#4 Injectivity and Surjectivity

Since \(f\) is injective, the equality gives \(f(x)+y=f(y)+x\). Hence \(f(x)-x=f(y)-y\) for all \(x,y\). Fixing \(y=0\), we get \(f(x)-x=f(0)\), so \(f(x)=x+f(0)\). Take \(c=f(0)\).

By surjectivity, for every \(t\in\mathbb R\) there is \(y\) such that \(t=f(y)\). Hence \(f(x+t)=f(x)+t\) for all \(x,t\). Taking \(x=0\), we get \(f(t)=f(0)+t\). Thus \(f(x)=x+c\), where \(c=f(0)\). Checking: \(f(x+f(y))=x+f(y)+c=f(x)+f(y)\). The answer is all \(f(x)=x+c\), \(c\in\mathbb R\).

If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\). By the condition, \(a+1=b+1\), so \(a=b\). Thus \(f\) is injective.

Now let \(T\in\mathbb Q\). Then \(f(f(T-1))=T\). Hence \(T\) lies in the image of \(f\). Since \(T\) was arbitrary, \(f\) is surjective. Therefore \(f\) is bijective.

Since \(f\) is surjective, \(t=f(y)\) runs through all real numbers. Hence \(f(x+t)=x+t+1\) for all \(x,t\). Every number \(u\) can be written as \(u=x+t\), so \(f(u)=u+1\). The check is immediate: \(f(x+f(y))=x+f(y)+1\). The answer is \(f(x)=x+1\).

By injectivity, \(f(x)+f(y)=x+y\). Taking \(y=0\), we get \(f(x)+f(0)=x\), so \(f(x)=x-c\), where \(c=f(0)\). Substitute this into \(f(x)+f(y)=x+y\): \((x-c)+(y-c)=x+y\), hence \(c=0\). Therefore \(f(x)=x\). The check is immediate.

With \(x=0\), \(f(f(y))=f(0)+y\). If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\), hence \(a=b\). Thus \(f\) is injective. With \(y=0\), \(f(x+f(0))=f(x)\), so \(f(0)=0\). Hence \(f(f(y))=y\).

Put \(y=f(t)\): \(f(x+f(f(t)))=f(x)+f(t)\), so \(f(x+t)=f(x)+f(t)\). On \(\mathbb Q\), an additive function has the form \(f(x)=cx\). From \(f(f(y))=y\), we get \(c^2=1\). The answers are \(f(x)=x\) and \(f(x)=-x\). Direct checking works.

With \(y=0\), \(f(f(x))=x+f(0)\), so \(f\) is injective and surjective. With \(x=0\), \(f(f(0)+y)=f(y)\). By injectivity, \(f(0)+y=y\), hence \(f(0)=0\). Thus \(f(f(x))=x\).

Put \(y=f(t)\): \(f(f(x)+f(t))=x+t=f(f(x+t))\). By injectivity, \(f(x)+f(t)=f(x+t)\). Hence \(f\) is additive on \(\mathbb Q\), so \(f(x)=cx\). The condition \(f(f(x))=x\) gives \(c^2=1\). The answer is \(f(x)=x\) or \(f(x)=-x\).

By surjectivity, \(t=f(y)\) is arbitrary. Hence \(f(x+t)=f(x)+2t\) for all \(x,t\). Taking \(x=0\), we get \(f(t)=f(0)+2t\). Therefore \(f(x)=2x+c\). Checking: \(f(x+f(y))=2x+2f(y)+c=f(x)+2f(y)\). All functions \(f(x)=2x+c\) work.

If \(f(a)=f(b)\), comparing the original equality for \(y=a\) and \(y=b\) gives \(3a=3b\), so \(a=b\). Thus \(f\) is injective. With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). With \(x=0\), \(f(f(y))=3y\), so \(f\) is surjective.

Let \(t=f(y)\). Then \(f(t)=3y\), and the original equation becomes \(f(x+t)=f(x)+f(t)\). Hence \(f\) is additive on \(\mathbb Q\), so \(f(x)=cx\) for some rational \(c\). The condition \(f(f(y))=3y\) gives \(c^2=3\), impossible for rational \(c\). Contradiction.

By surjectivity, \(t=f(y)\) is arbitrary, so \(f(x+t)=f(x)+t+1\). Taking \(x=0\), we get \(f(t)=f(0)+t+1\). Then the left-hand side of \(f(x+t)=f(x)+t+1\) equals \(f(0)+x+t+1\), while the right-hand side equals \(f(0)+x+1+t+1\). This gives \(0=1\), a contradiction. No such function exists.

If \(f(a)=f(b)\), comparing the equations for \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. Let \(c=f(0)\). With \(x=0\), \(f(f(y))=c+y+1\). Put \(t=f(y)\). Then \(f(t)=c+y+1\), so \(y+1=f(t)-c\). Hence the original equation gives \(f(x+t)=f(x)+f(t)-c\) for all \(x,t\) in the image. The function is surjective because \(f(f(T-c-1))=T\), so this holds for all \(t\).

Let \(g(x)=f(x)-c\). Then \(g(x+t)=g(x)+g(t)\), so on \(\mathbb Q\) we have \(g(x)=ax\). Also, from \(f(f(y))=c+y+1\) we get \(g(g(y)+c)=y+1\). Since \(g\) is additive, \(g(g(y))+g(c)=y+1\). At \(y=0\), \(g(c)=1\), hence \(g(g(y))=y\). Therefore \(a^2=1\) and \(ac=1\). If \(a=1\), then \(c=1\); if \(a=-1\), then \(c=-1\). The answer is \(f(x)=x+1\) and \(f(x)=-x-1\). Direct checking works.

If \(f(a)=f(b)\), comparing the equations for \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. Let \(c=f(0)\). With \(x=0\), \(f(f(y))=c+f(y)+y\). If \(t=f(y)\), then \(f(t)=c+t+y\), so \(t+y=f(t)-c\). Hence \(f(x+t)=f(x)+f(t)-c\). Thus \(g(x)=f(x)-c\) is additive, and on \(\mathbb Q\), \(g(x)=ax\).

Therefore \(f(x)=ax+c\). Substitute into the original equation: \(a(x+ay+c)+c=ax+(a+1)y+2c\). Comparing coefficients gives \(a^2=a+1\) and \(c(a-1)=0\). But \(a^2-a-1=0\) has no rational roots. Contradiction.

If \(f(a)=f(b)\), comparing the equalities for \(y=a\) and \(y=b\) gives \(2a=2b\), so \(a=b\). The function is injective. With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). With \(x=0\), \(f(f(y))=2y\), so \(f\) is surjective.

Let \(t=f(y)\). Then \(f(t)=2y\), hence \(f(x+t)=f(x)+f(t)\). The function is additive and increasing, so \(f(x)=cx\). From \(f(f(y))=2y\), we get \(c^2=2\), and increasing gives \(c>0\). The answer is \(f(x)=\sqrt{2}x\). Direct checking works.

With \(y=0\), \(f(f(x))=x+f(0)\). This implies injectivity: if \(f(a)=f(b)\), then \(a+f(0)=b+f(0)\). It also implies surjectivity. With \(x=0\), \(f(f(0)+y)=f(y)\), so injectivity gives \(f(0)=0\). Hence \(f(f(x))=x\).

Put \(y=f(t)\): \(f(f(x)+f(t))=x+t=f(f(x+t))\). By injectivity, \(f(x)+f(t)=f(x+t)\). The function is continuous and additive, so \(f(x)=cx\). The condition \(f(f(x))=x\) gives \(c^2=1\). The answer is \(f(x)=x\) and \(f(x)=-x\).

If \(f(a)=f(b)\), comparing the equations for \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. Let \(c=f(0)\). With \(x=0\), \(f(f(y))=c+f(y)+y\). If \(t=f(y)\), then \(f(t)=c+t+y\), so \(t+y=f(t)-c\). Hence \(f(x+t)=f(x)+f(t)-c\). Therefore \(g(x)=f(x)-c\) is additive and continuous, so \(g(x)=ax\).

Thus \(f(x)=ax+c\). Substituting into the original equation gives \(a(x+ay+c)+c=ax+(a+1)y+2c\). Hence \(a^2=a+1\) and \(c(a-1)=0\). Since the roots of the first equation are not \(1\), we get \(c=0\). Thus \(a=\frac{1+\sqrt{5}}{2}\) or \(a=\frac{1-\sqrt{5}}{2}\). Both functions \(f(x)=ax\) work.

Since \(f\) is surjective, \(t=f(x)\) runs through all real numbers. Then \(f(t+y)=t+f(y)\) for all \(t,y\). Taking \(y=0\), we obtain \(f(t)=t+f(0)\). Hence \(f(x)=x+c\). Checking: \(f(f(x)+y)=f(x)+y+c=x+c+y+c=f(x)+f(y)\). All functions \(f(x)=x+c\) work.

Since \(f\) is surjective, \(t=f(y)\) is arbitrary. Therefore \(f(x+t)+f(x-t)=2f(x)\) for all \(x,t\). Taking \(u=x+t\), \(v=x-t\), we get \(f\left(\frac{u+v}{2}\right)=\frac{f(u)+f(v)}{2}\). A continuous function preserving midpoints is affine: \(f(x)=ax+b\).

Surjectivity requires \(a\ne0\). Checking shows that every function \(f(x)=ax+b\), \(a\ne0\), satisfies the original equation. The answer is \(f(x)=ax+b\), where \(a\ne0\).

If \(f(a)=f(b)\), then using the equations with the same \(y\) gives \(a+y=b+y\), so \(a=b\). Thus \(f\) is injective. Also, every \(T\) equals \(f(f(T)+f(0))\), so \(f\) is surjective. Let \(g=f^{-1}\).

Putting \(y=0\), we get \(f(f(x)+f(0))=x\), hence \(g(x)=f(x)+f(0)\). Let \(c=f(0)\). Then \(g(x)=f(x)+c\). From the original equation with \(x=y=g(0)\), we obtain \(f(0)=2g(0)\). But from \(g(0)=f(0)+c\), we get \(g(0)=2c\). Hence \(c=4c\), so \(c=0\). Therefore \(g=f\) and \(f(f(x))=x\).

Now for arbitrary \(u=f(x)\), \(v=f(y)\), we have \(f(u+v)=x+y=f(u)+f(v)\), because \(g=f\). Thus \(f\) is additive. An increasing additive function has the form \(f(x)=ax\), where \(a>0\). The condition \(f(f(x))=x\) gives \(a^2=1\), hence \(a=1\). The answer is \(f(x)=x\).

With \(x=0\), \(f(f(y))=f(0)+y\). This implies injectivity in the usual way. With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). Thus \(f(f(y))=y\).

Putting \(y=f(t)\), we get \(f(x+t)=f(x)+f(t)\), so \(f\) is additive. A continuous additive function has the form \(f(x)=cx\). From \(f(f(y))=y\), we get \(c^2=1\), hence \(f(x)=x\) or \(f(x)=-x\). In the first case \(f(2)=2\), in the second \(f(2)=-2\). Neither gives \(3\). Contradiction.

If \(f(a)=f(b)\), comparing the equations for \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. Let \(c=f(0)\). With \(x=0\), \(f(f(y))=c+f(y)+y+1\). Put \(t=f(y)\). Then \(f(t)=c+t+y+1\), so \(t+y+1=f(t)-c\). Hence \(f(x+t)=f(x)+f(t)-c\). Therefore \(g(x)=f(x)-c\) is additive and continuous, so \(g(x)=ax\).

From \(f(c)=f(0+f(0))=c+c+0+1=2c+1\), we get \(g(c)=c+1\). Also, \(f(f(y))=c+f(y)+y+1\) becomes \(g(g(y)+c)+c=g(y)+y+2c+1\). Since \(g\) is additive and \(g(c)=c+1\), this gives \(g(g(y))=g(y)+y\). For \(g(x)=ax\), we get \(a^2=a+1\).

The remaining condition is \(g(c)=c+1\), that is \(ac=c+1\), or \(c(a-1)=1\). Thus \(a\) is one of the roots of \(a^2-a-1=0\), and \(c=\frac{1}{a-1}\). Hence the two functions are \(f(x)=ax+\frac{1}{a-1}\), where \(a=\frac{1+\sqrt{5}}{2}\) or \(a=\frac{1-\sqrt{5}}{2}\). Direct substitution confirms both.