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        riddles >> putnam exam (pure math) >> Solve functional equation involving a limit.
        
(Message started by: Aryabhatta on Aug 14^th, 2004, 10:55am)

Title: Solve functional equation involving a limit.
Post by Aryabhatta on Aug 14^th, 2004, 10:55am

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].

Title: Re: Solve functional equation involving a limit.
Post by Aryabhatta on Aug 16^th, 2004, 6:29pm

Hint:
[hide]
f(x) = x/2 is the only solution.
[/hide]

Title: Re: Solve functional equation involving a limit.
Post by Eigenray on Aug 16^th, 2004, 9:45pm

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.

Title: Re: Solve functional equation involving a limit.
Post by Aryabhatta on Aug 16^th, 2004, 10:24pm

on 08/16/04 at 21:45:42, Eigenray wrote:

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.