Operations preserving monotonicity
Task number: 2973
Which of the following operations performed on non-decreasing functions \(f\) a \(g\) result in functions that are also non-decreasing: \(f+g\), \(f-g\), \(f\cdot g\), \(\max\{f,g\}\), \(\min\{f,g\}\), \(f\circ g\)?
Resolution
The assumption that \(f\) and \(g\) are non-decreasing means that \(x<y \longrightarrow f(x)\le f(y)\land g(x)\le g(y)\).
Sum: \((f+g)(x)=f(x)+g(x)\le f(y)+g(y)=(f+g)(y)\).
The difference is not necessarily non-decreasing, because if we let \(g=2f\), then \(f-g=-f\), so we obtain a non-increasing function.
The product is not necessarily non-decreasing, because if we let \(f(x)=g(x)=x\), then \(f(x)g(x)=x^2\), which is not an increasing function.
Maximum: Assuming without loss of generality that \(f(x)\le g(x)\), we get: \(\max\{f(x),g(x)\}=g(x)\le g(y)\le \max\{f(y),g(y)\}\).
Minimum: Assuming without loss of generality that \(f(y)\le g(y)\), we get: \(\min\{f(x),g(x)\}\le f(x)\le f(y)= \min\{f(y),g(y)\}\).
Composition: \(x<y \longrightarrow g(x)\le g(y) \longrightarrow (f\circ g)(x)=f(g(x))\le f(g(y))=(f\circ g)(y)\).
Result
The functions \(f+g\), \(\max\{f,g\}\), \(\min\{f,g\}\), \(f\circ g\) are non-descreasing; only \(f-g\) a \(f\cdot g\) are not necessarily non-decreasing.
