Limits: If f(x+y)=f(x)+f(y), fcn ascending, prove f(x) cont.

[arman_phsa]

if f(x+y)=f(x)+f(y) and this function is ascending then prove f(x) is continues.

I solved this problem using the infinite sequences and series but the solution is long i want to know is there any other solution to prove this linear function is continues everywhere

thx

 

[arman_phsa]

Re: a limit problem

I finally come across a new solution method :
we know that the function is ascending so we have f(x+h)>=f(x) {h>0} so f(x)+f(h)>=f(x) so if h->0+ then limit f(h)>=0 "*"
and if h->0- then limit f(h)<=o "**"
from * and ** we have limit h->0, f(h)=f(0)
so the function is continues in x=0
know i will prove that if this function is continues at 0 it is continues everywhere :
we know that f(0)=0 { f(0)=2f(0) }
so f(0)=f(x-x)=f(x)+f(-x)=0 so f(-x)=-f(x) "1"
now we know that f is continues at 0 so we will use the definition :
for all e>0 (epsilon) we have d>0 (delta) that [ |x-0|<d ,|f(x)-f(0)|<e ] so [ |x|<d , |f(x)|<e ]
we know the above expression is correct and there is a relation between e and d
now we will pick "a" whether we want and set x=X-a and use in above :
|X-a|<d , |f(X-a)|<e so |f(X)+f(-a)|<e and from "1" we have :
|X-a|<d , |f(X)-f(a)|<e and this is the continues definition for x { limit x->a,f(x)=f(a) } and because we had no limit in choosing "a" the function is continues in every point of its domain
I don't know if my answer is correct or not so please send your comments about this answer or if you find another method please inform me
TY

 

[Unco]

Hi Arman,

Well done on your efforts.

I agree with most of what you have written. A couple of comments:
1) For consistency, you might like to also use the limit definition of continuity to prove continuity at 0 implies continuity everywhere. If a is a real number, let h = x-a; then \(\displaystyle \lim_{x\to a}f(x) = \lim_{h\to 0}f(a + h)\). The result then follows noting that \(\displaystyle \lim_{h\to 0}f(h)=f(0)\).

2) Having let X = x-a, and putting X for x in your "|x|<d , |f(x)|<e" result, you have |X|<d implies |f(X)|<e. Resubstituting x for X gives the desired result.

 

[arman_phsa]

thx Unco for your comments :
i agree with what you said and i see that your method in proving those 2 parts is more consistent and shorter of course, and my method is somehow longer and harder at those 2 points
Thx for your comments
and sry for my bad English