Let f:R → R satisfy f(x+y) = f(x) + f(y) for real numbers x and y

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Homework Help Overview

The problem involves a function f: R → R that satisfies the functional equation f(x+y) = f(x) + f(y) for all real numbers x and y. The original poster seeks to demonstrate that if f is continuous, then there exists a real number b such that f(x) = bx.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the functional equation and continuity, with some suggesting that f(x) can be expressed in terms of f(1) for rational x. Others explore the extension of this reasoning to all real numbers and question how to handle cases where x is not an integer.

Discussion Status

Some participants have offered hints and partial reasoning, particularly regarding the relationship between f(x) and f(1). There is an ongoing exploration of how to generalize findings from rational numbers to all real numbers, indicating a productive direction in the discussion.

Contextual Notes

Participants are navigating the constraints of continuity and the specific properties of the function as defined by the problem statement. There is a focus on the implications of the functional equation and the continuity condition without resolving the overall question.

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Homework Statement


Let f:R\rightarrowR satisfy f(x+y) = f(x) + f(y) for real numbers x and y. If we let f be continuous, show that \exists a real number b such that f(x) = bx.


Homework Equations


n/a


The Attempt at a Solution


Nooooo clue!
 
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Hint
Show that
f(x) = xf(1) when x is rational.

If f is continuous at c in R then f is continuous on R thus
f(x) = xf(1) for all x in R .
 
In which case b= f(1). :-)
 
hmmm.
so f(x) = f(x-1) + f(1) = 2f(1) + f(x-2) = ... = f(x-x) + xf(1) = xf(x)
but that only works when x is an integer. How would you do it for when x is rational?
 
x = m/n.

nx =m

f(nx) = f(x)+ ...f(x) = nf(x)

f(m) = mf(1).
Thus,

nf(x) = mf(1).

f(x) = (m/n)f(1) = xf(1).

:-)
 

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