Collection of Maths Problems

Operations preserving monotonicity

Task number: 2973

• 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.