Advanced Functional Equations

Let \(g(x)=f(x)-\frac a2x^2\). Then \(g(x+y)=g(x)+g(y)\). By continuity, \(g(x)=cx\). The answer is \(f(x)=\frac a2x^2+cx\), where \(c\in\mathbb R\). Direct checking works.

The quadratic correction gives \(f(x)=x^2+cx\). The condition \(f(1)=5\) gives \(1+c=5\), so \(c=4\). The answer is \(f(x)=x^2+4x\).

Let \(f(x)=cx\). Then \(cx^2=c^2x^2\) for all \(x\). We get \(c=0\) or \(c=1\). The answer is \(f\equiv0\), \(f(x)=x\).

Additivity on \(\mathbb Q\) gives \(f(x)=cx\). Then \(cxy=2cxy\) for all \(x,y\). Hence \(c=0\). The answer is \(f\equiv0\).

If \(f(x)=cx\), then \(f(f(x))=c^2x\). We need \(c^2=2\), but no rational \(c\) has this property. Contradiction.

Continuous additivity gives \(f(x)=cx\). Substitution: \(c^2xy=cxy+xy\). Hence \(c^2=c+1\). The answer is \(f(x)=\frac{1+\sqrt{5}}{2}x\) or \(f(x)=\frac{1-\sqrt{5}}{2}x\).

Let \(f(x)=cx\). Then \(cxy=2cxy\) for all \(x,y\), so \(c=0\). The answer is \(f\equiv0\).

Continuous additivity gives \(f(x)=cx\). Then \(cx^2=c^2x^2\), so \(c=0\) or \(c=1\). The answer is \(f\equiv0\), \(f(x)=x\).

We have \(f(x)=\frac a2x^2+cx\). From \(f(1)=1\): \(\frac a2+c=1\). From \(f(2)=6\): \(2a+2c=6\), i.e. \(a+c=3\). Subtracting gives \(\frac a2=2\), hence \(a=4\), \(c=-1\). The answer is \(f(x)=2x^2-x\).

On \(\mathbb Q\), \(f(x)=ax\), \(g(x)=bx\). The conditions give \(ab=1\). Hence \(a\in\mathbb Q\setminus\{0\}\), \(b=\frac1a\). The answer is \(f(x)=ax\), \(g(x)=\frac{x}{a}\).

If \(f(a)=f(b)\), comparing the equations for \(y=a,b\) gives \(2a=2b\), so \(a=b\). With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). With \(x=0\), \(f(f(y))=2y\). Let \(t=f(y)\). Then \(f(t)=2y\), and the original equation gives \(f(x+t)=f(x)+f(t)\). The function is additive and increasing, so \(f(x)=cx\). From \(c^2=2\) and increasingness, \(c>0\). The answer is \(f(x)=\sqrt{2}x\).

From the first condition, \(f(x)=\frac a2x^2+cx\). If \(a\ne0\), then \(f\) has degree \(2\), and \(f(f(x))\) has degree \(4\), impossible. Hence \(a=0\). Then \(f(x)=cx\), and \(c^2=1\). The answer is \(a=0\), \(f(x)=x\) or \(f(x)=-x\).

As before, \(f(x)=\frac a2x^2+cx\). If \(a\ne0\), the composition has degree \(4\), impossible. Thus \(a=0\), \(f(x)=cx\). Then \(f(f(x))=c^2x\), which cannot equal \(x+1\). Therefore there are no solutions.

Additivity on \(\mathbb Q\) gives \(f(x)=cx\). The left-hand side is \(\frac c x\), and the right-hand side is \(\frac{cx}{x^2}=\frac c x\). The condition does not restrict \(c\). The answer is all \(f(x)=cx\), \(c\in\mathbb Q\).

Let \(f(x)=cx\). Then \(\frac c x=c^2x^2\) for all \(x\ne0\). If \(c=0\), we get the solution \(f=0\). If \(c\ne0\), then \(1=cx^3\) for all \(x\), impossible. The only answer is \(f\equiv0\).

With \(y=0\), \(f(f(x))=x+f(0)\), which implies injectivity and surjectivity. With \(x=0\), \(f(f(0)+y)=f(y)\), so \(f(0)=0\). Thus \(f(f(x))=x\). Putting \(y=f(t)\), we get \(f(f(x)+f(t))=x+t=f(f(x+t))\), hence \(f(x)+f(t)=f(x+t)\). The function is additive and increasing, so \(f(x)=cx\). From \(f(f(x))=x\), \(c^2=1\), and increasingness gives \(c=1\). The answer is \(f(x)=x\).

Let \(g(x)=f(x)-a x^2\). Then \(g(x+y)+g(x-y)=2g(x)\). Continuous solutions of this equation are affine: \(g(x)=cx+d\). Therefore \(f(x)=a x^2+cx+d\). Direct checking works.

The first condition gives \(f(x)=ax+b\). Then \(f(f(x))=a^2x+b(a+1)\). We get \(a^2=1\), \(b(a+1)=0\). The answer is \(f(x)=x\) and \(f(x)=-x+b\), where \(b\in\mathbb R\).

We have \(f(x)=\frac a2x^2+cx\). If \(a\ne0\), then \(f(f(x))\) has degree \(4\), while the right-hand side has degree \(2\), impossible. Hence \(a=0\), \(f(x)=cx\). Then \(f(f(x))=c^2x\), which cannot equal \(x^2\) for all \(x\). Therefore there are no solutions.

From the first condition, \(f(x)=\frac a2x^2+cx\). If \(a\ne0\), then \(f\) is quadratic and \(f(f(x))\) has degree \(4\), while \(x+a\) has degree \(1\). Hence \(a=0\). Then \(f(x)=cx\), and the condition becomes \(c^2x=x\). Therefore \(c^2=1\). The answer is \(a=0\), \(f(x)=x\) or \(f(x)=-x\).