Function with multiplicative property

In summary, the question asks for a proof that f is continuous at x=0, which can be easily solved byassuming continuity and showing that f(x) = ax for all x in R.
  • #1
mattmns
1,128
6
Here is the question:
--------
Let [itex]f: \mathbb{R} \rightarrow \mathbb{R}[/itex] satisfy [itex]f(x+y) = f(x)f(y) \ \forall x,y \in \mathbb{R}[/itex].Let a = f(1) > 0. Show that [itex]f(r)=a^r \ \forall r \in \mathbb{Q}[/itex].
---------

This question actually has multiple parts to it. I have already proved the following:

f(0)=1
f(-x) = 1/f(x) for all x in R
f(x)>0 for all x in R.
[itex]f(n)=a^n \ \forall n \in \mathbb{N}[/itex]
[itex]f(n)=a^z \ \forall z \in \mathbb{Z}[/itex]
-----

Let r=p/q p,q in Z, q not 0.

To solve the problem I have been trying to show that [itex]f(1/q) = a^{1/q}[/itex] which I believe would lead to the solution, since we have the multiplicative property, and if not at least get a better idea of how to get the whole thing. However, this does not seem to be going anywhere, it feels like I need something else. I just had an idea of developing some type of composition property, and maybe that will get it. If you think this might work feel free to ignore this post, if not, any ideas are welcome. Not sure if that is good, basically I am thinking of trying to get some kind of identity involving f(xy) and then maybe something might happen? I suppose [itex]f(xy) = (f(x))^y[/itex].

Thanks!
 
Last edited:
Physics news on Phys.org
  • #2
What is f(1/q+1/q+...+1/q), q times?
 
  • #3
Awesome! Fantastic Idea, thanks!
 
Last edited:
  • #4
Is there any condition that f be continuous on R? If not then f(x)= ax may not be true.
 
  • #5
Yes, that is the next part actually :smile: We then have to prove that f is continuous, which I did a while ago, and then use this and the f(r) = ar for r in Q to show that f(x) = ax for x in R.
 
  • #6
You are given only that f(x+ y)= f(x)f(y) and asked to prove that f is continuous?

I am not familiar with f(x+ y)= f(x)f(y) but I know that the simpler equation
f(x+ y)= f(x)+ f(y) has non-continuous solutions. IF f is continuous then the general solution is f(x)= Cx but non-continuous solutions are very complicated.
 
  • #7
Well, we are given that f is continuous at x = 0, and from there we conclude that f is continuous for every point in R.

Here is my proof:

Since f is continuous at x=0 we have that [itex]\lim_{x\rightarrow x_0}f(x) = f(0) = 1[/itex].

From the multiplicative property of f and part (i) we get:

[tex]\lim_{x\rightarrow x_0}\dfrac{f(x)}{f(x_0)} = \lim_{x\rightarrow x_0}f(x)f(-x_0) = \lim_{x\rightarrow x_0}f(x-x_0) = \lim_{(x-x_0) \rightarrow 0}f(x-x_0) = f(0) = 1[/tex]

So, [tex]\lim_{x\rightarrow x_0}\dfrac{f(x)}{f(x_0)} = 1 \Rightarrow \lim_{x\rightarrow x_0}f(x) = f(x_0)[/tex]

Therefore f is continuous at every point in [itex]\mathbb{R}[/itex].
 
Last edited:
  • #8
mattmns said:
Well, we are given that f is continuous at x = 0
You were given that. You forgot to give that to the rest of us. :wink:
 
  • #9
Yes, it is a really good idea not to withold information.

If f(x+y)= f(x)+ f(y) (not YOUR equation!) and f is known to be continuous at x= 0, then you can prove that f is continuous at any point a:

In [itex]\lim{x\rightarrow a} f(x)[/itex], let h= x- a so that [itex]\lim_{x\rightarrow a}f(x)= \lim_{h\rightarrow 0}f(a+ h)[/itex]
[itex]= \lim_{h\rightarrow 0}(f(a)+ f(h))= f(a)+ \lim_{h\rightarrow 0}f(h)= f(a)[/itex]
since you will already have proved f(0)= 0.

You should be able to do something similar for f(x+ y)= f(x)f(y).
 
  • #10
Sorry for not posting that information :redface:
 
  • #11
Alright, I now have this question, continuation:

Assume that f is continuous at x = 0, use the fact that f is then continuous for all x in R, and that f(r) = ar for all r in Q, to conclude that f(x) = ax for all x in R.
------------

I think I understand why this is true (the rationals are dense in R, and we have the continuity to force the points to stay close together), but I just can't seem to get anywhere in the proof. It seems like I would like to grab some real number x, (and maybe another real y), then place some rationals as close as possible, and then somehow get that f(r) < f(x) < f(q), and then get that f(x) = ax (maybe with a limit or two and the continuity fact). However, I am not sure of how (or if what I said will work) to do this (and with proof). Any ideas for this one? Thanks!
 
  • #12
It depends how you define a^x for real x. It's usually defined simply as the limit of a^r as r goes over a rational sequence approaching x, which makes the question trivial.
 
  • #13
Ahh yes, that other (equivalent) definition of continuity, that is perfect! Thanks!
 

1. What is a function with multiplicative property?

A function with multiplicative property is a mathematical function that follows the rule of multiplication. This means that when two or more inputs are multiplied together, the output of the function is equal to the product of the individual outputs for each input. In other words, the output of a function with multiplicative property is determined by multiplying the inputs together.

2. How is a function with multiplicative property different from a regular function?

A regular function can have a variety of different properties, such as addition, subtraction, or exponentiation. A function with multiplicative property, on the other hand, only follows the rule of multiplication. This means that the output of the function is only affected by the inputs being multiplied together, and not by any other operations.

3. Can you give an example of a function with multiplicative property?

One example of a function with multiplicative property is f(x) = 2x, where x represents the input and 2 is the constant multiplier. This function follows the rule of multiplication, as the output is always equal to two times the input. For example, when x = 3, f(x) = 2(3) = 6.

4. What are the benefits of using a function with multiplicative property?

One benefit of using a function with multiplicative property is that it can simplify calculations, particularly when dealing with large numbers. By using multiplication, rather than other operations, the function can be more efficient and easier to work with. Additionally, functions with multiplicative property are often used in applications such as compound interest and exponential growth.

5. Can a function have both additive and multiplicative properties?

Yes, a function can have both additive and multiplicative properties. This type of function is known as a linear function, and it follows the rules of both addition and multiplication. An example of a linear function is f(x) = 2x + 3, where the output is determined by multiplying the input by 2 and then adding 3 to the result. However, it is important to note that not all functions have both additive and multiplicative properties, and some may only have one or the other.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
637
  • Calculus and Beyond Homework Help
Replies
1
Views
414
  • Calculus and Beyond Homework Help
Replies
3
Views
503
  • Calculus and Beyond Homework Help
Replies
3
Views
466
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
761
  • Calculus and Beyond Homework Help
Replies
2
Views
152
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
532
  • Calculus and Beyond Homework Help
Replies
1
Views
450
Back
Top