Functional Equation

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#1 h Thursday at 3:57 PM

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Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying: $f(xf(y)+2y)=f(f(y))+f(xy)+xf(y), \forall x, y \in \mathbb{R}$

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#2 h Thursday at 4:37 PM

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f is injective .Hint:P(2y/(y-f(y)),y)

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#3 h Thursday at 5:02 PM

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and finally P(0,x) we get f(2x)=ff(x)-->(injectivity)--->f(x)=2x

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#4 h Thursday at 5:04 PM

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AnhQuang_67 wrote:

Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying: $f(xf(y)+2y)=f(f(y))+f(xy)+xf(y), \forall x, y \in \mathbb{R}$

The only constant solution is $\boxed{\text{S1 : }f(x)=0\quad\forall x\in\mathbb R}$ . So let us from now look only for nonconstant solutions.
Let $P(x,y)$ be the assertion $f(xf(y)+2y)=f(f(y))+f(xy)+xf(y)$
Let $a=f(0)$

1) $f(x)=0$ $\iff$ $x=0$
Proof

2) $f(x)$ is injective
Proof

$P(0,x)$ $\implies$ $f(2x)=f(f(x))$ and injectivity implies $\boxed{\text{S2 : }f(x)=2x\quad\forall x\in\mathbb R}$ , which indeed fits

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#5 h Thursday at 5:06 PM

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MuradSafarli wrote:

f is injective

No.
Counter-example : $f\equiv 0$

MuradSafarli wrote:

Hint:P(2y/(y-f(y)),y)

And what if $f(y)=y$ for some $y$ ?

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#6 h Thursday at 5:35 PM

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The only solutions are $f\equiv 0$ and $f(x) = 2x$ , which clearly work. Now we show they're the only ones.

Let $P(x,y)$ denote the given assertion. Assume that $f$ isn't identically zero. We prove that $f(x) = 2x$ .

$P(0,0)$ compared with $P(1,0)$ gives $f(0) = 0$ .

$P(0,x): f(2x) = f(f(x))$ .

Claim: For all $y$ with $f(y) \ne y$ , we have $(y - f(y)) f(f(y)) + 2yf(y) = 0$
Proof: Setting $x = \frac{2y}{y - f(y)}$ gives $f(f(y)) + x f(y) = 0$ , so $f(f(y))+ \frac{2yf(y)}{y - f(y)} = 0$ , after which multiplying both sides by $y - f(y)$ gives the desired result. $\square$

Hence, for $y$ with $f(y) \ne y$ , we actually have $y(f(f(y)) + 2f(y)) - f(y) f(f(y)) = 0 \ \ \ \ \ (1)$
Claim: $f$ is injective at $0$ .
Proof: If $f(k) = 0$ for some $k \ne 0$ , $P(0,k)$ gives $f(2k) = 0$ .

Now, $P(x,k): f(2k) = f(xk)$ , so $f(xk) = 0$ , meaning $f$ is constant at $0$ , a contradiction. $\square$

Claim: $f(f(y)) + 2f(y) \ne 0$ for all $y \ne 0$ .
Proof: If $f(f(y)) + 2f(y) = 0$ , then $(1)$ gives $f(y) f(f(y)) = 0$ , which combined with the previous claim implies $y = 0$ . $\square$

Claim: If reals $a,b$ satisfy $f(a) \ne a$ , $f(b) \ne b$ , and $f(a) = f(b)$ , then $a = b$ .
Proof: Comparing $y = a$ and $y = b$ gives the desired result (because $a,b$ are nonzero and $f(f(a)) + 2f(a) \ne 0$ ). $\square$

Claim: $0$ is the only fixed point of $f$ .
Proof: Suppose some other real $c$ was a fixed point of $f$ .

$f(xc + 2c) = f(xc) + xc + c$ , so $f(x + 2c) = f(x) + x + c$ . Setting $x = 0$ gives $f(2c) = c$ and setting $x = -c$ gives $f(-c) = c$ , contradiction by setting $a = 2c$ and $b = -c$ in the previous claim. $\square$

Claim: $f$ is injective.
Proof: Suppose $f(a) = f(b)$ for some $a\ne b$ . Clearly one of $a,b$ is a fixed point, so $ab= 0$ , but this means $a = b = 0$ since $f$ is injective at $0$ . $\square$

Thus, from $f(2x) = f(f(x))$ , we can conclude $f(x) = 2x$ , as desired.

This post has been edited 1 time. Last edited by megarnie, Thursday at 5:35 PM

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