Let $Q_T=\mathbb{R}\times (0,T)$ and $X=\{u\in L^{\infty}(Q_T):\,\|u\|_{L^{\infty}(Q_T)}\leq1\}$. For $u\in X$, define $$N(u)(x,t)=\frac{1}{2}\int_0^t \int_{x-(t-s)}^{x+(t-s)}(u^2+g)(y,s)\,dy\,ds.$$ Suppose that $T<1$ and $\|g\|_{L^\infty(Q_T)}\leq1$. I proved that $N:X\rightarrow X$ is a contraction, and since $X$ is a complete metric space (because it is closed in $L^\infty(Q_T)$), by Banach fixed point theorem there exists a unique $u\in X$ such that $u=N(u)$.
I read that this implies that the nonlinear problem $$\begin{cases} u_{tt}-u_{xx}=u^2(x,t)+g(x,t),\;(x,t)\in Q_T \\ u(x,0)=0,\;x\in\mathbb{R} \\ u_t(x,0)=0,\; x\in\mathbb{R} \end{cases}$$ has a unique solution in $X$. By d'Alembert's formula I understand that if $u=N(u)$, then $u$ is the unique solution, but I think assuming that $f(x,t)=u^2(x,t)+g(x,t)$ satisfies $f$ and $f_x$ continuous (hypotheses for d'Alembert's formula). Thus, do we have to include some sort of continuity or regularity in the space $X$? Or from $u\in L^\infty(Q_T)$ and $u=N(u)$ we can deduce that $u\in C^2(Q_T)$, and therefore there are no regularity problems?
Let $f=u^2+g$. By d'Alembert's formula, we need $f$ and $f_x$ continuous on $\mathbb{R}\times[0,T]$.
Using $\|f\|_{L^{\infty}(Q_T)}<\infty$, $u=N(u)$ and dominated convergence, one has $u$ continuous on $\mathbb{R}\times[0,T]$.
Assume $g$ and $g_x$ continuous on $\mathbb{R}\times[0,T]$. As $f$ is continuous, $$\partial_x \int_{x-(t-s)}^{x+(t-s)}(u^2+g)(y,s)\,dy=(u^2+g)(x+(t-s),s)-(u^2+g)(x-(t-s),s).$$
Use the following result: if $H(x)=\int_a^b h(x,y)\,dy$, $h$ continuous and $h_x$ continuous on $[a,b]$, then there exists $H'(x)=\int_a^b h_x(x,y)\,dy$.
From the result and $u=N(u)$, there exists \begin{align*} u_x(x,t)= {} & \frac{1}{2}\int_0^t \left(\partial_x \int_{x-(t-s)}^{x+(t-s)}(u^2+g)(y,s)\,dy\right)\,ds \\= {} & \frac12\int_0^t \left((u^2+g)(x+(t-s),s)-(u^2+g)(x-(t-s),s)\right)\,ds \end{align*} continuous on $\mathbb{R}\times[0,T]$. Then there exists $f_x$ continuous on $\mathbb{R}\times[0,T]$.