Is composition of a continuous function a continuous operator? Let $g:\mathbb{R}\to\mathbb{R}$ be a continuous function and $(\Omega,\mathscr{f},\mu)$ a finite measure space. Consider the operator $\varphi:L^1(\Omega,\mathscr{f},\mu)\to L^1(\Omega,\mathscr{f},\mu)$ defined as
\begin{align*}
\varphi(f)= g \circ f 
\end{align*}
Is $\varphi$ a continuous operator ?
Edit: $g(f)$ belongs to $L^1(\Omega,\mathscr{f},\mu)$ for any $f \in L^1(\Omega,\mathscr{f},\mu)$.
 A: This operator isn't well-defined for many continuous functions $g: \mathbb R\to \mathbb R$. Let $\Omega = \mathbb R$ with the Lebesgue $\sigma$-algebra $\mathcal L$ and the Lebesgue measure $\lambda$. Take, for example, $$g(x) = \sqrt{\lvert x \rvert}, \,\,\,\,\,\,\, \text{ or } \,\,\,\,\,\, g(x) =e^x,$$ and let $$f(x) = \frac{1}{1+x^2}.$$ Now certainly $f \in L^1(\mathbb R, \mathcal L, \lambda),$ but $g\circ f \notin L^1(\mathbb R, \mathcal L, \lambda)$ for any of the above $g$.
A: Here's a proof sketch under the assumption that $(\Omega,\mu)$ is atomless.  Suppose without loss of generality that it is a probability measure.
I claim that the function $g$ must have linear growth, i.e. there is a constant $C$ such that $|g(t)| \le C(1+|t|)$ for all $t$.  Suppose not.  Because of the assumption that $\mu$ is atomless, we can find disjoint measurable sets $A_n$ with $\mu(A_n) = 2^{-n}$.  Then if $g$ does not have linear growth, we can find an increasing sequence of numbers $t_n$, $|t_n| \ge 1$, with $|g(t_n)| \ge 2^n |t_n|$.  Again by atomlessness, we can choose a measurable set $B_n \subset A_n$ with $\mu(B_n) = 1/(2^n|t_n|)$.  Now let $f = \sum_n t_n 1_{B_n}$.  We can easily check that $\int |f|\,d\mu = \sum_n |t_n| \mu(B_n) = \sum_n 1/2^n = 1$, but $\int |g(f)| \,d\mu = \sum_n |g(t_n)| \mu(B_n) \ge \sum_n 1 = \infty$.  Then $f \in L^1$ but $g(f) \notin L^1$, contradicting our hypothesis.
This being the case, suppose $f_n \to f$ in $L^1$.  Passing to a subsequence, we can also assume $f_n \to f$ almost everywhere.  Then by continuity of $g$ we have $g(f_n) \to g(f)$ almost everywhere.  Set $h_n = C(1+|f_n|)$ and $h = C(1+|f|)$, all of which are $L^1$, and note that $\int h_n = C(1 + \int |f_n|) \to C(1+\int |f|) = \int h$ by the assumption of $L^1$ convergence.  Now because of linear growth, $|g(f_n)| \le h_n$.  Using a variant of the dominated convergence theorem, we conclude that $\int |g(f_n)| \to \int |g(f)|$, and as a result, $g(f_n) \to g(f)$ in $L^1$.
