I haven't found a reference solution for this problem on the net, and I'm wondering whether my proof makes sense / is correct. Also I believe that it should be possible to find a simpler proof for the implication by using Vitali's Convergence theorem, but I'm not sure how. Any suggestions and comments are welcome.
Problem statement (adapted from original to make it self-contained; from R.G. Bartle 'The Elements of Integration', p.79, exercise 7.U):
Let $(X, \mathbf X, \mu)$ be a finite measure space and let $1 \leq p < \infty$. Let $\varphi$ be continuous on $\mathbb R$ to $\mathbb R$ and satisfy the condition: $(*)$ there exists $K > 0$ such that $|\varphi(t)| \leq K |t|$ for $|t| \geq K$. If $(f_n)$ converges to f in $L_p$ then $(\varphi \circ f_n)$ converges in $L_p$ to $\varphi \circ f$. Conversely, if condition $(*)$ is not satisfied, there exists a finite measure space and a sequence $(f_n)$ which converges in $L_p$ to $f$ but such that $(\varphi \circ f_n)$ does not converge in $L_p$ to $\varphi \circ f$.
Proof:
$\Longrightarrow$
(1. sequence and limit functions are in $L_p$ - implication in exercise 7.T)
Let $M = \max_{t \in [0, K]} |\varphi(t)|^p$; ($M < \infty$ as $[0, K]$ compact and $\varphi$ continuous)
$\Rightarrow$ $|\varphi \circ f|^p = |\varphi \circ f|^p (\chi_{|f|^{-1}([0, K[)} + \chi_{|f|^{-1}([K, \infty[)}) \leq M \chi_{|f|^{-1}([0, K[)} + K^p |f|^p \chi_{|f|^{-1}([K, \infty[)} \leq M + K^p |f|^p;$
$\Rightarrow$ $||\varphi \circ f||_p^p = \int |\varphi \circ f|^p \mathrm d \mu \leq \int M + K^p |f|^p \mathrm{d} \mu = M \mu(X) + K^p ||f||_p^p < \infty$; (by $\mu(X) < \infty$ and $f \in L_p$)$(\Rightarrow \varphi \circ f \in L_p)$
$(\mbox{by the same reasoning for } f_n \Rightarrow \varphi \circ f_n \in L_p, n \in \mathbb N)$(2. every subsequence has an $L_p$ convergent subsubsequence)
Let $\Lambda \subset \mathbb N$ be infinite;
$\Rightarrow (f_n)_{n \in \Lambda}$ converges in $L_p$ to $f$;
$\Rightarrow \exists \Lambda' \subset \Lambda$ infinite $ : f_n \rightarrow f \; \mu\mbox{-a.e.} \; (n \in \Lambda');$$\forall n \in \Lambda'$
$\Rightarrow \varphi \circ f_n \rightarrow \varphi \circ f \; \mu\mbox{-a.e.};$ (by continuity of $\varphi$)
$\Rightarrow \varphi \circ f_n \rightarrow \varphi \circ f$ almost uniformly; (by $\mu(X) < \infty$ and Egoroff's Theorem)Let $\delta > 0, E_{\delta} \subset X, \mu(E_{\delta}) < \delta : (\varphi \circ f_n |_{X \setminus E_{\delta}})$ converges uniformly to $\varphi \circ f |_{X \setminus E_{\delta}};$
on $X \setminus E_{\delta}$
$\Rightarrow \exists N \in \Lambda' : (n > N \Rightarrow |f_n - f| \leq 1)$;
Let $g = \sum_{n = 1}^{N} |f_n - f|^p + 1 \geq |f_n - f|^p$;
$\Rightarrow g \in L_1$; (by $\mu(X) < \infty$)
$\Rightarrow |f_n|^p = |f + (f_n - f)|^p \leq (2 \max\{|f|, |f_n - f|\})^p \leq 2^p(|f|^p + |f_n - f|^p) \leq 2^p(|f|^p + g)$; $\Rightarrow |\varphi \circ f_n|^p \leq M + K^p |f_n|^p \leq M + K^p 2^p (|f|^p + g) \in L_1;$ (again by $\mu(X) < \infty$)
$\Rightarrow ||\varphi \circ f_n||_p \rightarrow ||\varphi \circ f||_p$; (by Lebesgue Dominated convergence Theorem)$\Rightarrow ||\varphi \circ f_n||_p \rightarrow ||\varphi \circ f||_p \;$ on $X \setminus E$ where $E = \bigcap_{\delta > 0} E_{\delta}$;
$\Rightarrow \forall \delta > 0 \; E \subset E_{\delta} \Rightarrow \mu(E) \leq \mu(E_{\delta}) < \delta \Rightarrow \mu(E) = 0;$
$\Rightarrow ||\varphi \circ f_n||_p \rightarrow ||\varphi \circ f||_p \;$ on $X$;($\Rightarrow \forall \Lambda \subset \mathbb N $ infinite $\, \exists \Lambda' \subset \Lambda : (\varphi \circ f_n)_{n \in \Lambda'}$ converges in $L_p$ to $\varphi \circ f$);
(3. sequence converges in $L_p$)
Suppose $\varphi \circ f_n \nrightarrow \varphi \circ f$ in $L_p$;
$\Rightarrow \exists \epsilon >0 : \forall N \in \mathbb N \, \exists n > N : ||f_n - f||_p \geq \epsilon;$
Let $\Lambda = \{n \in \mathbb N \mid ||f_n - f||_p \geq \epsilon\};$
$\Rightarrow (\varphi \circ f_n)_{n \in \Lambda}$ is a subsequence which does not posses an $L_p$ convergent subsubsequence, contrary to what was proven before.
$(\Rightarrow \varphi \circ f_n \rightarrow \varphi \circ f$ in $L_p)$
$\Longleftarrow$
(1. construction of finite measure space and limit function - converse in exercise 7.T)
$\lnot (*) \Rightarrow \forall K > 0 \; \exists |t| \geq K: |\varphi(t)| > K |t|$;
Let $t_n \in \mathbb R, t_n > n, |\varphi(t_n)| > n |t_n|, n \in \mathbb N$;
Let $(X = \mathbb N, 2^{\mathbb N}, \mu)$ be a measure space with measure $\mu(E) = \sum_{n \in E} \frac 1 {{|t_n|}^p n^2}$;
$\Rightarrow \mu(X) = \sum^{\infty}_{n = 1} \frac 1 {{|t_n|}^p n^2} \leq \sum^{\infty}_{n = 1} \frac 1 {n^2} < \infty$ ($\Rightarrow$ the measure space is finite);
Let $f:\mathbb N \ni n \mapsto t_n$;
$\Rightarrow ||f||_p^p = \int |f|^p \mathrm d \mu = \sum_{n=1}^{\infty} \frac{|f(n)|^p} {{|t_n|}^p n^2} = \sum_{n=1}^{\infty} \frac 1 {n^2} < \infty;$
$\Rightarrow ||\varphi \circ f||_p^p = \int |\varphi \circ f|^p \mathrm d \mu = \sum_{n=1}^{\infty} \frac{|\varphi(f(n))|^p} {{|t_n|}^p n^2} = \sum_{n=1}^{\infty} \frac{|\varphi(t_n)|^p} {{|t_n|}^p n^2} \geq \sum_{n=1}^{\infty} \frac{1} {n^{(2-p)}} \geq \sum_{n=1}^{\infty} \frac{1} {n} = \infty;$
$(\Rightarrow f \in L_p \, \land \, \varphi \circ f \notin L_p)$(2. construction of sequence)
Let $f_n = (1 - \frac 1 n) f, \, n \in \mathbb N$;
$\Rightarrow ||f_n||_p = ||(1-\frac 1 n)f||_p = (1-\frac 1 n)||f||_p < \infty \Rightarrow f_n \in L_p \; \forall n \in \mathbb N;$
$\Rightarrow ||f_n - f||_p = \frac 1 n ||f||_p \rightarrow 0;$ ($f_n \rightarrow f$ in $L_p$)Suppose $\varphi \circ f_n \rightarrow \varphi \circ f$ in $L_p$;
$\Rightarrow \forall n \in \mathbb N \; ||\varphi \circ f||_p \leq ||\varphi \circ f_n||_p + ||\varphi \circ f_n - \varphi \circ f||_p < \infty;$(by Minkowski Inequality)
$\varphi \circ f \notin L_p \Rightarrow$ contradiction.
($\Rightarrow \varphi \circ f_n \nrightarrow \varphi \circ f$ in $L_p$)
$\square$