Easiest Functional Equation

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Easiest Functional Equation

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#1 h Mar 2, 2025, 4:10 PM • 1 Y

Y by MathLuis

Find all functions $f: \mathbb R \to \mathbb R$ such that $f(yf(x)+f(xy))=(x+f(x))f(y)$ Follows for all reals $x,y$ .

This post has been edited 1 time. Last edited by NCbutAN, Mar 2, 2025, 4:13 PM
Reason: .

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#2 h Mar 2, 2025, 4:44 PM • 1 Y

Y by Yihangzh

mb chat I goofed

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#3 h Mar 2, 2025, 4:53 PM

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InftyByond wrote:

literally plugging y=0 solves the problem immediately

How ?
If $f(0)=0$ plugging $y=0$ solves nothing

In my opinion

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#4 h Mar 2, 2025, 4:54 PM

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Plugging $y=0$ just gives $f(0) = 0$ . It doesn't immediately solve the problem but it looks like it helps.

Edit: After thinking further, I don't actually think it helps.

@below

Also, consider that $f(x) = 0$ and $f(x) = x$ are answers to the equation.

This post has been edited 4 times. Last edited by maromex, Mar 2, 2025, 6:39 PM

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#5 h Mar 2, 2025, 6:24 PM

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but plugging $y=0$ gives $f(x)= -x$ , is it right

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#6 h Mar 3, 2025, 1:19 AM • 3 Y

Y by maromex, scannose, KevinYang2.71

We claim the only solutions are $f\equiv 0$ and $f(x) = x$ . These work. Now we show they are the only ones. Suppose $f$ was not identically zero or the identity.

Claim: $f$ is nonlinear.
Proof: Suppose $f$ was linear.

Setting $y = 0$ gives $f(f(0)) = (x + f(x)) f(0)$ . If $f(0) \ne 0$ , then $f(x) + x$ is a constant, implying $f$ is an involution, so $f(f(0)) = 0$ , after which $P(0,0)$ gives $f(0) = 0$ , absurd. Hence $f(0)= 0$ . Let $f(x) = kx$ .

$P(1,1): f(2k) = k(k+1)$ , so $2k^2 = k^2 + k$ , which implies $k = k^2 \implies k \in \{0,1\}$ . $\square$

Let $P(x,y)$ be the given assertion and $g(x) = x + f(x)$ .

$P(x,0)$ gives that $f(0) = 0$ .

Claim: $f(1) \ne 0$ .
Proof: Suppose $f(1) = 0$ .

Setting $y = 1$ gives $f(2f(x)) = 0$ for all reals $x$ .

Setting $x =1$ gives $f(f(x)) = f(x)$ for all reals $x$ .

$P(f(x),2): 0 = 2f(x) f(2)$ . Since $f$ isn't linear, we also have $f(2) = 0$ .

$P(2, f(x)): 0 = 2f(f(x)) = 2f(x)$ , so $f$ is identically zero, a contradiction. $\square$

Claim: $f$ is injective.
Proof: If $f(c) = f(d)$ , then $P(c,1)$ compared with $P(d,1)$ gives that $(c + f(c)) f(1) = (d + f(d)) f(1)$ , so $c = d$ . $\square$

Claim: $g$ is injective.
Proof: Suppose for some reals $a,b$ , we have $g(a) = g(b)$ .

$P(a,1)$ compared with $P(b,1)$ gives $f(2f(a)) = f(2f(b))$ , so $2f(a) = 2f(b) \implies f(a) = f(b)\implies a= b$ . $\square$

Now, we note that $f(y (g(x) - x) + g( xy) - xy) = g(x) (g(y) - y)$ , so adding both sides by $y(g(x) - x) + g(xy) - xy$ gives that $g(y(g(x) - x) + g(xy) - xy) = g(x) (g(y) - y) + y(g(x) - x) + g(xy) - xy = g(x) g(y) + g(xy) - 2xy \ \ \ \ \ (1)$
Swapping $x,y$ and comparing gives $g(y(g(x) - x) + g(xy) - xy) = g(x(g(y) - y) + g(xy) - xy))$ , so $yf(x) = x f(y)$ , which implies for $x,y\ne 0$ that $\frac{f(x)}{x} = \frac{f(y)}{y}$ , so $\frac{f(x)}{x}$ is constant over reals $x\ne 0$ . Since $f(0) = 0$ , we have $f(x) = cx$ for some constant $c$ and all reals $x$ , a contradiction because $f$ is nonlinear.

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#7 h Mar 4, 2025, 1:52 AM • 2 Y

Y by megarnie, NCbutAN

The only solutions are $\boxed{f(x)=0}$ and $\boxed{f(x)=x}$ . These both work.

Let $P(x,\,y)$ denote the assertion. From $P(0,\,0)$ and $P(x,\,0)$ we get $f(f(0))=f(0)^2=f(0)(x+f(x))$ . If $f(0)\neq 0$ , we have $f(x)=f(0)-x$ . Plugging this into $P$ yields $f(0)^2=2xy$ , which is a contradiction for $x=y=0$ . Hence $f(0)=0$ .

Suppose $f(1)\neq 0$ . Then $x=\frac{f(2f(x))}{f(1)}-f(x)$ so $f$ is injective. $P(x,\,1)$ gives $x=f^{-1}\left(\frac{1}{2}f^{-1}(f(1)(x+f(x)))\right)$ so $x\mapsto x+f(x)$ is also injective. Comparing $P(x,\,y)$ and $P(y,\,x)$ gives
$yf(x)+f(xy)+f(yf(x)+f(xy))=xf(y)+yf(x)+f(xy)+f(x)f(y)=xf(y)+f(xy)+f(xf(y)+f(xy))$ so $xf(y)=yf(x)$ by injectivity. Thus $f(x)=cx$ for $c:=f(1)$ . A simple calculation shows that $c\in\{0,\,1\}$ .

Suppose now $f(1)=0$ . We claim that $f(x)=0$ . Assume FTSOC there exists $a$ such that $b:=f(a)\neq 0$ . $P(1,\,x)$ gives $f(f(x))=f(x)$ so $f(b)=b$ . $P(x,\,1)$ gives $f(2f(x))=0$ so $f(2b)=0$ . Then $P(2b,\,x)$ yields $f(2bx)=2bf(x)$ . From $P\left(b,\,\frac{1}{b}\right)$ we get $0=2bf\left(\frac{1}{b}\right)$ so $f\left(\frac{1}{b}\right)=0$ . Then $f(2)=f\left(2b\cdot\frac{1}{b}\right)=0$ so $P(2,\,x)$ gives $f(2x)=2f(x)$ . However, $0=f(2b)=2f(b)=2b$ , a contradiction. $\square$

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#8 h Mar 16, 2025, 7:12 AM

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may just be joever for me I misread the problem statement
this is about as easy as it is for me to not silly (impossible)

This post has been edited 1 time. Last edited by InftyByond, Mar 16, 2025, 8:15 AM
Reason: AAAAAAAAAA IM SORRY PCO IM SORRY EVERYONE

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