Question

Under what conditions is a linear function$\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{mx}\mathbf{+}\mathbf{b}\mathbf{,}\mathbf{m}\mathbf{\ne }\mathbf{0}\mathbf{,}$ a linear transform?

Short answer

$\mathrm{b}=0$

Step-by-step solution

Definition of Laplace Transform

Let f be a function defined for $\mathbf{t}\mathbf{⩾}\mathbf{0}$. Then the integral

$\mathcal{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}{\mathbf{e}}^{\mathbf{-}\mathbf{xt}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$

is said to be the Laplace transform of f, provided that the integral converges.

Use Laplace Transform

Original equation

$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{mx}+\mathrm{b},\mathrm{m}\ne 0$

A linear function is a linear transform if the two properties on the left-hand side hold true.

1. $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)$
   
2. $\mathrm{f}(\mathrm{a}·\mathrm{x})=\mathrm{a}·\mathrm{f}\left(\mathrm{x}\right)$
   

Looking at property 1 of the linear transform, evaluating $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)$results in $\mathrm{b}=0$

$\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{m}(\mathrm{x}+\mathrm{y})+\mathrm{b}$

$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)=(\mathrm{mx}+\mathrm{b})+(\mathrm{my}+\mathrm{b})$

$\mathrm{m}(\mathrm{x}+\mathrm{y})+\mathrm{b}=\mathrm{mx}+\mathrm{my}+2\mathrm{b}$

$\mathrm{b}=2\mathrm{b}$

$\mathrm{b}=0$

Looking at property 2 of the linear transform, evaluating $\mathrm{f}(\mathrm{a}·\mathrm{x})=\mathrm{a}·\mathrm{f}\left(\mathrm{x}\right)$results in $\mathrm{b}=0$.

$\mathrm{f}(\mathrm{a}·\mathrm{x})=\mathrm{m}\left(\mathrm{ax}\right)+\mathrm{b}$

$\mathrm{a}·\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}(\mathrm{mx}+\mathrm{b})$

$\mathrm{f}(\mathrm{a}·\mathrm{x})=\mathrm{a}·\mathrm{f}\left(\mathrm{x}\right)\to \mathrm{amx}+\mathrm{b}=\mathrm{amx}+\mathrm{ab}$

$\mathrm{b}=0$

After looking at both properties, the only restriction is $\mathrm{b}$must be equal to 0.

Therefore, $\mathrm{b}=0$.