If f(x+y)=f(x)+f(y) and f(x.y)=f(x)f(y) then f(x)=x , x in R

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1. Mar 20, 2017

issacnewton

1. The problem statement, all variables and given/known data

Suppose that the function $f$ satisfies the two properties. $f(x+y)=f(x)+f(y)$ and $f(x\cdot y)=f(x)\cdot f(y)$, but that $f$ is not always $0$. Prove that $\forall x~ f(x) = x$, as follows:
(a) Prove that $f(1)=1$
(b) Prove that $f(x)=x$ if $x$ is rational
(c) Prove that $f(x) > 0$ if $x>0$
(d) Prove that $f(x) > f(y)$ if $x > y$
(e) Prove that $f(x) =x$ for all $x$. (Hint: Y=Use the fact that between any two numbers there is a rational number.)

This is problem 17 from chapter 3 of Spivak's Calculus.

2. Relevant equations
Definition of a function, equality of two numbers

3. The attempt at a solution

(a) Now $f(x) = f(x\cdot 1) = f(x)\cdot f(1)$. Now if $f(1)=0$, then $f(x) = 0$ for all $x$. But we are told that this is not the case. So $f(1)\ne 0$. From the second property, $f(1) = f(1)\cdot f(1)$. Since $f(1)\ne 0$, we have $f(1)=1$

(b) First we will prove that $f(x) = x$ if $x \in \mathbb{N}$ using induction. We have already proven the Base case in (a). Now let $P(x)$ be the statement $f(x)=x$. So base case, $P(1)$, is true. Let $k \geqslant 1$ be arbitrary element in $\mathbb{N}$. Suppose $P(k)$ is true. Now $f(k+1) = f(k)+1=k+1$ because $P(k)$ is true. So $P(k+1)$ is true. Since $k \in \mathbb{N}$ is arbitrary, the result is true for all $x \in \mathbb{N}$. Hence $\forall x ~\in \mathbb{N} f(x) =x$. Now let $c$ be an arbitrary rational number. So $c = \frac{m}{n}$, for some $m \in\mathbb{Z}$ and for some $n \in \mathbb{N}$. So $n \ne 0$. Hence $\frac{1}{n} \ne 0$. Now $f(1) = f(n \cdot \frac{1}{n}) = f(n)\cdot f(\frac{1}{n})$. This leads to $f(\frac{1}{n}) = \frac{1}{f(n)} = \frac{1}{n}$, since $n \in \mathbb{N}$. Now since $m \in\mathbb{Z}$, we can have $m$ to be nonpositive. There might be cases where $m \leqslant 0$. Now $f(0) = f(0+0) = f(0) + f(0)$, so $f(0) = 0$. So if $m=0$, $f(m)=m$. Next if $m<0$, then $-m >0$ and $-m \in \mathbb{N}$. Using our earlier result for natural numbers, $f(-m)=-m$. Now $0 =f(0) = f(m+(-m)) = f(m) + f(-m)$, hence $f(-m) = -f(m)$. This means that $-m = f(-m) = -f(m)$. So if $m<0$, we have $f(m)=m$. So we proved that in cases where $m\leqslant 0$, we have $f(m) = m$. Now since $c=\frac{m}{n}$, we have $f(c) = f(m\cdot \frac{1}{n}) = f(m)\cdot f(\frac{1}{n})$. Using the results we have proven for $m$ and $n$, we get $f(c) = \frac{m}{n}=c$. So proved that if $x$ is a rational number, $f(x)=x$

(c) Suppose $x>0$. Now $\sqrt{x}$ is the unique positive square root of $x$, such that $x = (\sqrt{x})^2$. Hence $f(x) = f((\sqrt{x})\cdot (\sqrt{x})) =\left[ f(\sqrt{x})\right]^2$. But $f(\sqrt{x}) \in \mathbb{R}$, so $\left[ f(\sqrt{x})\right]^2 \geqslant 0$. This means that $f(x)\geqslant 0$. Now we will prove that if $f(x) = 0$, we reach a contradiction. Suppose $f(x)=0$. Since $x>0$,we have $\frac{1}{x} >0$. Now $1=f(1) = f(x\cdot \frac{1}{x}) = f(x)\cdot f(\frac{1}{x})$. Now since $f(x)=0$, this leads to $1=0$, which is a contradiction. So $f(x)\ne 0$ and since $f(x)\geqslant 0$, we have $f(x) > 0$ if $x>0$.

(d) Suppose $x>y$. Then $(x-y) > 0$. Using the previous result, we have $f(x-y) > 0$. Now $f(x-y) = f(x) + f(-y)$. Now $0=f(0) = f(y + (-y)) = f(y) + f(-y)$. Hence $f(-y) = -f(y)$ and $f(x-y) = f(x)-f(y)$. As $f(x-y) > 0$, this leads us to $f(x) > f(y)$.

(e) Let $x$ be arbitrary. Let $\varepsilon >0$ be arbitrary. There exists a rational number $q$ in $(x, x+\varepsilon)$. Now $x < q < x + \varepsilon$. Using the result in (d) and (b), we have $f(x) < q < f(x) + f(\varepsilon)$. Hence we have $f(x) - x < q-x$. But $q < x + \varepsilon$, this means that $q-x < \varepsilon$. So we have $f(x) - x < \varepsilon$. Similarly, there exists a rational number $p$ in $(x-\varepsilon, x)$. So we have $x-\varepsilon < p < x$. Again using the results of (d) and (b), we have $f(x) - f(\varepsilon) < p < f(x)$. So we have, $p - x < f(x) -x$. But we have $-\varepsilon < p-x$. So this leads to $-\varepsilon < f(x) - x$. Combining the two inequalities, we get $-\varepsilon < f(x) - x < \varepsilon$. This is same as $|f(x) - x| < \varepsilon$. Since $\varepsilon >0$ is arbitrary, this means that $f(x)=x$. And since $x$ is arbitrary, the result is true for all $x$.

Does this sound a good proof ? It took me 2 days to think through all of this. But solving Spivak's problems is rewarding exercise. But he assumes some knowledge of analysis it seems. For example, I had to use the definition of equality of two numbers in part (e), which is given in some another Analysis book I have. Spivak doesn't mention this in first 3 chapters.

2. Mar 20, 2017

haruspex

Looks fine, but I'm not sure what you mean by
Seems to me you used the axiom of completeness, i.e. that the reals include their own limit points.

3. Mar 20, 2017

issacnewton

haruspex, This definition was given in some other Analysis book. Is it OK to use it here ? And is the proof correct ?

4. Mar 20, 2017

Buffu

Actually if you can prove (e) without using (d), (b) then you can easily prove rest of them.

for eg :-
(d) :-
$$\lim_{h\to 0} {f(x+h) - f(x)\over h} = \lim_{h\to 0} {f(h) \over h} = \lim_{h\to 0} {h\over h} > 0$$

Therefore the function is increasing.

But this is a circular argument.

5. Mar 20, 2017

Buffu

--I may be wrong but $f(x) = 0$ satisfy both conditions but does contradict the rest of the question.--

EDIT:
Did not read the rest of the question. Sorry.

6. Mar 20, 2017

haruspex

I'm still unclear what that definition is and where exactly you used it.

7. Mar 20, 2017

issacnewton

haruspex, Two numbers , $a$ and $b$ are equal iff $\forall \varepsilon > 0$, $|a-b| < \varepsilon$. This is the definition I am talking about and I used it to prove that $f(x) = x$ for arbitrary $x \in \mathbb{R}$

8. Mar 20, 2017

haruspex

Ok. I find that rather strange as a definition. The axiomatic system with which I am familiar builds up to the reals, first using the axioms of an Abelian group for 0, addition and subtraction, then bringing in multiplication etc. If a≠b then a-b is nonzero, (a-b)/2 is nonzero, and we can construct an ε>0 s.t. |a-b|>ε. So if what you quote is a definition in some system then it must be in place of one of the axioms I know.

Anyway, having solved part d, I think there is an easier way using the hint. If f(x)≠x then between them .....

9. Mar 21, 2017

issacnewton

Ok, I now see the use of that hint. Suppose $\exists ~x$ such that $f(x)\ne x$. Then $f(x) > x$ or $f(x)<x$. If $f(x) < x$, then there is a rational number $c$ such that $f(x) < c < x$. Using result from (d), we have $f(c) < f(x)$. But using the result of (b), $f(c)=c$, so we have $c < f(x)$. Hence we have $f(x) < c < f(x)$, which is a contradiction. Similarly, we reach contradiction for $f(x) > x$. Hence, $\forall x ~ f(x) = x$.

I found the exact theorem I was using to show the equality of $f(x)$ and $x$. Theorem :- If $a\in\mathbb{R}$ is such that $0 \le a < \varepsilon$ for every $\varepsilon >0$, then $a=0$. Now I let $a=|f(x)-x|$ in this theorem. I concluded that $a = |f(x)-x| = 0$, which means that $f(x) = x$. I hope that my use of this theorem is correct.

10. Mar 21, 2017

haruspex

Ok, so it is a theorem, not a definition. That makes more sense.