Composition operator

Remarks

Linearity is automatic: $(af + bg) \circ \varphi = a (f \circ \varphi) + b (g \circ \varphi)$.

Example: let $X$ be the Hardy space $H^2$ of the unit disc $\mathbb{D}$ and $\varphi(z) = z/2$. For $f(z) = \sum_{n \ge 0} a_n z^n$ in $H^2$, $f \circ \varphi = \sum_{n \ge 0} 2^{-n} a_n z^n$, whose coefficients are again square-summable; so $C_\varphi$ is defined, and it is the diagonal operator with entries $2^{-n}$ on the basis $z^n$.

Non-example: $\varphi(z) = 2z$ does not map $\mathbb{D}$ into $\mathbb{D}$. For $f(z) = \sum_{n \ge 0} z^n / (n + 1)$, which lies in $H^2$, the function $f(2z)$ is not defined on all of $\mathbb{D}$, since the series diverges for $\abs{z} > 1/2$.

When $X$ is a normed space one usually asks in addition that $C_\varphi$ be bounded.