Consider a nonlinear and continuos function $f:\mathbb{R} \rightarrow \mathbb{R}$ and we define the functional
\begin{equation} F(u) = \int_{[0,1]^2} f(u(x,y)) dxdy \end{equation} where $u$ is an element of $L_p([0,1]^2)$.
I wonder if $F:L_p([0,1]^2) \rightarrow \mathbb{R}$ is a continuos function for $1 \leq p \leq \infty$.
Thanks in advance.
My answer here (when composition of continuous and Lebesgue integrable function Lebesgue integrable) shows that if a measurable function $g$ satisfies that $$ \Phi_g : L^1 \to L^1, h \mapsto g \circ h $$ is well-defined (i.e., if $g \circ h$ is integrable for every integrable $h$), then $|g(x)| \leq a + b|x|$ for certain $a,b \geq 0$ and all $x \in \Bbb{R}$.
Now, if $p < \infty$ and if your functional $F$ is well-defined, we can apply the above to $$ g(x) := f({\rm{sgn}}(x) \cdot |x|^{1/p}). $$ Indeed, if $h \in L^1(\Omega)$ (with $\Omega := [0,1]^2$), then ${\rm{sgn}}(h) \cdot |h|^{1/p} \in L^p(\Omega)$, so that (since $F$ is well-defined), we have $g \circ h \in L^1(\Omega)$.
By the property from above, this implies that there are $a,b \geq 0$ such that $|g(x)| \leq a + b|x|$ for all $x$. This easily yields $|f(x)| \leq a + b |x|^p$ for all $x \in \Bbb{R}$.
Now, since you also assume $f$ to be continuous, this actually implies that $F$ is continuous, too. To see this, assume that $F$ were not continuous. Thus, there is a sequence $g_n \to g$ (with convergence in $L^p$) and such that $F(g_n) \not \to F(g)$. By switching to a subsequence, we can assume that $|F(g_n) - F(g)| \geq \epsilon$ for all $n$ and some $\epsilon > 0$. By taking a further subsequence, we also get $g_n \to g$ almost everywhere.
By continuity, $f \circ g_n \to f \circ g$ almost everywhere. Now, we want to use the general form of the dominated convergence theorem as stated here (General Lebesgue Dominated Convergence Theorem). Note $$ |f(g_n(x))| \leq a + b |g_n(x)|^p =: h_n. $$ Furthermore, $h_n (x) \to h(x) := a + b|g(x)|^p$ pointwise a.e. and the $L^p$ conergence $g_n \to g$ easily implies $\int h_n \to \int h$. Hence, by the general dominated convergence theorem, we get $$ F(g_n) = \int f(g_n(x)) \, dx \to \int f(g(x)) \, dx = F(g), $$ a contradiction to $|F(g_n) - F(g)| > \epsilon$ for all $n$.$\square$
In case of $p=\infty$, the proof is easier. If $g_n \to g$ in $L^\infty$, we can change $g_n, g$ on a null-set to get $g_n \to g$ uniformly. Because of $g \in L^\infty$, we get $|g_n(x)| \leq C$ for all $n$ and (almost) all $x$. But $f$ is uniformly continuous on the set $[-C,C]$, so that we get $f \circ g_n \to f\circ g$ uniformly. Since $\Omega = [0,1]^2$ has finite measure, this entails $F(g_n) = \int f \circ g_n \to \int f \circ g = F(g)$. Thus, apart from continuity, we do not need to assume anything for $p=\infty$.
