Can You Solve This Challenging Functional Equation?

[LCSphysicist]

Homework Statement

"Find all functions f: R in R such that x and y belong to R" (R is obviously real space) (continuation in the blank frame

Relevant Equations

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$$f(xf(y) + f(x)) + f(y^2) = f(x) + yf(x + y)$$

A tricky question, i think.
First fact i found was:
f(f(0)) = 0
So i separate it in two types of functions
f(0) = 0 and f(0) = u.

I was trying to analyzing both cases, with the cases where x = y and x = -y but is is rather extended way, so i believe there is a better attempt to solve the question
Anyway i am not sure if we have a solution with explicit functions (f(x) = x²) or if the answer end to be something like (f(x) + f(y) + ...)

Any tips?

@fresh_42

 

[anuttarasammyak]

I see an obvious example of solution
$$f(x)=x$$.

 

[LCSphysicist]

I see an obvious example of solution
$$f(x)=x$$.

And how do you prove this is the only function possible?

 

[anuttarasammyak]

I see another solution $$f(x)=0$$.

There may be more but I am not sure. How about writing f(x) in Taylor expansion form and taking a look at the coefficients of the same $x^n y^m$ term ,i.e
$$f(x)=f(0)+xf'(0)+...$$
Hence we know $f(0)=0, f'(0)=0,1$, f"(0)=0,...

 

[member 587159]

I see another solution $$f(x)=0$$.

There may be more but I am not sure. How about writing f(x) in Taylor expansion form and taking a look at the coefficients of the same $x^n y^m$ term ,i.e
$$f(x)=f(0)+xf'(0)+...$$
Hence we know $f(0)=0, f'(0)=0,1$, f"(0)=0,...

Who says a solution must be differentiable?

 

[LCSphysicist]

I see another solution $$f(x)=0$$.

There may be more but I am not sure. How about writing f(x) in Taylor expansion form and taking a look at the coefficients of the same $x^n y^m$ term ,i.e
$$f(x)=f(0)+xf'(0)+...$$
Hence we know $f(0)=0, f'(0)=0,1$, f"(0)=0,...

I imagined in open in Taylor, but i believe it would be so troublesome, we have not just x, we have f as function of y and other variables too.

I was thinking if the only possible answer happens to be just a lot of guess, i would be disappointed

. Maybe @PeroK can think in something to help us.

 

[PeroK]

Maybe @PeroK can think in something to help us.

Not immediately!

 

[haruspex]

So i separate it in two types of functions
f(0) = 0 and f(0) = u.

Follow that path. What general equation does x=0 give you? What about y=0 instead?

 

[LCSphysicist]

Follow that path. What general equation does x=0 give you? What about y=0 instead?

Yes, i tried a lot and i found that the only functions which satisfy all the conditions is $f(\gamma) = \gamma$
I checked and confirmed that the angular coefficiente need to be 1, and can not be any linear coefficient.
@anuttarasammyak made a nice guess #2.

 

[haruspex]

i found that the only functions which satisfy all the conditions is $f(\gamma) = \gamma$

Are you saying you have proved that, or just that it is the only solution you have found?

 

[anuttarasammyak]

Say x=0
$$f(f(0))+f(y^2)=f(0)+yf(y)$$.
Say y=0
$$f(xf(0)+f(x))=f(x)$$,further say x=0
$$f(f(0))=f(0)$$
So
$$f(y^2)=yf(y)$$
f(0)=0. For $y \neq 0$
$$\frac{f(y^2)}{y^2}=\frac{f(y)}{y}=c$$
We can easily know allowed values of c by inputting f(y)=cy, which satisfy f(0)=0, to the original equation or the above second equation.

EDIT
I add on the features
$$f(f(x))=f(x)$$
which shows nature of projection.

 

[stevendaryl]

Say x=0
$$f(f(0))+f(y^2)=f(0)+yf(y)$$.
Say y=0
$$f(xf(0)+f(x))=f(x)$$,further say x=0
$$f(f(0))=f(0)$$
So
$$f(y^2)=yf(y)$$
f(0)=0. For $y \neq 0$
$$\frac{f(y^2)}{y^2}=\frac{f(y)}{y}=c$$
We can easily know allowed values of c by inputting f(y)=cy, which satisfy f(0)=0, to the original equation or the above second equation.

From
$$\frac{f(y^2)}{y^2}=\frac{f(y)}{y}$$

we can't conclude that both sides are constant, can we? We can define a new function
$$g(x) = f(x)/x$$

Then we have:
$$g(x) = g(x^2)$$

Does that imply that g is a constant function?

 

[anuttarasammyak]

Hi.
$$\frac{f(x^2)}{x^2}=\frac{f(x)}{x},\ f(0)=0$$
So f(x) is odd function of x. Let us expand it in series
$$f(x)=\sum_{n=0}^{+\infty}a_{2n+1}\ x^{2n+1}$$
The above relation shows
$$\sum_{n=0}^{+\infty}a_{2n+1}\ (x^{4n}-x^{2n})=0$$
All the coefficients except $a_1$ are zero.

 

[haruspex]

Hi.
$$\frac{f(x^2)}{x^2}=\frac{f(x)}{x},\ f(0)=0$$
So f(x) is odd function of x. Let us expand it in series
$$f(x)=\sum_{n=0}^{+\infty}a_{2n+1}\ x^{2n+1}$$
The above relation shows
$$\sum_{n=0}^{+\infty}a_{2n+1}\ (x^{4n}-x^{2n})=0$$
All the coefficients except $a_1$ are zero.

As has been pointed out, this approach is utterly invalid. We are not given that f is continuous, let alone differentiable.

 

[anuttarasammyak]

We are not given that f is continuous, let alone differentiable.

Thanks. I admit my proof #13 works only if f(x) can be expressed in series.

For example
f(1)=1, f(-1)=-1, f(x)=0 otherwise
is a solution.

 

[haruspex]

f(1)=1, f(-1)=-1, f(x)=0 otherwise
is a solution.

Are you sure? What about x=1, y=-2?

 

[anuttarasammyak]

Thanks. #15 is a solution of equation f(x)/x=f(x^2)/x^2, f(x)=0. I see it does not satisfy the equation of post #1 , as you pointed out. As I should find the allowed values of c in post #11, not all the solutions of the equation f(x)/x=f(x^2)/x^2, f(x)=0 are the solution of the equation of post #1.

Now I am much interested to know whether we have more solutions other than analytic f(x)=cx, c={0,1}.

 

[haruspex]

Thanks. #15 is a solution of equation f(x)/x=f(x^2)/x^2, f(x)=0. I see it does not satisfy the equation of post #1 , as you pointed out. As I should find the allowed values of c in post #11, not all the solutions of the equation f(x)/x=f(x^2)/x^2, f(x)=0 are the solution of the equation of post #1.

Now I am much interested to know whether we have more solutions other than analytic f(x)=cx, c={0,1}.

So far I have only managed to show
f(0) = 0
f(f(x)) = f(x)
f(x²) = x f(x)
If f(u) = u then f(nu) = nu for all integers n.

Edit:
Oh, and I forgot
f(-x)=-f(x)

 

[William Crawford]

So far I have only managed to show
f(0) = 0
f(f(x)) = f(x)
f(x²) = x f(x)
If f(u) = u then f(nu) = nu for all integers n.

That is also tremendous progress. This allow you to simplify the functional equation to $$xf(y) + yf(y) = yf(x+y)$$

 

[anuttarasammyak]

Re:#19
Your result toghether with the equation suggest the relation
$$f(xf(y)+f(x))=xf(y)+f(x)$$
I am not sure it stands.

 

[haruspex]

That is also tremendous progress. This allow you to simplify the functional equation to $$xf(y) + yf(y) = yf(x+y)$$

If so, doesn't it immediately follow that yf(x )=xf(y)?

 

[William Crawford]

Re:#19
Your result toghether with the equation suggest the relation
$$f(xf(y)+f(x))=xf(y)+f(x)$$
I am not sure it stands.

No, you are right. What I wrote in 19 wasn't correct.

 

[William Crawford]

From $f(x^2)=xf(x)$ conclude that $f$ is an odd function i.e. $f(-x)=-f(x)$. If we take $(x,y) = (-x,y)$ then we get that
$$ f(-xf(y)+f(-x)) + yf(y) = f(-x) + yf(y-x) $$
or, by using the odd property of $f$, that
$$ -f(xf(y)+f(x)) + yf(y) = -f(x) + yf(y-x). $$
Now, add this equation with the original functional equation, to get that
$$ 2f(x) = f(x+y) + f(x-y).$$
Next, use that $f(x+y)$ is symmetric under $(x,y)\mapsto(y,x)$ to derive that
$$ f(x)-f(y) = f(x-y)$$
or equivalently that
$$ f(x)+f(y) = f(x+y).$$
This is nothing more than Cauchy's functional equation. Thus, if it is assumed that $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous, then $f(x) = ax$ for $a\in\mathbb{R}$ are all the solutions.

 

[anuttarasammyak]

Excellent! but
$$f(f(x))=f(x)$$
puts us
$$a^2=a$$?

 

[William Crawford]

Excellent! but
$$f(f(x))=f(x)$$
puts us
$$a^2=a$$?

True, however that just imply that either $a=1$ or $a=0$.

 

[anuttarasammyak]

Thanks. That meet with my #17. Can you tell me if other non analytic solutions exist?

 

[William Crawford]

I believe that $f(x) = ax$ for $a\in\{0,1\}$ are the entire family of continuous solutions (consisting of elementary functions) to the original functional equation. As any continuous solution also has to be a solution to the Cauchy's functional equation. However there exist non-continuous solutions to the Cauchy's functional equation and there might therefore also exist non-continuous solutions to the original functional equation.

 

[anuttarasammyak]

If f(u) = u then f(nu) = nu for all integers n.

f(x)+f(y)=f(x+y) tells us f(nx)=f(x)+f((n-1)x)=... = n f(x)

Is "If f(u) = u" is necessary to say f(nu) = nf(u) and why n is limited to integer not any real number ?

 

[William Crawford]

Thanks. That meet with my #17. Can you tell me if other non analytic solutions exist?

The function $f(x) = \mathrm{max}(x,0)$ (i.e. the positive part) is also a solution to Cauchy as well and though it is continuous it isn't differentiable at $x=0$ and thus not analytic. Check for yourself if also satisfy the original functional equation.

 

[anuttarasammyak]

Hi. Say f(x)=max(x,0)
$$f(x)+f(-x)=|x|$$and

$$f(x)+f(-x)=0$$
seem incompatible.

 

[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.