Inequality Conditions in Functional Equations

Since \(f\) is increasing and additive, \(f(x)=cx\). From \(f(1)=4\), \(c=4\). The answer is \(f(x)=4x\).

An additive function bounded on an interval is continuous. Therefore \(f(x)=cx\). From \(f(1)=3\), we get \(f(x)=3x\).

If \(x

If \(f(a)>0\), then \(f(na)=nf(a)\) exceeds \(100\) for large \(n\), contradiction. If \(f(a)<0\), then \(f((-n)a)=-nf(a)\) exceeds \(100\). Hence \(f(a)=0\) for every \(a\).

If \(f(a)>a\), then \(f(f(a))>f(a)\), so \(a>f(a)\), contradiction. If \(f(a)

Boundedness on an interval implies linearity: \(f(x)=cx\). Since \(f(2)=10\), \(2c=10\), so \(c=5\). The answer is \(f(x)=5x\).

For \(x\), \(f(x)\ge x\). For \(-x\), \(f(-x)\ge -x\), i.e. \(-f(x)\ge -x\), so \(f(x)\le x\). Therefore \(f(x)=x\) for all \(x\).

Let \(g(x)=f(x)-f(0)\). Then \(g(0)=0\), \(g\) satisfies the same Jensen equation and is bounded above on an interval. The standard result gives continuity and linearity \(g(x)=ax\). Hence \(f(x)=ax+b\).

The function is strictly increasing and additive, hence linear: \(f(x)=cx\). From \(f(1)=1\), \(c=1\). The answer is \(f(x)=x\).

Nonnegativity on the positive half-line gives nondecreasingness. Hence the additive function is linear: \(f(x)=cx\). From \(0\le f(1)\le1\), we get \(0\le c\le1\).

For \(q>0\), \(qf(x)=f(qx)\le q^2x^2\). Divide by \(q\): \(f(x)\le qx^2\). Letting \(q o0+\), we get \(f(x)\le0\). Applying this to \(-x\), we get \(-f(x)\le0\), i.e. \(f(x)\ge0\). Hence \(f(x)=0\).

Let \(x e0\). For \(q>0\), \(qf(x)=f(qx)\ge q^2x^2\), so \(f(x)\ge qx^2\). As \(q o\infty\), the right-hand side grows without bound, impossible for fixed \(f(x)\). Contradiction.

A nondecreasing additive function is linear: \(f(x)=cx\), \(c\ge0\). Then \(cxy=c^2xy\) for all \(x,y\), so \(c=0\) or \(c=1\). The answers are \(f\equiv0\) and \(f(x)=x\).

If \(f(a)>0\), then for large negative \(n\), \(nf(a)<-1\), i.e. \(f(na)<-1\), contradiction. If \(f(a)<0\), take large positive \(n\). Therefore \(f(a)=0\) for every \(a\).

Let \(g(x)=f(x)-f(0)\). Then \(g(0)=0\), \(g\) satisfies Jensen's equation and is nondecreasing. By the standard fact, a Jensen function with monotonicity is linear: \(g(x)=ax\). Hence \(f(x)=ax+b\).

The estimate \(|f(x)|\le2|x|\) gives boundedness on \([-1,1]\), hence \(f(x)=cx\). The condition becomes \(|c||x|\le2|x|\), so \(|c|\le2\). The answer is all \(f(x)=cx\), where \(|c|\le2\).

If \(x0\), and \(f(y)-f(x)=f(y-x)>0\). Thus \(f\) is strictly increasing. A strictly increasing additive function is linear: \(f(x)=cx\). From \(f(1)=1\), we get \(c=1\).

From the upper bound, as before, for \(q>0\): \(qf(x)=f(qx)\le q^2x^2\), so \(f(x)\le qx^2\). Letting \(q o0+\), we get \(f(x)\le0\). Apply the same to \(-x\): \(f(-x)\le0\), i.e. \(-f(x)\le0\), so \(f(x)\ge0\). Therefore \(f(x)=0\).

With \(y=x\), \(f(x)^2\ge x^2\), which does not give boundedness. But with \(y=-x\), \(f(x)f(-x)\ge -x^2\), so \(-f(x)^2\ge -x^2\), hence \(f(x)^2\le x^2\). Thus \(f\) is bounded on \([-1,1]\), so \(f(x)=cx\). The condition becomes \(c^2xy\ge xy\) for all \(x,y\). If \(xy>0\), then \(c^2\ge1\); if \(xy<0\), then \(c^2\le1\). Hence \(c^2=1\). The answer is \(f(x)=x\), \(f(x)=-x\).

With \(y=x\), \(f(x)^2\le x^2\). Hence \(|f(x)|\le1\) on \([-1,1]\), so the additive function is bounded on an interval. Therefore \(f(x)=cx\). Substitute: \(c^2xy\le xy\) for all \(x,y\). If \(xy>0\), then \(c^2\le1\). If \(xy<0\), division reverses the sign and gives \(c^2\ge1\). Thus \(c^2=1\). The answer is \(f(x)=x\) and \(f(x)=-x\). The check is immediate.