#5 Cauchy-Type Equations

The statement is true for \(n=1\). If \(f(n)=7n\), then \(f(n+1)=f(n)+f(1)=7n+7=7(n+1)\). Hence \(f(n)=7n\) for all \(n\in\mathbb N\).

From \(f(2)=f(1)+f(1)\), we get \(f(1)=5\). For positive \(n\), \(f(n)=5n\). Also \(0=f(0)=f(n+(-n))=f(n)+f(-n)\), so \(f(-n)=-5n\). The answer is \(f(n)=5n\) for all \(n\in\mathbb Z\).

By additivity on \(\mathbb Q\), \(f(q)=qf(1)\). From \(f(3)=3f(1)=12\), we get \(f(1)=4\). The answer is \(f(q)=4q\).

Let \(g(x)=f(x)+6\). Then \(g(x+y)=f(x+y)+6=f(x)+f(y)+12=g(x)+g(y)\). Hence \(g(x)=ax\), where \(a\in\mathbb Q\). Therefore \(f(x)=ax-6\). All such functions work.

If \(n=p_1^{a_1}\cdots p_k^{a_k}\), then by multiplicativity \(f(n)=f(p_1)^{a_1}\cdots f(p_k)^{a_k}=p_1^{2a_1}\cdots p_k^{2a_k}=n^2\). The answer is \(f(n)=n^2\).

Additivity gives \(f(x)=cx\). Then \(f(xy)=cxy\), while \(f(x)f(y)=c^2xy\). Hence \(cxy=c^2xy\) for all \(x,y\), so \(c=0\) or \(c=1\). The answers are \(f(x)=0\) and \(f(x)=x\).

Since \(f\) is continuous and additive, \(f(x)=cx\). From \(c\sqrt{3}=6\), we get \(c=\frac{6}{\sqrt{3}}=2\sqrt{3}\). The answer is \(f(x)=2\sqrt{3}x\).

By the standard fact, an additive function bounded on an interval is continuous at zero and hence everywhere. Therefore \(f(x)=cx\). Since \(f(1)=2\), we get \(c=2\). Thus \(f(x)=2x\).

As in the theory, continuous solutions have the form \(f(x)=ax+b\). From \(f(0)=3\), we get \(b=3\). From \(f(2)=9\), \(2a+3=9\), so \(a=3\). The answer is \(f(x)=3x+3\).

The recursion gives \(f(n+1)-f(n)=2n+1=(n+1)^2-n^2\). Since \(f(1)=1=1^2\), telescoping yields \(f(n)=n^2\). Multiplicativity checks: \(f(mn)=m^2n^2=f(m)f(n)\). The answer is \(f(n)=n^2\).

By additivity, \(f((x+y)^2)=f(x^2)+2f(xy)+f(y^2)\). On the other hand, by the condition this equals \((x+y)f(x+y)=(x+y)(f(x)+f(y))\). After cancelling \(x f(x)+y f(y)\), we get \(2f(xy)=x f(y)+y f(x)\). Put \(y=1\). Then \(2f(x)=x f(1)+f(x)\), hence \(f(x)=x f(1)\). Take \(c=f(1)\).

Let \(g(x)=f(x)-x^2\). Then \(g(x+y)+g(x-y)=2g(x)\). On \(\mathbb Q\), this midpoint equation gives \(g(x)=ax+b\). Hence \(f(x)=x^2+ax+b\), where \(a,b\in\mathbb Q\). Direct checking works.

Since \(f\) is additive on \(\mathbb Q\), \(f(x)=cx\), where \(c\in\mathbb Q\). Then \(f(f(x))=c^2x\). The condition gives \(c^2=4\), hence \(c=2\) or \(c=-2\). The answers are \(f(x)=2x\) and \(f(x)=-2x\).

Let \(g(x)=f(x)-1\). The first condition gives \(g(x+y)=g(x)+g(y)\). The second condition, after substituting \(f=g+1\), becomes \(g(xy)=g(x)g(y)\). On \(\mathbb Q\), an additive function has the form \(g(x)=cx\), and multiplicativity gives \(c=0\) or \(c=1\). Thus \(f(x)=1\) or \(f(x)=x+1\). Both functions work.

From the midpoint equation and continuity, \(f(x)=ax+b\). Then \(f(f(x))=a(ax+b)+b=a^2x+b(a+1)\). This must equal \(x\) for all \(x\). Hence \(a^2=1\) and \(b(a+1)=0\). If \(a=1\), then \(b=0\), giving \(f(x)=x\). If \(a=-1\), then \(b\) is arbitrary, giving \(f(x)=-x+b\). All these functions work.

With \(y=0\), we get \(f(0)=0\). Continuous solutions of the quadratic Cauchy equation have the form \(f(x)=cx^2\). From \(f(2)=12\), \(4c=12\), so \(c=3\). The answer is \(f(x)=3x^2\). Checking follows by expanding squares.

Let \(g(x)=f(x)-\frac{x^3}{3}\). Then \(g(x+y)=f(x+y)-\frac{(x+y)^3}{3}=f(x)+f(y)+xy(x+y)-\frac{x^3+y^3+3xy(x+y)}{3}=g(x)+g(y)\). On \(\mathbb Q\), \(g(x)=ax\). The answer is \(f(x)=\frac{x^3}{3}+ax\), where \(a\in\mathbb Q\). The check follows from the cubic identity.

Continuous additivity gives \(f(x)=cx\). The second condition becomes \(c^2xy=cxy+xy\) for all \(x,y\). Hence \(c^2=c+1\). We get \(c=\frac{1+\sqrt{5}}{2}\) or \(c=\frac{1-\sqrt{5}}{2}\). Both functions work.

Let \(g(x)=f(x)-1\). Then \(g(x+y)=g(x)+g(y)\). The second condition gives \(g(xy)+1=(g(x)+1)(g(y)+1)-(g(x)+1)-(g(y)+1)+2=g(x)g(y)+1\), hence \(g(xy)=g(x)g(y)\). On \(\mathbb Q\), additivity gives \(g(x)=cx\). Multiplicativity gives \(c=c^2\), so \(c=0\) or \(c=1\). The answer is \(f(x)=1\) and \(f(x)=x+1\).

Putting \(x=y\), we get \(f(2x)+f(0)=4f(x)\), so \(f(2x)=4f(x)\). A standard induction using the quadratic Cauchy formula gives \(f(nx)=n^2f(x)\) for integers \(n\). With \(x=1\), this gives \(f(n)=n^2\). Then from \(f(n\cdot \frac{m}{n})=n^2 f(\frac{m}{n})\), we obtain \(f(\frac{m}{n})=\frac{m^2}{n^2}\). Hence \(f(q)=q^2\) for all rational \(q\).

For an arbitrary real \(x\), choose rationals \(q_k\to x\). By continuity, \(f(x)=\lim f(q_k)=\lim q_k^2=x^2\). Therefore \(f(x)=x^2\) for all \(x\).