Can You Solve This Challenging Functional Equation?

[stevendaryl]

Okay, so we almost have come to the conclusion that $f(x) = x$ for all $x$, but not quite. Instead, we have the facts that:

1. $f(p) = p$ for all rational numbers $p$
2. $f(ax + by) = af(x) + bf(y)$ for all rational numbers $a$ and $b$
3. $f(f(x)) = f(x)$

These are all consistent with $f(x) = x$, but I don't see that they imply it, either. Is it possible, for example, to have 1-3 hold and to also have $f(\pi) = 1$?

 

[anuttarasammyak]

Hi.

Okay, so we almost have come to the conclusion that f(x)=x for all x, but not quite.

Another conclusion is f(x)=0 for all x, Anyway I will follow you.

Instead, we have the facts that:

1. $f(p) = p$ for all rational numbers $p$
2. $f(ax + by) = af(x) + bf(y)$ for all rational numbers $a$ and $b$
3. $f(f(x)) = f(x)$

I have questions on 1. and 2.
As for 2. first, $f(x)+f(y)=f(x+y)$ that William Crawford identified tells $f(nx+my)=nf(x)+mf(y)$ for integer n,m. How would I develop it so that they are rational numbers ?
2. and $f(1)=1$ gives 1. How can I get $f(1)=1$?

 

[stevendaryl]

Another conclusion is f(x)=0 for all x, Anyway I will follow you.

If you look at the original functional equation, $f(x) = x$ is a solution.

I have questions on 1. and 2.
As for 2. first, $f(x)+f(y)=f(x+y)$ that William Crawford identified tells $f(nx+my)=nf(x)+mf(y)$ for integer n,m. How would I develop it so that they are rational numbers ?

Well, $f(m * x/m) = m * f(x/m)$. So $f(x/m) = 1/m f(x)$.

2. and $f(1)=1$ gives 1. How can I get $f(1)=1$?

I'm saying that that is one solution, $f(x) = x$. I'm asking if there are other solutions besides that one (and the trivial one, $f(x) = 0$).

 

[anuttarasammyak]

Thanks. Now I understand :

The relation $f(x+y)=f(x)+f(y)$ gives $f(ax+by)=af(x)+bf(y)$ for rational numbers a,b, not for irrational numbers. We add $f(1)=1$ to the conditions of the solution. These two give $f(x)=x$ for all rational numbers x.

I assume @stevendaryl suggests this f(x) for rational number x together with a different definition of f(x) for irrational number x would satisfy the original equation in post #1.

From the relation $f(x^2)=xf(x)$
f(2)=\sqrt{2} f(\sqrt{2})=2,
f(\sqrt{2})=\sqrt{2}
Similarly
f(a\sqrt{2}+b\sqrt{3})=a\sqrt{2}+b\sqrt{3} for rational numbers a,b. We observe f(x)=x for some irrational numbers.

Inputting x=1 into the equation in post #1, as f(1)=1, we get f(y)=y for any real number y including irrational numbers in spite of the above expectation.

 

[William Crawford]

The relation f(x+y)=f(x)+f(y) gives f(ax+by)=af(x)+bf(y) for rational numbers a,b, not for irrational numbers.

Let $c$ be a rational number (possibly irrational), then there exist a sequence $\{q_n\}_{n\in\mathbb{N}}$ of purely rational numbers that converges to $c$. Therefore (assuming $f$ is continuous)
$$
\begin{align*}
f(cx) &= \lim_{n\rightarrow\infty}f(q_nx) \\
&= \lim_{n\rightarrow\infty}q_nf(x) \\
&= cf(x)
\end{align*}
$$
and thus $f(ax+by) = af(x) + bf(y)$ is true even when $a$ and $b$ are irrational.

 

[stevendaryl]

Let $c$ be a rational number (possibly irrational), then there exist a sequence $\{q_n\}_{n\in\mathbb{N}}$ of purely rational numbers that converges to $c$. Therefore (assuming $f$ is continuous)
$$
\begin{align*}
f(cx) &= \lim_{n\rightarrow\infty}f(q_nx) \\
&= \lim_{n\rightarrow\infty}q_nf(x) \\
&= cf(x)
\end{align*}
$$
and thus $f(ax+by) = af(x) + bf(y)$ is true even when $a$ and $b$ are irrational.

I guess I didn’t make it explicit, but I was asking about the case where we drop the assumption of continuity.

 

[haruspex]

I guess I didn’t make it explicit, but I was asking about the case where we drop the assumption of continuity.

If we plug f(x)+f(y)=f(x+y) back into the original equation,
$f(xf(y))+f(f(x))+yf(y)=f(x)+yf(x)+yf(y)$
$f(xf(y))=yf(x)$
We know that f(1) is either 0 or 1. If it is 1, setting x=1:
$f(f(y))=y$
But we already know f(f(y))=f(y), so f(y)=y.
If f(1)=0, f(f(y))=0, so f(y)=0.

So either f(x)=x for all x or f(x)=0 for all x,