# Thread: Function 0 or x

1. ## Function 0 or x

Let f:R->R be such that f(x+y)=f(x)+f(y) and f(x.y)=f(x).f(y), $\displaystyle \forall$x,y$\displaystyle \in$R. Prove that f(x)=0 or f(x)=x, $\displaystyle \forall$x$\displaystyle \in$R.

2. Originally Posted by rhemo
Let f:R->R be such that f(x+y)=f(x)+f(y) and f(x.y)=f(x).f(y), $\displaystyle \forall$x,y$\displaystyle \in$R. Prove that f(x)=0 or f(x)=x, $\displaystyle \forall$x$\displaystyle \in$R.
It doesn't have to be continuous or something?
Let $\displaystyle x\in\mathbb{R}$. Then $\displaystyle f(x)=f\left(x\cdot 1\right)=f(x)f(1)$
$\displaystyle f(x)=f\left(x\cdot 1\right)=f(x)f(1)\implies f(x)\left(f(1)-1\right)=0$. If $\displaystyle f(x)=0$ we are done, so assume that $\displaystyle f(x)\ne 0$ then $\displaystyle f(1)=1$..etc.

3. It doesn't have to be continuous. It has already been shown that f(1)=1. Let x be a positive rational number. Then there is an integer p and a positive integer q such that x = p/q.

Therefore f(x) = f(p/q) = f(p)*f(1/q)

But $\displaystyle p=\sum_{i=1}^{p}1 \Rightarrow f(p)=f(\sum_{i=1}^{p}1)=\sum_{i=1}^{p}f(1)=\sum_{i =1}^{p}1 = p$

Furthermore, we can use a trick to show that f(1/q) = 1/q.

$\displaystyle f(1) = 1 = \sum_{i=1}^{q}\frac{1}{q} \Rightarrow q \cdot f(1/q) = \sum_{i=1}^{q}f(1/q) = f(\sum_{i=1}^{q}1/q)=f(1)=1 \Rightarrow f(1/q)=\frac{1}{q}$

Ok. Now it is your turn:

i) Why is f(x) positive for all positive x?
ii) Why is f(x) monotonically increasing?
iii) Why is f(x)=x even for x in R\Q?

4. Originally Posted by JoachimAgrell
It doesn't have to be continuous. It has already been shown that f(1)=1. Let x be a rational number. Then there is a positive integer p and an integer q such that x = p/q.

Therefore f(x) = f(p/q) = f(p)*f(1/q)

But $\displaystyle p=\sum_{i=1}^{p}1 \Rightarrow f(p)=f(\sum_{i=1}^{p}1)=\sum_{i=1}^{p}f(1)=\sum_{i =1}^{p}1 = p$

Furthermore, we can use a trick to show that f(1/q) = 1/q.

$\displaystyle 1 = \sum_{i=1}^{q}\frac{1}{q}$

(I posted this by accident, i'm still finishing)
It's clearly true for rational values, but you haven't said a thing for non rational values. IF we knew it were supposed to be continuous we could finish.

5. Originally Posted by Drexel28
It's clearly true for rational values, but you haven't said a thing for non rational values. IF we knew it were supposed to be continuous we could finish.
We don't need to assume the continuity of f.

i) f(x) is positive whenever x is positive.

Proof. Since x is positive, x has a positive square root, say, y = sqrt(x). Then
f(x) = f(y)*f(y) > 0 for all positive x.

ii) f(x) is monotonically increasing.

Proof. In fact, if y > x then y - x > 0, and, by proposition i), f(y-x) = f(y)-f(x) > 0 => f(y)>f(x).

iii) f(x)=x for all x in R.

Proof. Suppose f(x) < x for some x in R. There is, however, a rational number y such that f(x) < y < x. But y < x implies y = f(y) < f(x), i.e., f(x)<y<f(x), a contradiction.

6. Originally Posted by JoachimAgrell
We don't need to assume the continuity of f.

i) f(x) is positive whenever x is positive.

Proof. Since x is positive, x has a positive square root, say, y = sqrt(x). Then
f(x) = f(y)*f(y) > 0 for all positive x.

ii) f(x) is monotonically increasing.

Proof. In fact, if y > x then y - x > 0, and, by proposition i), f(y-x) = f(y)-f(x) > 0 => f(y)>f(x).

iii) f(x)=x for all x in R.

Proof. Suppose f(x) < x for some x in R. There is, however, a rational number y such that f(x) < y < x. But y < x implies y = f(y) < f(x), i.e., f(x)<y<f(x), a contradiction.
You misunderstood me. I am not saying we need to have that $\displaystyle f$ is continuous. I am saying that we can show that $\displaystyle f$ is continuous at $\displaystyle 0$ and thus with a little work it is continuous at all points. And thus, by the fact that $\displaystyle f(1)=f\left(\frac{p}{p}\right)=pf\left(\frac{1}{p} \right)\implies f\left(\frac{1}{p}\right)=\frac{f(1)}{p}=\frac{1}{ p}$ and then $\displaystyle f\left(\frac{p}{q}\right)=pf\left(\frac{1}{q}\righ t)=\frac{p}{q}$ we see that $\displaystyle f(x)=x,\text{ }x\in\mathbb{Q}$ and since $\displaystyle \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}$ and a continuous function is determined on a dense subset the conclusion follows.

Your proof is equally good thought, good job!