Question 3.44: For mappings f , g : R → R and every λ ∈ R define the mappin......
For mappings f,\,g:\mathbb{R}\to\mathbb{R} and every λ ∈ \mathbb{R} define the mappings f + g, f • g and λ f from \mathbb{R} to \mathbb{R} in the usual way, namely by setting
(f + g)(x)=f(x) +g(x), (f• g)(x)=f(x)g(x),
(λf)(x) = λf(x)
for every x ∈ \mathbb{R}.
(a) Show that there are bijections f, g with f + g not a bijection. Show also that there are bijections f, g with f•g not a bijection. Do there exist bijections f,g such that neither f + g nor f•g is a bijection?
(b) Prove that if λ ≠ 0 then λf is a bijection if and only if f is a bijection.
(c) Define [fg] : \mathbb{R}\to\mathbb{R} by [f,\ g]=f\circ g-g\circ f. Do there exist bijections f,g with [fg] a bijection?
(d) If 0 denotes the mapping from \mathbb{R}\ {\mathrm{~to~}}\ \mathbb{R} described by x → 0, prove that, for all mappings f,g,h : \mathbb{R}\to\mathbb{R},
[\,[f g]h]+[\,[g h]f\,]+\,[\,[h f]g\,]=0.(a) Take, for example, f=\mathrm{id}_{\mathbb{R}}\ \mathrm{and}\ g=-\mathrm{id}_{\mathbb{R}}. Then (f + g)(x) = 0 for every x\in\mathbb{R} and so f + g is not a bijection.
Consider now f=\operatorname{id}_{\mathbb{R}} and define g:\mathbb{R}\to\mathbb{R} by
g(x)=\left\{\begin{array}{c l l}{{\frac{1}{x}}}&{{\mathrm{~if~}}}&{{x\not=0;}}\\ {{0}}&{{\mathrm{~if~}}}&{{x=0.}}\end{array}\right.Then g is a bijection; its graph is shown in Fig. S3.42.
Now
(f\cdot g)(x)=f(x)g(x)={\left\{\begin{array}{l l}{x \cdot\frac{1}{x}={{{1}}}}&{{\mathrm{if}}}&{{\mathrm{~}x\neq0;}}\\ {0 \cdot 0=0 \quad\mathrm{if}\quad x=0,}\end{array}\right.}so f·g is not a bijection. For this f and g we also have
(f+g)(x)=\left\{\begin{array}{ccc}x+\frac{1}{x} & \text { if } & x \neq 0 ; \\ 0 & \text { if } & x=0 .\end{array}\right.The equation x+1/x=1 has no solution in \mathbb{R}_{*} so there is no x\in\mathbb{R} such that ({{f}}+g)(x)=1\,. Thus f + g is not a bijection.
(b) Suppose that f is a bijection. Then for \lambda\neq0 we have
(\lambda f)(x_{1})=(\lambda f)(x_{2})\Rightarrow\lambda f(x_{1})=\lambda f(x_{2})\\ \begin{array}{r}{\Rightarrow f(x_{1})=f(x_{2})}\\ {\Rightarrow x_{1}=x_{2}}\end{array}and so \lambda f is injective. Also, since f is surjective, given any y\in\mathbb{R}, there exists t\in\mathbb{R} with f(t)=y/\lambda. Then (\lambda f)(t)=y and hence \lambda f is also surjective. This shows that if f is a bijection then so is \lambda f for every \lambda\neq0. Suppose now that \lambda f is a bijection with \lambda\neq0. Applying the above result to \lambda f and \mu=1/\lambda, it follows that \mu(\lambda f)=(1/\lambda)(\lambda f)=f is also a bijection.
(c) Consider, for example, the mappings f,g:\mathbb{R}\to\mathbb{R}{\mathrm{~given~by}}f(x)=x^{2}, g(x)=x+1. We have
[f g](x)=f\left[g(x)\right]-g[f(x)]\\ \begin{array}{c}{{=(x+1)^{2}-(x^{2}+1)}}\\ {{=2x,}}\end{array}\\so \textstyle[f g] is a bijection.
(d) We have
\left[[f g]h\right]=\left[(f\circ g-g\circ f)h\right]\\ \begin{array}{c}{=(f\circ g-g\circ f)\circ h-h\circ(f\circ g-g\circ f)}\\ {=f\circ g\circ h-g\circ f\circ h-h\circ f\circ g+h\circ g\circ f}\end{array}\\and similarly
Adding these together, we obtain the required result.
