Author |
Topic: Solve functional equation involving a limit. (Read 916 times) |
|
Aryabhatta
Uberpuzzler
Gender: 
Posts: 1321
|
 |
Solve functional equation involving a limit.
« on: Aug 14th, 2004, 10:55am » |
 Quote  Modify
|
Find all functions f which satisfy the following:
1) f : [0,1] [to] [bbr] such that Limit x [to] t f(x) = g(t) exists [forall] t [in] [0,1]
2) f(x) + g(x) = x [forall] x [in] [0,1].
|
|
 IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: Solve functional equation involving a limit.
« Reply #1 on: Aug 16th, 2004, 6:29pm » |
Quote Modify
|
Hint:
f(x) = x/2 is the only solution.
|
|
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Solve functional equation involving a limit.
« Reply #2 on: Aug 16th, 2004, 9:45pm » |
Quote Modify
|
Fix t [in] [0,1], and let [epsilon] > 0.
There exists [delta] > 0 such that for | t - x | < [delta], | g(t) - f(x) | < [epsilon]. It follows that for any such x, | g(t) - g(x) | [le] [epsilon]. Therefore g is continuous.
Since f(x) = x - g(x) is continuous, we must have g(t) = f(t), so f(x) = g(x) = x/2.
|
|
IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: Solve functional equation involving a limit.
« Reply #3 on: Aug 16th, 2004, 10:24pm » |
Quote Modify
|
on Aug 16th, 2004, 9:45pm, Eigenray wrote:Fix t [in] [0,1], and let [epsilon] > 0.
There exists [delta] > 0 such that for | t - x | < [delta], | g(t) - f(x) | < [epsilon]. It follows that for any such x, | g(t) - g(x) | [le] [epsilon]. Therefore g is continuous.
|
|
Right! Well done. It can be shown that g is a continuous function.
To clarify Eigenray's point:
Let h(z) = |g(t) - f(z)| for |t - z | < [delta].
We have that h(z) < [epsilon].
Taking limits as z [to] x, we get |g(t) - g(x)| [le] [epsilon].
It can be shown that any f which has a limit at every point has a countable number of discontinuities. (Could be zero or even finite)
A classical example of such a function is the "ruler" function (or dirichlet function, dont know which)
f : (0,1]-> [bbr]
f(x) = 0 if x is irrational
f(p/q) = 1/q where p/q is in lowest terms.
It can be shown that f is continuous at each irrational, discontinuous at each rational and Limit of f at each point is 0.
|
| « Last Edit: Aug 16th, 2004, 11:09pm by Aryabhatta » |
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print