If $\hat O f(x) = f(x^2+1)$, is $\hat O$ a linear operator?

Question

If $\hat O f(x) = f(x^2+1)$, is $\hat O$ a linear operator? It seems this follows the condition being linear operator. But think in following way;

$\hat O f(x) = f(x^2+1) = g(x) \Rightarrow$

$\hat O f(x) = g(x)$.

Now, this seems to me $\hat O$ is not linear. If my proposition is wrong, please correct me and try to explain with simple and plain examples.

By the way, Gasiorowicz in his Quantum Physics book (3rd ed.) claims it is a linear operator.

Thank you in advance for your helps.

Answer 1

Let $D_\hat{O}$ be the set over which the operator $\hat{O}$ is defined.

Let $f(x) = h(x) + g(x)$ with $h,g \in D_\hat{O}$ and thus $f \in D_\hat{O}$. Then, it is :

$$\hat{O}f(x) = f(x^2+1) \Leftrightarrow \hat{O}[(h(x)+g(x)] = h(x^2+1) + g(x^2+1) $$

$$\Leftrightarrow$$

$$\hat{O}[h(x)+g(x)] = \hat{O}h(x) + \hat{O}g(x)$$

Now, for $\lambda \in \mathbb R$ and $f(x) \in D_\hat{O}$, let $g(x) = \lambda f(x)$ which also is $g(x) \in D_\hat{O}$. Then, it is :

$$\hat{O}g(x) = g(x^2+1) =\lambda f(x^2+1) $$

$$\Leftrightarrow$$

$$\hat{O}(\lambda f(x)) = \lambda \hat{O}f(x)$$

Thus, the operator $\hat{O}$ respects the operation of addition and the operation of scalar multiplication $\implies$ $\hat{O}$ is a linear operator.

Regarding your proposition : It doesn't really make any sense. You just let $g(x)$ be the function yielded by the operator. This doesn't say anything about linearity.