## Operations preserving monotonicity

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.  