The following question was part of an advanced calculus of variations class. A solution was presented, being different and way more simple then the proof being presented in the literature.
Let $\Omega \subset \mathbb{R}^n$ be an open, bounded and connected set. Let $g: \bar{\Omega} \times \mathbb{R} \to \mathbb{R}$ be a function such that
$$
g\in C^0(\bar{\Omega} \times \mathbb{R})
$$
as well as a growth estimate for all $x \in \bar{\Omega}$
$$
|g(x,t)|\leq C_1+C_2|t|^{r/p}
$$
Proof that the function $h: L^r(\Omega)\to L^p(\Omega), \phi \to g(\cdot,\phi(\cdot))$ is continuous.
The proof presented included the following steps:
For an arbitrary $\phi_0 \in L^r(\Omega)$ and $z \in L^r(\Omega)$, consider the function $f:L^r(\Omega)\to L^p(\Omega), z \to g(\cdot,\phi_0+z)-g(\cdot,\phi_0)$ and study its continuity at $z(\cdot)=0$. Note that $f(\cdot,0)=0$. Using the growth estimate, one can derive a similar estimate for $f$ of the form:
$$
|f(x,z(x)|\leq A_1+A_2 |\phi_0(x)|^{r/p}+A_3|z(x)|^{r/p}
$$
So far, so good. Up until this point, I am confident the solution is correct.
In the last steps, I have my doubts if it is really that simple:
Now, taking any sequence $z_n \xrightarrow{L^r}0$, we have that $z_n \to 0$ pointwise a.e. and using the estimate for $f$ from above
$$
\int_{\Omega} |f(x,z_n(x))|^p \leq \int_{\Omega}D_1+D_2 |\phi_0(x)|^{r}+D_3|z_n(x)|^{r} \leq \infty
$$
the dominated convergence theorem can be applied. This should yield the result by using the continuity of $g$:
$$
\lim_{n \to \infty} \int_{\Omega} |f(x,z_n(x))|^p=\int_{\Omega} \lim_{n \to \infty}|f(x,z_n(x))|^p=\int_{\Omega} |f(x,0)|^p
$$
Can somebody please double check the last steps if the solution is actually correct?
The original solution suggested to split up the integral $\int_{\Omega} |f(x,z_n(x))|^p$ in 2 parts: One part of $\Omega$, where the integrand is small due to $z_n$ being small in the $L^r$ sense, and the other part of $\Omega$ ,where the grwoth estimates could be used, but the set where $|z_n| \geq \delta$ was small.
Appendix: As pointed out in the answer, the argument was flawed in the sense that we assumed $z_n \to 0$ a.e pointwise. However, this is only correct for a subsequence. There seems to be no straight-forward way to proof that $\int_{\Omega} \lim_{n \to \infty}|f(x,z_n(x))|^p=\int_{\Omega} |f(x,0)|^p$ without splitting up $\Omega$ into different parts. Convergence in measure does require such a split.
in one part of your proof you have said:
$z_n \to 0$ in $L^r$ then $z_n \to 0$ pointwise.
That's not the case unfortunately, there is one standard counter example to this statement called the "Typewriter sequence":
consider the sequence of indicator functions defined on the interval $[0,1]$ $f_n=I[\frac{n-2^k}{2^k},\frac{n-2^k+1}{2^k}]$ whenever $k \geq 0$ and $2^k \leq n \leq 2^{k+1}$.
You have that $f_n$ converges to $0$ in $L^1$ norm but $f_n$ doesn't converges to $0$ pointwise almost everywhere.
So maybe that's the why the author took that long way for the proof.
Best regards,
Andrea