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Proof of continuous. f(x+y)=f(x)+f(y)

  1. Dec 3, 2008 #1
    Let f be a real-valued function on R satisfying f(x+y)=f(x)+f(y) for all x,y in R.

    If f is continuous at some p in R, prove that f is continuous at every point of R.

    Proof: Suppose f(x) is continuous at p in R. Let p in R and e>0. Since f(x) is continuous at p we can say that for all e>0, there exists a d>0 such that if 0<|x-p|<d implies that |f(x)-f(p)|< e/2.


    I know that |f(x)-f(y)| = |f(x)-f(p)+f(p)-f(y)| <= |f(x)-f(p)|+|f(y)-f(p)|.

    I also know that |x-y|=|(x-y+p)-p| <= |x-p|+|y-p|.

    Help: I am confused on how to show that f is cont. at every point in R. I appreciate any help little or big! Thank you
     
  2. jcsd
  3. Dec 3, 2008 #2
    So if |x - y| < d, then |(x - y + p) + p| < d. Then...

    Now use the fact that f(x + y) = f(x) + f(y).
     
  4. Dec 4, 2008 #3
    Let f be a real-valued function on R satisfying f(x+y)=f(x)+f(y) for all x,y in R.

    If f is continuous at some p in R, prove that f is continuous at every point of R.

    Proof: Suppose f(x) is continuous at p in R. Since f(x) is continuous at p we can say that for all e>0, there exists a d>0 such that if 0<|x-p|<d implies that |f(x)-f(p)|< e.

    Let y be a real number and fix 0<|x-y|<d.
    Then |(x-y+p)-p|<d
    So |f(x-y+p)-f(p)|<e.
    or in other words, |f(x-y)+f(p)-f(p)|<e.
    or, |f(x)-f(y)|<e.


    I have a concern that I used f(x-y)=f(x)-f(y). Does this work because f(-y)= -f(y) ?

    I also changed my e/2 into just e.
    I also included a "let" before the manipulating.

    Does this look better? I appreciate the help mutton.
     
    Last edited: Dec 4, 2008
  5. Dec 4, 2008 #4

    HallsofIvy

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    You don't need to go back to the basic definition of limit. A function, f, is continuous at x= a if and only if:
    f(a) is defined
    [itex]\lim_{x\rightarrow a} f(x)[/itex] is defined
    [itex]\lim_{x\rightarrow a} f(x)= f(a)[/itex]

    Suppose f is continuous at x= a. Then [itex]\lim_{x\rightarrow a} f(x)= f(a)[/itex]

    Let h= x- a. Then x= a+ h and as x goes to a, h goes to 0:
    [itex]\lim_{x\rightarrow a} f(x)=\lim_{h\rightarrow 0} f(a+ h)= \lim{h\rightarrow 0}f(a)+ f(h)[/itex]

    What does that tell you about f at x= 0?
    (It is important to recognize that if f(x+ y)= f(x)+ f(y), then f(x)= f(x+ 0)= f(x)+ f(0) so f(0)= 0.)

    Now let b be any other value and go the opposite way.
     
  6. Dec 4, 2008 #5
    It works because f(x) = f(x - y + y) = f(x - y) + f(y).

    But what you say is also true because f(-y) = f(0 - y) = f(0) - f(y) = -f(y), as HallsofIvy showed that f(0) = 0.
     
  7. Oct 18, 2009 #6

    DDW

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    I was just wondering what you meant by "let b be any other value and go the opposite way."
    do you mean let h=x+b?
    Thanks.
     
  8. Oct 18, 2009 #7

    HallsofIvy

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    Actually, h= x- b. If b is any real number, then then [itex]\lim_{x\to b} f(x)= \lim_{h\to 0}f(b+ x)[/itex][itex]= \lim_{h\to 0}f(b)[/itex][itex]+ \lim_{h\to 0}f(h)[/itex][itex]= f(b)+ \lim_{h\to 0}f(h)= f(b)[/itex].
     
  9. Oct 18, 2009 #8

    DDW

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    Thanks!
     
  10. Sep 26, 2010 #9
    What is the exact difference between the h = x-a and h =x-b step? What is the point of doing both steps?
     
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