Find F: R->R satisfying F(x+y)=F(x)+F(y) and F(xy)=F(x)F(y)

[iN10SE]

Homework Statement

Let F:R->R be a function such that, for all x,y belonging to R, we have F(x+y)=F(x)+F(y) and F(xy)=F(x)F(y). Prove that F is one of the following two functions:
i> f(x)=0
ii> f(x)=x
(Hint : At some point in your proof, the fact that every positive real number is the sqaure of a real number will be valuable)

Homework Equations

The Attempt at a Solution

Let f(x)=x^n. Then f(x+y)=f(x)+f(y) is not satisfied. So f(x) can't be a polynomial. Similarly f(x) can't contain e^x, logx, any trigonometric function.
Now let f(x)=ax, where a=!0,1.
then f(x+y)=f(x)+f(y)=2ax is satisfied but not f(xy)=f(x).f(y) as f(x^2)=a.x^2 but f(x).f(x)=a^2.x^2. NOw a = a^2 iff a= 0 or 1. hence the proof.

My confusion is (as I am new to prrof oriented maths.), I have actually never used the hint. So, did I skip any argument?

Thanks in advance.

 

[Elucidus]

The proffered proof shows some examples of what f(x) cannot be and gives two examples of what f(x) could be, but it does not prove that f(x) cannot be something else other than $(x \mapsto 0) \text{ or } (x \mapsto x)$.

--Elucidus

 

[iN10SE]

thanks euclidian
ok...second try :

F(x+y)=F(x)+F(y)
put y=0
F(x)=F(x)+F(0)
so either F(x)=0 for all x
or F(0)=0 => F(x) has the form F(x)=xg(x) where g(x)=!0 for all x

now I have to show g(x)=1 for all x.

From the relations of F(x) and g(x), it can be derived
g(x+y)=(xg(x)+yg(y))/(x+y) and g(xy)=g(x)g(y)
put y=0 in second relation
g(x)=g(x).g(0)

g(x)=!0 for all zero, g(0)=1.

now I am stuck... but i GUESS i AM PRETTY NEAR?

 

[Elucidus]

The following piece of putative logic is troubling:

$$\text{If } F(0) = 0 \text{ then either } F(x) = 0 \text{ for all } x \text{ or there exists } g(x) \text{ such that } g(x) \neq 0 \text{ for all } x \text{ and } F(x) = x \cdot g(x).$$

Take F(x) = sin(x). F(0) = 0 but F(x) is not always zero, nor does there exist g(x) so that g(x) is always nonzero and F(x) = xg(x).

It is possible to show that F(x) must be odd and F(0) = 0, but I don't think that implies that an x can be factored out of F(x).

--Elucidus

 

[iN10SE]

Thanks. but any suggestion on how to proceed then?

 

[Caesar_Rahil]

What is we substitute x=y;
then F(2x)=2F(x)
also F(x^2)=(F(x))^2
i think something to do with square may come from here

 

[Elucidus]

Not true, i can say F(x)=x . (sinx / x )
your g(x)= sinx / x

Yes but sin(x)/x isn't always nonzero.

--Elucidus

 

[Caesar_Rahil]

Yes but sin(x)/x isn't always nonzero.

--Elucidus

Well when is it zero??

 

[Caesar_Rahil]

F(0)=0
F(1)=1 or F(1)=0
also F(2x)=2F(x)
Put x=1;
F(2)=2 or F(2)=0
you can try going on like that
Also if F(x) is differentiable (it is not given but still)
F'(x)=lim(h->0) F(x+h)-F(x) / h
= lim (h->0) F(h) / h
= constant
since F'(x)=constant
=> F(x) is linear => F(x) = kx +c
Put x=0,1;
you get F(x) = x or F(x) = 0

Proof still not valid though

 

[Elucidus]

Well when is it zero??

At any nonzero multiple of pi.

--Elucidus

 

[Elucidus]

For any integer k, F(kx) = kF(x) and specifically F(0) = 0 (as mentioned) and F(-x) = -F(x) (which makes it odd).

If one let's F(x) = x + G(x) then

xy + G(xy) = F(xy) = F(x)F(y) = [x + G(x)][y + G(y)] = xy + xG(y) + yG(x) + G(x)G(y)

indicates that G(xy) = xG(y) + yG(x) + G(x)G(y) and when y = 1

G(x) = xG(1) + G(x) + G(x)G(1) from which we get

0 = xG(1) + G(x)G(1) = G(1)[x + G(x)] = G(1)F(x).

So either G(1) = 0 or F(x) = 0 for all x.

If F(x) is not identically 0 then G(1) = 0 and since F(0) = 0 then G(0) = 0.

If it could be shown that G(x) is either 0 or -F(x) then we'd be in business.

Additioinally for $$x \geq 0, F(x) = F(\sqrt{x} \cdot \sqrt{x}) = F(\sqrt{x})^2 \geq 0.$$

So F(x) is nonnegative to the right of 0 and by radial symmetry nonpositive to the left of zero.

I'm not sure how to leverage this to a solution but I feel it's nearby but eluding me. I'm not even sure if any of this is relevant.

--Elucidus

 

[Elucidus]

F(0)=0
F(1)=1 or F(1)=0
also F(2x)=2F(x)
Put x=1;
F(2)=2 or F(2)=0
you can try going on like that
Also if F(x) is differentiable (it is not given but still)
F'(x)=lim(h->0) F(x+h)-F(x) / h
= lim (h->0) F(h) / h
= constant
since F'(x)=constant
=> F(x) is linear => F(x) = kx +c
Put x=0,1;
you get F(x) = x or F(x) = 0

Proof still not valid though

I explored this approach as well, but F is not known to be differentiable (let alone continuous).

Secondly

$$\lim_{h \rightarrow 0} \frac{F(h)}{h} \text{ is not necessarily a constant.}$$

--Elucidus

 

[Office_Shredder]

Try to start off by showing for any rational number of the form p/q (p,q integers), F(p/q*x) = p/qF(x). Then if F(1)=c, we have that F(p/q) = cp/q for any rational number. Verify what F(1) can be.

At that point you either assume F is continuous, or you make a very savvy argument (which eludes me to this point) to include the irrational numbers

 

[Dick]

For any integer k, F(kx) = kF(x) and specifically F(0) = 0 (as mentioned) and F(-x) = -F(x) (which makes it odd).

If one let's F(x) = x + G(x) then

xy + G(xy) = F(xy) = F(x)F(y) = [x + G(x)][y + G(y)] = xy + xG(y) + yG(x) + G(x)G(y)

indicates that G(xy) = xG(y) + yG(x) + G(x)G(y) and when y = 1

G(x) = xG(1) + G(x) + G(x)G(1) from which we get

0 = xG(1) + G(x)G(1) = G(1)[x + G(x)] = G(1)F(x).

So either G(1) = 0 or F(x) = 0 for all x.

If F(x) is not identically 0 then G(1) = 0 and since F(0) = 0 then G(0) = 0.

If it could be shown that G(x) is either 0 or -F(x) then we'd be in business.

Additioinally for $$x \geq 0, F(x) = F(\sqrt{x} \cdot \sqrt{x}) = F(\sqrt{x})^2 \geq 0.$$

So F(x) is nonnegative to the right of 0 and by radial symmetry nonpositive to the left of zero.

I'm not sure how to leverage this to a solution but I feel it's nearby but eluding me. I'm not even sure if any of this is relevant.

--Elucidus

It's eluding me too. I think I actually looked at this before. By using my usual recourse when a problem is way too tough for me. I hunted through the sci.math forums. I THINK i found an entry by one of the brainiacs there that there ARE such functions which are not 0 or the identity. But they have to be constructed by axiom of choice type arguments and you really can't find a concise example of one accessible to a finite brain. That's just a vague recollection, so I could be wrong. But I think the OP might be omitting a condition on the function F that might make this a lot easier.

 

[Elucidus]

Consider $F(\frac{p}{q} \cdot x)$ when x = q (p, q integers). This shows that either $F(x) \equiv 0 \text{ or } F(\frac{p}{q}) = \frac{p}{q}$ (why?).

So F is either identically 0 or the identity on the rationals and if F is known to be contunuous, then F must either be identically 0 or the identiy on R.

The condition of continuity I believe is necessary to prove the claim. As Dick mentioned, I vaguely recollect that there are bizarre discontinuous functions that satisfy the premises that are neither identically 0 nor the identity.

--Elucidus

EDIT: The first sentence should say $F(1) = 0 \text{ not }F(x) \equiv 0$.

 

[Office_Shredder]

It's eluding me too. I think I actually looked at this before. By using my usual recourse when a problem is way too tough for me. I hunted through the sci.math forums. I THINK i found an entry by one of the brainiacs there that there ARE such functions which are not 0 or the identity. But they have to be constructed by axiom of choice type arguments and you really can't find a concise example of one accessible to a finite brain. That's just a vague recollection, so I could be wrong. But I think the OP might be omitting a condition on the function F that might make this a lot easier.

By the linearity and multiplicity condition, once you've prove F is the identity on the rationals (assuming it's non-zero, since that case is trivial), you can basically extend it to any algebraic number - F(solution to p(x)) must be a solution to p(x). The transcendentals are a bit more of a mystery.

 

[Dick]

By the linearity and multiplicity condition, once you've prove F is the identity on the rationals (assuming it's non-zero, since that case is trivial), you can basically extend it to any algebraic number - F(solution to p(x)) must be a solution to p(x). The transcendentals are a bit more of a mystery.

There's a LOT more transcendentals than there are algebraic numbers, aren't there? Wish I could find a reference on this.

 

[Elucidus]

I've managed to prove that if F(1) = 0 then F is periodic with period 1 and is therefore F identically 0.

I also have that if F(1) is not 0 then F(x) = x for all rationals. But I haven't made the final jump to all real numbers yet (it may not be possible).

--Elucidus

 

[Dick]

I've managed to prove that if F(1) = 0 then F is periodic with period 1 and is therefore F identically 0.

I also have that if F(1) is not 0 then F(x) = x for all rationals. But I haven't made the final jump to all real numbers yet (it may not be possible).

--Elucidus

If F(1)=0 then F(x)=F(1*x)=F(1)*F(x)=0*F(x)=0. Why do you need periodic?

 

[Office_Shredder]

Using the properties given, F(1) non-zero means F(1)=1 and from that you can prove all algebraic numbers are mapped to themselves. I'm almost 100% sure you can't glean anything from the transcendentals this way, because all you can do is form rational polynomials with addition and multiplication, and the algebraic numbers are exactly those that you can find as roots to rational polynomials.

New idea: Suppose F(pi) = c. What other values of x have we determined F(x) for? Obviously any rational multiple of pi, and any linear combination of integral powers of pi. I think that's it. So you can probably choose a basis of R over Q (here's the axiom of choice), make an equivalence relation based on which guys are multiples of each other, and then on the set of equivalence classes define F to be whatever the hell you want by defining F(basis vector)

 

[Elucidus]

If F(1)=0 then F(x)=F(1*x)=F(1)*F(x)=0*F(x)=0. Why do you need periodic?

Because I was missing the obvious looking too deeply at a trivial case.

--Elucidus

 

[Elucidus]

New idea: Suppose F(pi) = c. What other values of x have we determined F(x) for? Obviously any rational multiple of pi, and any linear combination of integral powers of pi. I think that's it. So you can probably choose a basis of R over Q (here's the axiom of choice), make an equivalence relation based on which guys are multiples of each other, and then on the set of equivalence classes define F to be whatever the hell you want by defining F(basis vector)

I believe this is how those whacked out discontinuous extensions alluded to earlier are made.

--Elucidus

 

[Dick]

Because I was missing the obvious looking too deeply at a trivial case.

--Elucidus

Nice to know smart people have been looking at this and haven't found some obvious proof I've been completely blind to.

 

[Hurkyl]

Oh bah, I should have been following this thread.

Additioinally for $$x \geq 0, F(x) \geq 0.$$

I'm not sure how to leverage this to a solution but I feel it's nearby but eluding me. I'm not even sure if any of this is relevant.

Okay, so you've proven F maps positive numbers to positive numbers.

Since you can say something about positive numbers, doesn't that suggest you can you say something about the ordering of the real numbers?

 

[Dick]

Oh bah, I should have been following this thread.


Okay, so you've proven F maps positive numbers to positive numbers.

Since you can say something about positive numbers, doesn't that suggest you can you say something about the ordering of the real numbers?

Ooops. Guess I had a false recollection. Thanks for getting this thread on the right track again.

 

[Elucidus]

Sheesh. I've been so wrapped up in looking for a simple proof by contradiction that I completely missed monotonicity. This was that last piece I needed to complete the proof.

--Elucidus

 

[Caesar_Rahil]

It has been proven that F(0)=0 and F(1)=1 or 0
put x=-y;
F(x)+F(-x)=0
so F(x) is odd

BELOW PROOF WORKS ONLY IF X IS RATIONAL:

now, if x is Natural number
F(x)
=F(1+1+1+... x times)=F(1)+F(1)+...x times
=x F(1)
=x or 0
since F(x) is odd; so is true for all negatives natural numbers
let x=p/q
p and q are integers
F(p)
=F(p/q*q)
=F(p/q)+F(p/q)+...q times
=q * F(p/q)
F(p/q) = F(p)/q
F(p/q)=p/q or 0
so F(x) = x or 0

I am trying for irrationals

 

[Hurkyl]

Ooops. Guess I had a false recollection. Thanks for getting this thread on the right track again.

That's because it's a sneaky problem! If you know the pattern of generating counterexamples in similar situations, then it's just so utterly obvious that it works here. It just happens to be utterly false.

 

[JG89]

Spivak gives the following approach for the problem:

1) Prove f(1) = 1
2) Prove f(x) = x for rational x
3) Prove f(x) > 0 if x > 0
4) Prove if x > y then f(x) > f(y)
5) Prove f(x) = x for irrational x (Use the fact that the rationals are dense on R)

The question isn't insanely hard if you follow the above procedure.

 

[Elucidus]

Spivak gives the following approach for the problem:

1) Prove f(1) = 1
2) Prove f(x) = x for rational x
3) Prove f(x) > 0 if x > 0
4) Prove if x > y then f(x) > f(y)
5) Prove f(x) = x for irrational x (Use the fact that the rationals are dense on R)

The question isn't insanely hard if you follow the above procedure.

Yeah. I knew I needed (5) but blew it on (4). Monotonicity was staring at me, taunting me. I went and looked back at my scratch work and right there on page 2 is

$$u \leq v \Rightarrow f(u) \leq f(v)$$

I will mention though that (1) and (2) should be

1) Prove that f(1) = 0 or 1.
2) Prove that if f(1) = 0 then f(x) = 0 for all x, and
that if f(1) = 1 then f(x) = x for all rational x.

--Elucidus

 

[Dick]

I screwed up most of all by throwing a monkey wrench into peoples thinking by suggesting it might not be true. Mea culpa again.

 

[ashok vardhan]

intuitively, the solution is f(kx)=kf(x). i am able to prove it for all integers k.but for a rational number k i am facing problem.so i assumed k to be p/q and later that i can't proced.would you give me a small hint

 

[HallsofIvy]

Given that f(x+ y)= f(x)+ f(y), you can prove, by induction on the positive integer n, that f(nx)= nf(x), x any real number. From that, for n a non-zero integer, f(n/n)= nf(1/n)= f(1)= 1. Thus, f(1/n)= 1/f(n). You can get the case for f(m/n) from that.

This is an entirely different problem from before, isn't it? Requiring that f be continuous at x= 0 rather than that f(xy)= f(x)f(y).