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william wu
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f(x) + f(y) = f(x + y)  
« on: May 18th, 2007, 12:10pm »
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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.
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Re: f(x) + f(y) = f(x + y)  
« Reply #1 on: May 28th, 2007, 6:56pm »
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(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)
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Obob
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Re: f(x) + f(y) = f(x + y)  
« Reply #2 on: May 28th, 2007, 7:53pm »
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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
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Re: f(x) + f(y) = f(x + y)  
« Reply #3 on: Jun 3rd, 2007, 1:36pm »
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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.)
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Re: f(x) + f(y) = f(x + y)  
« Reply #4 on: Jun 3rd, 2007, 2:32pm »
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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.
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Re: f(x) + f(y) = f(x + y)  
« Reply #5 on: Jun 3rd, 2007, 2:37pm »
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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.
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Re: f(x) + f(y) = f(x + y)  
« Reply #6 on: Jun 4th, 2007, 5:15pm »
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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.
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