#3 Linear Functional Equations

For integers \(n\), \(f(n)=5n\). If \(q=m/n\), then \(n f(q)=f(m)=5m\), so \(f(q)=5q\).

Then \(g\) is additive and \(g(2)=6\), so \(g(1)=3\). Hence \(g(q)=3q\), and \(f(q)=3q+1\).

With \(x=0\), \(f(y)=f(0)\) for all \(y\). Every constant function works.

From \(x=y=0\), \(f(0)=0\). For any \(u,v\in\mathbb Q\), take \(x=u/2\), \(y=v/3\): then \(f(u+v)=f(u)+f(v)\). Hence \(f(q)=7q\).

Let \(g(x)=f(x)-4\). Then \(g(x+y)=g(x)+g(y)\), and \(g(1)=5\). On \(\mathbb Q\), \(g(q)=5q\), so \(f(q)=5q+4\).

From \(x=0\) in the second condition, \(f(1)=5\). Now \(g=f-2\) is additive and \(g(1)=3\). Hence \(g(q)=3q\), so \(f(q)=3q+2\).

For \(g=f-1\), \(g((x+y)/2)=(g(x)+g(y))/2\), \(g(0)=0\), \(g(1)=3\). On rationals, the midpoint relation gives additivity: \(g(u+v)=g(u)+g(v)\) after taking \(x=2u\), \(y=2v\) and using \(g(2t)=2g(t)\). Hence \(g(q)=3q\), and \(f(q)=3q+1\).

We have \(f(f(x))=a(ax+b)+b=a^2x+b(a+1)\). Thus \(a^2=4\), \(b(a+1)=3\). If \(a=2\), then \(b=1\); if \(a=-2\), then \(b=-3\). Answer: \(2x+1\) and \(-2x-3\).

From \(f(x+1)=f(x)+2\) at \(x=1\), \(f(2)=f(1)+2\), so \(f(1)=5\). From the first equation at \(x=y=1\), \(f(2)=2f(1)-f(0)\), hence \(7=10-f(0)\), so \(f(0)=3\). Then \(g=f-3\) is additive and \(g(1)=2\), so \(g(q)=2q\). Answer: \(f(q)=2q+3\). Direct verification checks all three conditions.

On rationals, \(f(q)=qf(1)\). Let rational sequences \(r_n\uparrow x\), \(s_n\downarrow x\). By monotonicity, \(f(r_n)\le f(x)\le f(s_n)\), so \(r_nf(1)\le f(x)\le s_nf(1)\), with the sign handled by two-sided approximation. Passing to the limit gives \(f(x)=xf(1)\).

If \(h\) is small, choose \(n\) so that \(0\le nh\le1\). Then \(|f(h)|=|f(nh)|/n\le10/n\), which tends to zero as \(h\to0\). Thus \(f\) is continuous at zero. A continuous additive function is linear: \(f(x)=xf(1)\).

If \(x\le y\), then \(y-x\ge0\), so \(f(y)-f(x)=f(y-x)\ge0\). Hence \(f\) is nondecreasing. By the previous result, \(f(x)=cx\), with \(c=f(1)\ge0\).

Additivity gives \(f(q)=cq\). Substitute into the second condition: \(cxy=x\cdot cy+y\cdot cx=2cxy\). With \(x=y=1\), \(c=2c\), so \(c=0\). Answer: \(f\equiv0\).

If \(x>0\), then \(x=t^2\) for some \(t\ne0\). Thus \(f(x)=f(t)^2\ge0\). Moreover \(f(t)\ne0\), otherwise \(1=f(1)=f(t\cdot1/t)=0\). Hence \(f(x)>0\) for \(x>0\), so \(f\) is increasing. An increasing additive function is linear: \(f(x)=cx\). From \(f(1)=1\), \(c=1\).

We have \(f(f(x)+1)=a(ax+b+1)+b=a^2x+ab+a+b\). Thus \(a^2=9\) and \(b(a+1)+a=5\). If \(a=3\), then \(4b+3=5\), so \(b=1/2\). If \(a=-3\), then \(-2b-3=5\), so \(b=-4\). Both work.

Let \(c=f(0)\), \(g=f-c\). Then \(g\) is additive, and \(f(3x)=g(3x)+c=3g(x)+c\). The right side of the second condition is \(3(g(x)+c)-4=3g(x)+3c-4\). Hence \(c=3c-4\), so \(c=2\). The additive part on \(\mathbb Q\) is \(g(q)=kq\). Answer: \(f(q)=kq+2\), \(k\in\mathbb Q\).

Let \(g(x)=f(x)-3\). Then \(g\) is additive. On \([0,1]\), \(g(x)\ge -8\). Also \(g(x)=g(1)-g(1-x)\le g(1)+8\), so \(g\) is bounded on \([0,1]\). Therefore \(g(x)=cx\), and \(f(x)=cx+3\).

From \(f(0)=0\), \(b=0\), so \(f(x)=ax\). Then \(f(f(x))=a^2x\), which cannot equal \(x+1\) for all \(x\), because the right side has constant term \(1\). No solutions.

By monotonicity, the additive function is linear: \(f(x)=cx\). Then \(f(f(x))=c^2x=x\), so \(c^2=1\). Since \(f\) is nondecreasing, \(c\ge0\), hence \(c=1\). Answer: \(f(x)=x\).

Let \(A(x)=f(x)-2\). Then \(A(x+y)=A(x)+A(y)\). Without extra conditions, all solutions are \(f(x)=A(x)+2\), where \(A\) is arbitrary additive. If \(f\) is nondecreasing, then \(A\) is nondecreasing, so \(A(x)=cx\). The condition \(f(1)=5\) gives \(c+2=5\), hence \(c=3\). Thus \(f(x)=3x+2\).