Zhidong Leong

Posted on Monday, 07 April, 2003 - 03:40 pm:

Does anyone have any simple worked examples of functional equation? For example, how do I go about solving this simple equation:
f(x + y) = f(x) + f(y) for all real x, y

Marcos

Posted on Monday, 07 April, 2003 - 04:06 pm:

I think the function is f(x) = kx for some k.
If you can't see how I got this (or disagree) then please say so...

Marcos

Zhidong Leong

Posted on Monday, 07 April, 2003 - 04:11 pm:

It is quite obvious that the answer is f(x) = kx. But how you got this?

David Turton

Posted on Monday, 07 April, 2003 - 05:53 pm:

Zhidong,

Try solving it for natural numbers x,y first, then go to integers, then rationals, then reals, building up information as you go.

To get you started, think about what f(0) is, and f(1), f(2), and see if you can spot a pattern that you can use for induction.

Dave

Zhidong Leong

Posted on Monday, 07 April, 2003 - 07:23 pm:

Now I got f(x)= xf(1)
Is it logical to say that the answer is f(x)=kx?
From what I have got, how do I know that the value of f(1) is not fixed?

Michael Doré

Posted on Monday, 07 April, 2003 - 07:59 pm:

If you have got to

f (x) = x f (1)

then certainly it is logical to say

f (x) = k x

. Simply set

k = f (1)

f (1)

is just a number.

Can you see why $f (x) = x f (1)$ for all rational $x$ ? Like David says it's best to look at the integers first, try and prove any patterns you spot by induction, then move onto the rationals.

Unfortunately you'll find there is no obvious way to go from rationals to reals. In fact if we assume the axiom of choice then there are other functions which satisfy the equation. It is a well known fact that if you assume the axiom of choice then every vector space has a Hamel basis. What this means in this particular case is that there exists a (uncountable) collection of reals $r_{i}$ such that every real $x$ can be written uniquely as a finite sum:

$\sum a_{i} r_{i}$

where $a_{i}$ are rationals and all but finitely many of $a_{i}$ are zero.

Using this fact, can you see how to construct other solutions to $f (x + y) = f (x) + f (y)$ other than $f (x) = k x$ ?

Marcos

Posted on Tuesday, 08 April, 2003 - 11:26 am:

I did what David said, basically...

Michael, couldn't you use the fact that the reals are dense on the rationals (is this correct terminology? :) ) and the fact that $f (x + ε) > f (x)$ (for $k > 0$ , for negative $k$ the opposite is true)

Marcos

Kerwin Hui

Posted on Tuesday, 08 April, 2003 - 12:11 pm:

Why should f be monotone? The whole point here is that, assuming a Hemel basis exists, you can write some everywhere discontinuous functions f satisfying f(x+y)=f(x)+f(y), by choosing what f does to each member of the basis.

Kerwin

Marcos

Posted on Tuesday, 08 April, 2003 - 12:37 pm:

Good point! So f(x) = kx is the only continuous f then?

Zhidong Leong

Posted on Tuesday, 08 April, 2003 - 01:12 pm:

I don't quite understand. What exactly is Hamel basis? And how does it help me find the other solution?

Michael Doré

Posted on Tuesday, 08 April, 2003 - 10:17 pm:

In the present situation, a Hamel basis just means what I wrote above. Namely it is a set

X

of reals such that every real

x

can be written uniquely in the form

\sum a_{i} r_{i}

where

r_{i}

are members of

X

a_{i}

are rationals and all but finitely many of

a_{i}

are zero.

You can prove a Hamel basis exists if you assume the axiom of choice. I won't go into the proof since the proof uses Zorn's lemma, and that in turn is quite complicated to prove and doesn't really have anything to do with functional equations per se.

So can you see how to use the fact a Hamel basis exists to construct other functions satisfying the functional equation? Hint: look at Kerwin's post.

Demetres Christofides

Posted on Wednesday, 09 April, 2003 - 08:52 am:

DO NOT try to find a Hamel basis! Just assume that there is one.

Demetres