Showing $\{u_n\}$ converge uniformly to solution of heat equation with lower order term 
Given a bounded continuous function $f(x),$ define the sequence of functions $\{u_n\}$ by
  $$ (u_n)_t - \Delta u_n = u_{n-1}^2, \;\;\; u_n(0,x) = f(x).
$$
  Show that for $T$ sufficiently small, this sequence converges uniformly to the solution of $u_t = \Delta u + u^2$ with $u(0,x) = f(x)$ on $x \in \mathbb{R}$ and $0 \leq t \leq T.$
Here, it may be useful to let $K(t,x)$ be the fundamental solution for the heat equation with the estimate
  $$\sup_{x\in\mathbb{R}}|u(x,t)| \leq \sup_{x\in\mathbb{R}}|u(x,0)| \;\text{ where } \; u(x,t) = \int_\mathbb{R} K(t,x-y)u(0,y) \;dy.
$$ 

I've been pretty stuck on how to begin the problem.  The only similar looking problems I've seen like this have been for ODEs, but so far none of that intuition has helped me here.  Any suggestions on how to start such a problem, or what methods could be used to show this result? 
Any suggestions, outlines of how to approach the problem, full solutions, anything would be greatly appreciated. Thanks in advance.
 A: The idea here is to treat the non-linear term $u^2$ as if it was a forcing term (rather than use $u^2$, we can use $g(u)$ for any locally Lipschitz $g$). Indeed, the fundamental solution for the heat equation gives $$u_n(x,t) = \int_0^t\int_{\mathbb R^n} K(s-t,x-y)g(u(s,y))dyds + \int_{\mathbb R^n} K(t,x-y)f(y) dy.$$ Let $V = \{v \in C([0,T]\times \mathbb R^n) : \| v - f \|_{\infty,\infty} \le \epsilon\}$ and define \begin{align*} &H:V\to V \\ u &\mapsto \int_0^t \int_{\mathbb R^n} K(s-t,y-x)u(s,y)^2 dyds + \int_{\mathbb R^n} K(t,y-x) f(y) dy. \end{align*} The goal is to show that $H$ is a contraction. Then by the Banach Fixed point theorem, the above iteration converges to some fixed point of $H$ which is a solution of the original equation. For $u,v \in V$, $x \in \mathbb R^n$ and $t \in [0,T_0]$ for some $T_0 > 0$, we see
\begin{align*}
\lvert (Hu)(x,t) - (Hv)(x,t)\rvert &= \left \lvert \int^t_0 \int_{\mathbb R^n} K(s-t,x-y) [g(u(s,y)) - g(v(s,y))] dy\,ds 
\right \rvert\\
&\le \int^t_0 \int_{\mathbb R^n} K(s-t,y-x)\lvert g(u(s,y)) - g(v(s,y)) \rvert dy\,ds \\
&\le L\int^t_0 \int_{\mathbb R^n} K(s-t,y-x) \lvert u(s,y) - v(s,y) \rvert dy ds\\
&\le  L \| u -v \|_{\infty,\infty} \int^t_0 \int_{\mathbb R^n} K(s-t,y-x) dy \,ds \\ &\le LT_0 \| u -v \|_{\infty,\infty}
\end{align*} where $L$ is the Lipschitz constant for $g$ on the interval $[\| f\|_{\infty} - \epsilon, \| f\|_{\infty} + \epsilon]$. Thus, passing to the supremum in the above, we have $$\|Hu - Hv \|_{\infty,\infty} \le LT_0 \| u - v \|_{\infty,\infty}.$$  Now if we take $T_0$ small enough (e.g. $T_0 = \frac 1{2L}$), we get that $H$ is a contraction and thus has a unique fixed point. This fixed point will solve the equation. However, we aren't quite done because the original problem asked us to prove existence and uniqueness for arbitrary $T \ge 0$ and we have only prove this up to some small $T_0$. However, you can repeat the process starting at $T_0$ and get the same result on $T_0 + \delta$ which shows that the set $\{T_0\in [0,T] : H \text{ is a contraction so long as } t \in [0,T_0] \}$ is open. It is not difficult to show that this set is also closed and thus since it is non-empty $$\{T_0\in [0,T] : H \text{ is a contraction so long as } t \in [0,T_0] \} = [0,T].$$ This gives existence and uniqueness up to $T$.
