Question

Question: Find the square; operate on numbers or on functions.

Short answer

The square of any function which operates on the function $f\left(x\right)$ by squaring it does not represent a linear operator.

Step-by-step solution

Definition of the linear operator:

An operator is said to be a linear operator if the following relations are satisfied if$\mathit{O}\mathbf{(}\mathit{A}\mathbf{+}\mathit{B}\mathbf{)}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{+}\mathit{O}\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{O}\mathbf{\left(}\mathit{k}\mathit{A}\mathbf{\right)}\mathbf{=}\mathit{k}\mathit{O}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{.}\mathbf{}$ .

Where, K is a number, and A and, B are numbers, functions, vectors, etc.

Operation on square function for O(A+B)=O(A)+O(B) :

The square function which operates on the function is given by

$Q\left(x\right)=\left[f\right(x){]}^2$

It is known that

$$\begin{gathered} {\left(f\left(x\right)+g\left(x\right)\right)}^2=f^2\left(x\right)+2f\left(x\right)g\left(x\right)+g^2\left(x\right) \\ ={\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2+2f\left(x\right)g\left(x\right)+g^2\left(x\right) \\ ={\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2 \end{gathered}$$

Operation on square function for  O(kA)=kO(A) :

Evaluate the square of $\left(kf\right(x){)}^2$.

$$\begin{gathered} \left(kf\right(x){)}^2=k^2{f}^2\left(x\right) \\ =k^2{\left[f\left(x\right)\right]}^2 \\ \ne k^2{\left[f\left(x\right)\right]}^2 \end{gathered}$$

Therefore, it is shown that the square of any function which operate on the function $f\left(x\right)$ by squaring it does not represent a linear operator.