Find all functions f:R→R, where R is the set of real numbers, such that f(x)f(yf(x)−1)=x2f(y)−f(x)∀x,y∈R
Problem
Source: SMO 2015 open
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31.03.2018 12:53
31.03.2018 12:59
Let P(x,y) be the assertion. First,it is easy to see that f(x)≡0 is a solution. Now,we will assume that f(x)≢. P(1,1)\implies f(1)f(f(1)-1)=0\implies \exists\, a\in\mathbb{R},f(a)=0. P(a,0)\implies a^2f(0)=0\implies f(0)=0. P(x,0)\implies f(x)f(-1)=-f(x)\implies f(-1)=-1. P(-1,-y)\implies f(y-1)=-f(-y)-1^{(*)}. Hence the assertion become P(x,y);f(x)[-f(-yf(x))-1]=x^2f(y)-f(x)\implies \boxed{f(x)f(-yf(x))=-x^2f(y)}. If f(d)=0,then P(d,y)\implies d=0.Hence d\neq 0\implies f(d)\neq 0. Now,for any x\neq 0,we have P(x,x)\implies f(-xf(x))=-x^2.Hence all nonpositive real numbers are in the range of f(x). From the new assertion,it is easy to see that f(a)=f(b)\implies a=\pm b. But if f(a)=f(-a)\,\exists a\neq 0,then (*)\implies f(a-1)=f(-a-1)\implies a-1=\pm (-a-1) which is a contradiction.Hence f(x) is injective. Thus P(x,x)\implies f(-xf(x))=-x^2\,\forall x\neq 0\implies f(-xf(x))=f(xf(-x))=-x^2\implies -f(x)=f(-x)\,\forall x\neq 0. Hence f(x) is surjective and f(y-1)=f(y)-1. Thus the assertion now become P(x,y);\boxed{f(x)f(yf(x))=x^2f(y)}. P(x,1)\implies f(x)f(f(x))=x^2f(1)=-x^2f(-1)=x^2. Hence f(x)f(yf(x))=f(x)f(f(x))f(y)\implies f(yf(x))=f(f(x))f(y)\,\forall x\neq 0\implies f(yf(x))=f(f(x))f(y)\,\forall x,y\in\mathbb{R}. But since f(x) is surjective,we get that f(x) is multiplicative. P(x,y+1)-P(x,y)\implies f(x)[f(yf(x)+f(x))-f(yf(x))]=x^2=f(x)f(f(x))\implies f(z+a)-f(z)=f(a)\,\forall a\neq 0. Now it is easy to see that f(x) is addictive,too.Hence \boxed{f(x)\equiv x}.(Since we have exclude f(x)\equiv 0.) Hence all solution are f(x)\equiv 0 and f(x)\equiv x which clearly satisfied the assertion.
31.03.2018 13:01
31.03.2018 13:25
AnArtist wrote: Claim 2) f(\frac{1}{x})= \frac{f(x)}{x^2} \forall x \neq 0. Proof P( x, \frac{1}{f(x)}). P(x,\frac {1}{f(x)}) gives f(\frac {1}{f(x)})=\frac {f(x)}{x^2}
02.11.2020 08:19
Here is a bit of improvement on the solutions above. If f is non-constant, P(0,y): f(0)=0. If f(c)=0, P(c,y): c=0 P(1,1): f(1)f(f(1)-1)=0 Thus, f(1)=1. P(1,y): f(y+1)=f(y)+1 Hence the original equation becomes f(x)f(yf(x))=x^2f(y) We get f(x)f(f(x))=x^2 and f(x+1)=f(x)+1, we can also do: x^2+2x+1=(x+1)^2=f(x+1)f(f(x+1))=(f(x)+1)(f(f(x))+1)=f(x)f(f(x))+f(x)+f(f(x))+1=x^2+f(x)+f(f(x))+1 And so f(x)+f(f(x))=2x. Lwt f(x)=a, f(f(x))=b, ab=x^2, a+b=2x so b=2x-a \implies a(2x-a)=x^2 \implies (x-a)^2=0 Thus, f(x)=a=x
02.11.2020 08:43
dominicleejun wrote: Find all functions f : \mathbb{R} \rightarrow \mathbb{R}, where \mathbb{R} is the set of real numbers, such that f(x)f(yf(x) - 1) = x^2 f(y) - f(x) \quad\forall x,y \in \mathbb{R} let p(x,y)=f(x)f(yf(x)-1)=x^2f(y)-f(x) clearly p(0,y) either gives f(x)=0 or f(0)=0 so assume f(0)=0 then p(x,0)=f(x)f(-1)=-f(x) so since f(x) is not the zero constent f(-1)=-1 now p(-1,y)=f(-y-1)=-f(y)-1 so f(y-1)=-f(-y)-1 so f(yf(x)-1)=-f(-yf(x))-1 so rewrite as p(x,y)=-f(x)f(-yf(x))=x^2f(y). now p(x,-1)=f(x)f(f(x))=x^2. now if f(x)=0 then p(x,y)=x^2f(y)=0 so x=0. now rewrite as p(x,y)=f(-yf(x))=-f(y)f(f(x)) now if f(a)=f(b) then a^2=b^2 also as we had f(-y-1)=-f(y)-1 we have f(-a-1)=f(-b-1) so (b+1)^2=(a+1)^2 which gives us a=b. now p(x,x)=f(-xf(x))=-x^2=p(-x,-x) so -xf(x)=xf(-x) which gives -f(x)=f(-x) so rewrite equations as f(y+1)=f(y)+1 p(x,y)=f(yf(x))=f(y)f(f(x)) as x \in D_f for x<0 and clearly if a \in D_f then -a \in D_f we have f is surgective. so rewrite as p(x,y)=f(xy)=f(x)f(y) so f((x+1)y)=f(x+1)f(y)=f(x)f(y)+f(y)=f(xy)+f(y) so f(a+b)=f(a)+f(b) . these give f(x)=x for all real x.
11.08.2021 03:02
Let P(x,y) be the assertion f(x)f(yf(x)-1)=x^2f(y)-f(x). P(1,1)\Rightarrow f(1)f(f(1)-1)=0 If f(1)=0, then P(1,x)\Rightarrow\boxed{f(x)=0}. Otherwise, there is a j with f(j)\ne0 and f(f(1)-1)=0. P(f(1)-1,j)\Rightarrow f(1)=1 P(1,x+1)\Rightarrow f(x+1)=f(x)+1 P(x,1)\Rightarrow f(x)f(f(x))=x^2 Taking x\mapsto x+1 in this, we see that f(x)+f(f(x))=2x. Then f(x)(2x-f(x))=x^2, so \boxed{f(x)=x}.
19.06.2022 21:02
As per usual, we use P(x,y). Assume not f\equiv 0 (which works.) We get f(1)(f(1)-1)=0 from P(1,1). Take f(z)=0, then P(z,y) implies z=0. In particular f(1)=1. \begin{align*} P(1,x)\implies &f(x-1)=f(x)-1 \\ P(a,b)\implies &f(a)f(bf(a))=a^2f(b) \\ P(a,1)\implies & f(a)f(f(a))=a^2f(1) \\ P(a+1,1)\implies & f(f(a))=2a-f(a) \\ P(a,1)\implies & f(a)\equiv a \quad \text{which fits.} \end{align*}
20.06.2022 14:53
Let P(x,y) denote the given assertion. P(0,x): f(0)f(xf(0)-1)=-f(0). So either f(0)=0 or f(xf(0)-1)=-1 for any x. If f(0)\ne 0, then xf(0)-1 can take on any real value, which implies f\equiv -1, which is not a solution. Thus, we have f(0)=0. Now, noting \boxed{f\equiv 0} works, we can assume f is non-constant. Claim: f is injective. Proof: Suppose f(a)=f(b) with a\ne b. P(a,x): f(a)f(xf(a)-1)=a^2f(x)-f(a). P(b,x): f(b)f(xf(b)-1)=b^2f(x)-f(b). This implies a^2f(x)=b^2f(x). If we set x such that f(x)\ne 0, then a^2=b^2\implies a=\pm b\implies a=-b, since a\ne b. Then f(a)=f(-a). In fact, this implies f is injective at 0. P(a,-a): f(a)f(-af(a)-1)=a^2f(a)-f(a). P(-a,a): f(a)f(af(a)-1)=a^2f(a)-f(a). If f(a)\ne 0, then we have f(-af(a)-1)=f(af(a)-1). However, this implies either af(a)+1=af(a)-1, or -af(a)-1=af(a)-1, both are absurd. So f(a)\ne f(-a). If f(a)=0, then a=0. Since f is injective at 0, f is injective. \blacksquare P(x,0): f(x)f(-1)=-f(x). If we set x\ne 0, then we get f(-1)=-1. P(1,1): f(1)f(f(1)-1)=0, so f(1)=1. P(1,x): f(x-1)=f(x)-1\implies f(x+1)=f(x)+1. Now we rearrange the FE. We have f(x)(f(yf(x))-1)=x^2f(y)-f(x), sof(x)f(yf(x))=x^2f(y)Let Q(x,y) be the assertion here. Q(x,1): f(x)f(f(x))=x^2\implies f(f(x))=\frac{x^2}{f(x)}. P(x+1,1): (f(x)+1)f(f(x))=(x+1)^2-f(x)-1=x^2+2x-f(x)\implies f(f(x))=\frac{x^2+2x-f(x)}{f(x)+1}. This implies\begin{align*} \frac{x^2}{f(x)}=\frac{x^2+2x-f(x)}{f(x)+1} \\ \implies x^2(f(x)+1)=f(x)(x^2+2x-f(x)) \\ \implies x^2f(x)+x^2=x^2f(x)+2xf(x)-f(x)^2 \\ \implies x^2=2xf(x)-f(x)^2 \\ \implies f(x)^2-2xf(x)+x^2=0 \\ \implies (f(x)-x)^2=0 \\ \implies \boxed{f(x)=x} \\ \end{align*}which clearly works.
02.09.2024 19:28
The answers are f \equiv 0 and f \equiv x , which clearly work. Let P(x, y) denote the assertion. We start by making the following claim: Claim: For all x \in \mathbb{R} , we have f(x)f(f(x)) = x^2. Proof We first note that P(x, 0) gives us f(x)f(-1) = -f(x) \implies f(x)(f(-1) + 1) = 0. So either f(x) = 0 for all x \in \mathbb{R} or f(-1) = -1 . Note that f \equiv 0 is indeed a solution, so assume f(-1) = -1 . Now we observe that: \bullet P(-1, y) \implies f(-y - 1) = -(f(y) + 1) . \bullet P(x, -1) \implies f(x)f(-f(x) - 1) = -x^2 - f(x) . Comparing the first and second bullets, we get that: f(x)[f(f(x)) + 1] = x^2 + f(x) \implies f(x)f(f(x)) = x^2, as claimed. Now, we show that if there exists a z \neq 0 such that f(z) = 0 , then f is zero everywhere. Indeed, by P(z, x) , z^2f(x) = 0 \implies f(x) = 0, as needed. Now P(0, x) gives us that f(0)f(xf(0) - 1) = -f(0) \implies f(0)[f(f(0)x - 1) + 1] = 0. If f(0) \neq 0 , then f(x) = -1 for all x , which is not a solution. So indeed f(0) = 0 . Thus we can write f(f(x)) = \frac{x^2}{f(x)}, for all x \neq 0 . We assume x \neq 0 from now on. We also realize that P(f(x), f(x)) gives us f(f(x))f(f(f(x))f(x) - 1) = f(x)^2f(f(x)) - f(f(x)), \frac{x^2}{f(x)}f(x^2 - 1) = x^2f(x) - \frac{x^2}{f(x)}, f(x^2 - 1) = f(x)^2 - 1. This gives us f(1) = \pm 1 . Note that f(1) = -1 does not work as P(-1, 1) gives f(-2) = 0 . So f(1) = 1 . Now, P(1, x) gives us that f(x - 1) = f(x) - 1 for all natural n . Therefore, the equation becomes f(x)f(yf(x)) = x^2f(y), which we now denote by P(x, y) . Now, P(x, x) gives f(xf(x)) = x^2. Finally, P(xf(x), y) implies f(x^2y) = f(x)^2f(y). Note that y = 1 gives f(x^2) = f(x)^2, so the equation becomes f(x^2y) = f(x^2)f(y). Holding y non-negative, we get that f is multiplicative in \mathbb{R}^+ , but as f is odd, we have f is multiplicative everywhere. To finish, as we have f(x + 1) = f(x) + 1, let x \rightarrow x/y for y \neq 0 , you get f(x + y)f\left(\frac{1}{y}\right) = f(x)f\left(\frac{1}{y}\right) + 1, f(x + y) = f(x) + f(y), as f(1/y) = 1/f(y) . Therefore, f is additive as well. But as f(x^2) = f(x)^2 \geq 0 , by Cauchy, we have f(x) = cx for all x \in \mathbb{R} . Plugging it back in, we get c \in \{0, 1\} , giving us the solutions.
31.10.2024 21:38
f(x)(f(yf(x) - 1)+1)=x^2f(y)Answers are f\equiv 0 and f(x)=x. Assume that f is non-constant. Plugging x=0 yields f(0)(f(yf(0)-1)+1)=0. f(0)\neq 0 contradicts with our assumption hence f(0)=0. y=0 implies f(x)f(-1)+f(x)=0 thus, f(-1)=-1. Note that if f(a)=0, then x=a gives a^2f(y)=0 which is impossible for a\neq 0. x=y implies f(xf(x)-1)=x^2-1 which gives that f is surjective over positive reals. Also for x=1, f(f(1)-1)=0 or f(1)=1. Plugging x,f(y), by symmetry we get \frac{x^2f(f(y))}{f(x)}=\frac{y^2f(f(x))}{f(y)} thus, \frac{f(f(x))f(x)}{x^2} is constant. For x=1, it's equal to 1 hence x^2=f(x)f(f(x)). Plug y=1 to see that f(x)(f(f(x)-1)+1)=x^2\iff f(f(x)-1)=f(f(x))-1\overset{\text{surjectivity for positive reals}}{\implies} f(x^2-1)=f(x^2)-1P(-1,-y^2) gives -f(y^2)=-f(y^2-1)-1=f(-y^2) hence f is odd which proves the surjectivity for all reals. Now we can choose f(x) as any real on equation f(f(x)-1)=f(f(x))-1 so f(x-1)=f(x)-1\iff f(x+n)=f(x)+n. f(x)f(yf(x))=x^2f(y)Compare P(x,y) with P(x,y+1) to observe f(x)f(yf(x)+f(x))=x^2f(y)+x^2=f(x)f(yf(x))+x^2. After plugging x,\frac{y}{f(x)} we get f(x)(f(y+f(x))-f(y))=x^2 or f(y+f(x))-f(y)=f(f(x)) Since f is surjective, f(x+y)=f(x)+f(y) hence f is additive. (f(x)+f(z))(f(yf(x))+f(yf(z)))=(x+z)^2f(y)\iff 2xzf(y)=f(x)f(yf(z))+f(z)f(yf(x))Plug y=z=1 to verify 2x=f(x)+f(f(x)). Also f(x)f(f(x))=x^2 hence x^2=f(x)(2x-f(x))\iff (x-f(x))^2=0 which implies f(x)=x for all reals as desired.\blacksquare