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william wu       Gender: Posts: 1291 f(x) + f(y) = f(x + y)   « on: May 18th, 2007, 12:10pm » Quote Modify

Suppose that f is any extended real-valued function which for eveyr x and y satisfies:

f(x) + f(y) = f(x+y).

(i) Show that f is either eveyrwhere finite or everywhere infinite.

(ii) Show that if f is both measurable and finite, then f(x) = x f(1) for each x. IP Logged

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KjSlag
Newbie  Posts: 4 Re: f(x) + f(y) = f(x + y)   « Reply #1 on: May 28th, 2007, 6:56pm » Quote Modify

(i)
i may be wrong, but I don't see how f could be infinite anywhere (except for f(inf) or f(-inf))
f(x) + f(y) = f(x+y)
f(x) + f(0) = f(x+0)
f(0) = f(x)-f(x)
f(0) = 0

f is finite everywhere because:
take the derivative dx
f'(x) + 0 = f'(x+y)
thus, for some c,
f'(x) = c
i don't know how to prove that c != inf

(ii)
because f'(x) is constant, f must be of the form c*x
c is obviously equal to f(1) IP Logged
Obob
Senior Riddler     Gender: Posts: 489 Re: f(x) + f(y) = f(x + y)   « Reply #2 on: May 28th, 2007, 7:53pm » Quote Modify

No, f can in fact be infinite everywhere.  For if we define f(x)=infty for all x, then sure enough f(x)+f(y)=f(x+y) for all x and y.   The problem is that in extended real arithmetic, it is not true that infty - infty = 0.

You also can't take the derivative of an arbitrary function.

(i) If f(x)=infty for some x, then we cannot have f(y)=-infty for any y.  For the equation f(x)+f(y)=f(x+y) is nonsense, for every possible value of f(x+y).  But then for any number z, f(z)=f(z-x)+f(x) = infty since f(x)=infty and f(z-x) is not -infty.  Therefore f(z)=infty for all z.  A similar proof shows that if f is -infty somewhere then f=-infty.

 « Last Edit: May 28th, 2007, 7:57pm by Obob » IP Logged
Icarus
wu::riddles Moderator
Uberpuzzler      Boldly going where even angels fear to tread.

Gender: Posts: 4863 Re: f(x) + f(y) = f(x + y)   « Reply #3 on: Jun 3rd, 2007, 1:36pm » Quote Modify

So, Obob, what if you define - = 0 in the extended reals?

(The definition means that certain additions involving both and - are no longer associative, but while inconvenient, that doesn't stop us from being able to make the definition. Though I have seen it before, I admit that this definition is not commonly used. A much more common definition of indefinite forms is 0 * = 0. This one is very helpful in stating many measure theory theorems.) IP Logged

"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
Obob
Senior Riddler     Gender: Posts: 489 Re: f(x) + f(y) = f(x + y)   « Reply #4 on: Jun 3rd, 2007, 2:32pm » Quote Modify

Suppose that f(x)=infty and f(y)=-infty.  Then f(x+(y+y))=f(x)+f(y+y)=f(x)+(f(y)+f(y))=infty+(-infty+-infty)=infty-inft y=0,
but f((x+y)+y)=f(x+y)+f(y)=(f(x)+f(y))+f(y)=(infty-infty)-infty=-infty, a contradiction.  So with this definition of infty-infty=0, it cannot happen that f takes on both infty and -infty as values.

Essentially, the problem is that addition is associative in R but not in extended real arithmetic. IP Logged
Obob
Senior Riddler     Gender: Posts: 489 Re: f(x) + f(y) = f(x + y)   « Reply #5 on: Jun 3rd, 2007, 2:37pm » Quote Modify

Does anybody have any thoughts for part (b)?  It seems fairly strange.  Without the axiom of choice, there are no nonmeasurable subsets of the real numbers.  So then the statement says that every Q-linear mapping R->R is a constant multiple of the identity.  Of course, its pretty rare to do measure theory without the axiom of choice, though. IP Logged
Icarus
wu::riddles Moderator
Uberpuzzler      Boldly going where even angels fear to tread.

Gender: Posts: 4863 Re: f(x) + f(y) = f(x + y)   « Reply #6 on: Jun 4th, 2007, 5:15pm » Quote Modify

on Jun 3rd, 2007, 2:32pm, Obob wrote:
 Suppose that f(x)=infty and f(y)=-infty.  Then f(x+(y+y))=f(x)+f(y+y)=f(x)+(f(y)+f(y))=infty+(-infty+-infty)=infty-inft y=0, but f((x+y)+y)=f(x+y)+f(y)=(f(x)+f(y))+f(y)=(infty-infty)-infty=-infty, a contradiction.  So with this definition of infty-infty=0, it cannot happen that f takes on both infty and -infty as values.   Essentially, the problem is that addition is associative in R but not in extended real arithmetic.

Yep. I had been planning on posting your solution to (i) myself if no one had done anything on this thread yet, just to bring it back to people's attention. But since you had provided, I gave this minor objection instead.

Of course, the reason is that I haven't made progress on the real problem, which is (ii). I had thought a week ago that I had an "almost everywhere" solution, but when I tried to post it, I got to the end and couldn't find that contradiction I thought I had anywhere. (Anyone finding a lost contradiction lying around, please return it to me!)

on Jun 3rd, 2007, 2:37pm, Obob wrote:
 Does anybody have any thoughts for part (b)?  It seems fairly strange.  Without the axiom of choice, there are no nonmeasurable subsets of the real numbers.  So then the statement says that every Q-linear mapping R->R is a constant multiple of the identity.  Of course, its pretty rare to do measure theory without the axiom of choice, though.

Not having the axiom of choice does not mean that there are no unmeasurable subsets of the real numbers. All it means is that you cannot prove their existence. You also cannot prove their non-existence.

Of course, you could then take the non-existence of unmeasurable sets as an axiom. But it is this strong axiom that allows you to say that -linear implies -linear.

Perhaps the way to approach this is to assume the existence of a -linear, -nonlinear function, and show that it must be unmeasurable. IP Logged

"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous

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