Re: functional equation

From: □[► Јo∫h ◄ ]□ (nomail_at_noaddress.not)
Date: 10/30/04 

Date: Sat, 30 Oct 2004 14:28:12 +0200

Robert Israel wrote:
> In article <2ueq5hF28sd6jU1@uni-berlin.de>,
> â ¡[â º Ð oâ «h â ]â ¡ <nomail@noaddress.not> wrote:
>
>>We now that is possible to define the base $a>1$ logarithmic function as
>>the only function f:(0,+inf) --> R such that:
>
>
>>1. x<y => f(x)<f(y)
>>2. f(a) = 1
>>3. f(xy) = f(x)+f(y)
>
>
>>Now using only this theorem (thinks as nobody knows the proof), is it
>>possible to prove the existence and uniqueness of the restriction of the
>>log_a to [1,+inf). i.e. that exist and is unique the function f:[1,+inf)
>>--> R such that 1. 2. and 3. are satisfied?
>
>
> I don't know what "thinks as nobody knows the proof" means.
>
> You seem to be asking, if there is a unique f on (0,infty) satisfying
> 1,2 and 3, is there a unique f on [1,infty) satisfying the same
> conditions (but for x,y in [1,infty))?
>
> Hint: Given any such function f on [1,infty), define g on (0,infty) by
> g(x) = f(x) for x >= 1, g(x) = -f(1/x) for 0 < x < 1. Prove that
> g satisfies 1, 2 and 3...
>

Ok, let me try to use your hint.

Let f1,f2 be to functions satisfying 1,2 and 3. Then I can consider the
two extensions g1 and g2 putting:

g1(x) = f1(x) for x >= 1,
g1(x) = -f1(1/x) for 0 < x < 1

and

g2(x) = f2(x) for x >= 1,
g2(x) = -f2(1/x) for 0 < x < 1

and these exstension satisfies 1,2 and 3; so the guaranteed unicity:
g1 = g2
and by this equality also the equality of their restriction to [1,inf);
i.e. f1=f2. qed.

Is this proof right?

Thanks, Giò.

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