Functional equation in reals

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#1 h Mar 31, 2019, 5:59 PM • 3 Y

Y by tiendung2006, Adventure10, Mango247

Find all functions $f:\mathbb R\to\mathbb R$ such that
$4xyf(x^2-y^2)=(x^2-y^2)f(2x)f(2y)$ for all real $x$ and $y$ .

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#2 h Mar 31, 2019, 9:20 PM • 2 Y

Y by Adventure10, Mango247

By $P(1,1),$ we obtain that $f(0) = 0.$

If $f(2y) = 0,$ for some $y \neq 0,$ then $P(x, y)$ gives that $f(x^2 - y^2) = 0$ for all $x \neq 0.$ As a result, we know that $f(a) = 0$ for all $a > -y^2.$

But then $P(x, a)$ for some huge positive $a$ gives that $f(b) = 0$ for all $b > -a^2,$ and so it's easy to see that we can get that $f \equiv 0$ from this, which works.

Otherwise, suppose that $f(y) \neq 0,$ for all $y \neq 0.$

Write $S = \mathbb{R}/\{0\}.$

Let's define the function $g: S \rightarrow S$ by $g(r) = \frac{f(r)}{r},$ for $r \in S.$

Then, the condition rewrites as $g(x^2 - y^2) = g(2x)g(2y),$ for all $x, y \in S$ with $x^2 \neq y^2.$

As a result of the above, whenever we have $a^2 + c^2 = 2b^2$ for $a^2, b^2, c^2$ distinct, we know that $g(2a)g(2b) = g(a^2 - b^2) = g(b^2 - c^2) = g(2b)g(2c),$ so in view of $g(2b) = 0$ we have $g(2a) = g(2c).$

Notice that this works whenever $a, c \in S$ with $a^2 \neq c^2,$ and so this clearly is enough to show that $g$ is constant.

Hence, by the definition of $g,$ we know that $f(x) \equiv cx, \forall x \in \mathbb{R},$ and plugging this back shows that $f(x) \equiv 0, x$ are the only two solutions.

These two clearly both work, so we're done.

$\square$

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pco

#3 h Apr 1, 2019, 8:30 AM • 2 Y

Y by Pluto1708, Adventure10

Vlados021 wrote:

Find all functions $f:\mathbb R\to\mathbb R$ such that
$4xyf(x^2-y^2)=(x^2-y^2)f(2x)f(2y)$ for all real $x$ and $y$ .

Let $P(x,y)$ be the assertion $4xyf(x^2-y^2)=(x^2-y^2)f(2x)f(2y)$

$P(1,1)$ $\implies$ $f(0)=0$

If $f(u)=0$ for some $u\ne 0$ :
Let $x>0$ :
$P(x,\frac u2)$ $\implies$ $f(x^2-\frac{u^2}4)=0$ and so $f(x)=0$ $\forall x\ge 0$
$P(\frac u2,x)$ $\implies$ $f(\frac{u^2}4-x^2)=0$ and so $f(x)=0$ $\forall x\le 0$
And so $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ which indeed is a solution

So let us from now consider $f(x)\ne 0$ $\forall x\ne 0$
Let $x\ne 0$
$P(x,\sqrt{x^2+1}-1)$ $\implies$ $f(2x)=2x$ $\forall x\ne 0$ , still true when $x=0$

And so $\boxed{\text{S2 : }f(x)=x\quad\forall x}$ which indeed is a solution

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#4 h Oct 17, 2022, 1:51 PM

Y by

Let $P(x,y)$ denote the assertion $4xyf(x^2-y^2)=(x^2-y^2)f(2x)f(2y).$ Clearly $f(0)=0.$ If $f(x_0)=0$ for $x_0\neq 0$ then, $P(x,x_0)$ and $P(x_0,x)$ gives $f\equiv 0.$ So $f(x)=0$ iff $x=0.$ Then $P(\sqrt{x^2+1}+1,x)$ shows $f\equiv \text{Id},$ which fits.

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maolus

#7 h Oct 21, 2022, 6:27 AM

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$f(x) \equiv 0$ is a solution. Consider $f(x) \neq 0, \forall x \in \mathbb{R}$
It is obvious that: $f(0) = 0$ and $f(-x) = -f(x) \forall x \in \mathbb{R}$

Let $x \leq 0 \Rightarrow x^2 - 2x \geq 0$ .

$P(x, \sqrt{x^2 - 2x}) \Rightarrow 4x.\sqrt{x^2-2x}.f(2x) = 2x.f(2x).f(2\sqrt{x^2-2x}) \Rightarrow 2.\sqrt{x^2-2x}=f(2.\sqrt{x^2-2x}), \forall x \leq 0$

also, $2.\sqrt{x^2-2x}$ receive all value in $\mathbb{R+} \Rightarrow f(x) = x, \forall x \in \mathbb{R+}$ .
According to $f(x) = -f(-x)$ for all $x \in \mathbb{R}$ and $f(0) = 0 \Rightarrow f(x) = x ,\forall x \in \mathbb{R}$

So, $f(x) = 0 \forall x \in \mathbb{R}$ and $f(x) = x \forall x \in \mathbb{R}$ are solutions

Sorry for my worst English. Thanks for reading

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