 -4alpha > 0.$$
Hence, the injectivity of $f$ is established. Finally, by letting $y=0$ in the FE, it holds
$$f(-f(x)) =-f(x) -1,$$ which is saying that $f(t) = t-1$ if $(-t)$ belongs to the range of $f$. Let $f(u) = u'$. By the above FE, $-(1+u')$ also belongs to the range of $f$. Thus, $f(1+u') = u'$. By injectivity, we get $u= 1+u'$, hence $f(u) = u-1$ for all $uin mathbf{R}$, as desired.
Note: exactly the same argument shows that $f(u) = 1-u$ in the case $gamma =1.$
answered 2 days ago
Song
5239
We can prove the solution is injective and from this fact, solve the equation completely, getting the solution $f(x) = 1-x,forall xinmathbf{R}$ or $f(x)= x-1, forall x$. I'll start from the scratch.
Let $gamma := f(0)$. Then, from the FE, we have $f(gamma^2) =0$. Our first claim is that
$$f(y) = 0 Rightarrow y = 1.$$
Assume to the contrary that $yneq 1$, but $f(y)=0$. Then by setting $x=frac{y}{y-1}$, we should have $$0 = f(f(x)f(y)) = f(0) =gamma$$, contradicting $gamma neq 0$.
From the above argument we get that $gamma^2 = 1$ and for all $y neq 1$, it holds that $$f(y)f(frac{y}{y-1})=1 cdots(*)$$
So far we have $gamma =pm1$, and we will see if $gamma =1$, then $f(x) = 1-x$ and otherwise, $f(x) = x-1$.
Assume that $gamma = -1$. From the functional equation, we have
$$f(x+1) = f(x) + 1,quadforall xin mathbf{R},$$
$$f(x+n) = f(x) + n,quadforall xin mathbf{R}, ;nin mathbf{Z},$$
$$f(n) = n-1, quad forall nin mathbf{Z}.$$
Using this fact, after a change of variable $y-1 mapsto y$, we get
$$left(f(y)+1right)left(f(frac{1}{y})+1right)=1,quad forall yneq 0 cdots(**)$$
Notice this implies $f^{-1}(-1) = {0}$. Our next claim is that
$$f(alpha) =f(beta) Rightarrow f(qalpha) = f(qbeta), f(alpha+q) = f(beta+q),quadforall qin mathbf{Q}.$$
We may assume $alpha neq 0, beta neq 0$. To prove this, note that $f(alpha) =f(beta)$ implies $f(nalpha)=f(nbeta)$ for all $ninmathbf{Z}$ by the FE. Then, by $(**)$, for non-zero $n$, we have $f(frac{1}{nalpha})=f(frac{1}{nbeta}).$ Hence, it holds $f(frac{m}{nalpha})=f(frac{m}{nbeta})$ and again by $(**)$, $$f(frac{n}{m}alpha)=f(frac{n}{m}alpha), quadforall ninmathbf{Z}, min mathbf{Z} setminus {0}.$$ This prove the fisrt half. For the second half, note that by the FE,
$$f(alpha x) -f(beta x) = f(alpha +x )-f(beta +x) quadforall xin mathbf{R}.$$
So, in particular, $f(alpha +q ) -f(beta +q) =0,forall qin mathbf{Q}$.
Our (almost) final claim is that $f$ is injective. Assume to the contrary that $f(alpha) = f(beta) $ for $alphaneqbeta$. We proceed by finding the solution $(x,y,q)in mathbf{R}timesmathbf{R}timesmathbf{Q}$ satisfying:
$$xy = alpha +q,; x+y = beta +1 +q.$$ Once this is done, then by the FE,
$$f(f(x)f(y)) = f(xy) -f(x+y) = f(alpha +q)-f(beta +1 +q) = f(alpha +q)-f(beta +q)-1 = -1, $$ getting $f(x)f(y) = 0$. Without loss of generality, $f(x) = 0$ and hence $x=1$. So we end up with $y -q=alpha = beta ,$ contradiction!
We notice that the real solution $(x,y)$ exists if and only if its quadratic discriminant
$$(x-y)^2 = (x+y)^2 -4xy = (beta+q+1)^2 -4(alpha+q)= q^2 +2(beta-1)q +(beta+1)^2 -4alpha geq 0.$$ Of course we can choose $q_0inmathbf{Q}$ such that
$$q_0^2 +2(beta-1)q_0 +(beta+1)^2 -4alpha > 0.$$
Hence, the injectivity of $f$ is established. Finally, by letting $y=0$ in the FE, it holds
$$f(-f(x)) =-f(x) -1,$$ which is saying that $f(t) = t-1$ if $(-t)$ belongs to the range of $f$. Let $f(u) = u'$. By the above FE, $-(1+u')$ also belongs to the range of $f$. Thus, $f(1+u') = u'$. By injectivity, we get $u= 1+u'$, hence $f(u) = u-1$ for all $uin mathbf{R}$, as desired.
Note: exactly the same argument shows that $f(u) = 1-u$ in the case $gamma =1.$
answered 2 days ago
Song
5239
answered 2 days ago
Song
5239
answered 2 days ago
Song
5239
answered 2 days ago
Song
5239
5239
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If anyone would like to post as a comment any function that satisfies this, that'd be very helpful.
– DreamConspiracy
Nov 12 at 22:58
1
Because you have nowhere explicitly expresed $x$ (or $y$). They are only implicitly given in $f$.
– greedoid
Nov 13 at 8:34
1
@EricWofsey The OP posted this functional equation here before (see math.stackexchange.com/questions/2968359/…). There is a correct and complete answer that includes a proof of injectivity. A similar proof was given here by JavaTeachMe2018, but the answerer deleted the answer. The OP never specified what is not clear about the injectivity proofs in these answers (by Zvi in the link I gave, and by JavaTeachMe2018 here).
– Batominovski
Nov 14 at 21:57
1
There is also another old link: math.stackexchange.com/questions/2362996. Riemann gave basically the same proof. If the OP wants a different injectivity proof, that may be difficult. From my search, the similar solutions given by these three users are the only known ways, with some small twists, to solve the problem.
– Batominovski
Nov 14 at 22:03
4
I don't see any way to prove injectivity except to catalog all the functions that satisfy the requirement and then to prove each one injective. Your example of $f(x)=x-1$ does work and is injective by the same proof as you gave for $x-10$ at the start. The solution $f(x)=0$ is not injective, which shows that not all solutions are injective. One therefore cannot prove injectivity for all solutions.
– Ross Millikan
Nov 15 at 0:10