Additive functions

  • Thread starter Ed Quanta
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Ok, so suppose f is a function which takes us from the Reals in p space to the Reals in m space. And f(x+y)=f(x) + f(y) for all x and y 's in the Reals in p space.

Now if f(0)=0, how do I now show f(x-y)= f(x) - f(y)?
 

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  • #2
jcsd
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f(y-y) = f(0) = f(y) + f(-y) = 0 therfore f(-y) = -f(y).
 
  • #3
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Ahhhh, so now if f is continuous at some c is an element of the Reals in p space, how does this imply f is continuous throughout the Reals in p space?
 
  • #4
matt grime
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Because you can translate everything back to c.
 
  • #5
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How?

f(x-y)=f(x)-f(y)

so were f(x-y)=f(n), we can always write this in terms of c

f(c-t)=f(c)-f(t) where t=c-n

We know at c, f is continuous, but how do we then conclude that for f(t)?
 
  • #6
matt grime
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Let x= c+t, y=c+s for some s,t consider f(x-t) and the fact that if |x-y| < d then |t-s| < d

or if you prefer, an interval of width d centred on x can be translated back to an interval of width d centred on c.
 
  • #7
HallsofIvy
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Use 0 (the 0 vector in Rn of course).

Knowing that f is continuous at q means
lim(x->q)f(x)= f(q). Let h= x-q so that x= q+h and x= q is equivalent to h= 0. The limit becomes lim(h->0)f(q+h)= lim(h->0) f(q)+ f(h)= f(q)+ lim(h->0)f(h)= f(q) so
lim(h->0)f(h)= 0.

Now, if p is any other point, lim(x->p)f(x)= lim(h->0)f(p+ h) (taking h= x- p)
= lim(h->0) f(p)+ f(h)= f(p)+ lim(h->0)f(h)= f(p)+ 0= f(p) so that f is continuous at p.

In fact, it is possible to prove (but much harder, I understand) that if f(x+y)= f(x)+ f(y) and f is bounded on any interval, no matter how small, then f is continuous.
 
  • #8
matt grime
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Is it hard? I've not seen this one before like this (though it's close to the bounded implies cont at 0)

Let x_n be a cauchy seqence tending to 0. Let y_n be the largest integer such that |x_ny_n| is still less than 1. Passing to a s subsequence, we may assume that y_n is strictly monotone increasing and y_n =>n, but then

|f(x_n)| < M/n where M bounds the value on the unit ball.

so f(x_n) is cauchy, hence converges to zero, so it's continous at 0 and the previous result applies.
 

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