Let $$u_t -\Delta u = f$$ $$u(0) = u_0$$ where $f \in L^2(0,T;V^*)$ and $u_0 \in L^2(\Omega)$ with $V=H^1(\Omega)$.
Define the map $S:H \to \{ u \in L^2(0,T;V) : u_t \in L^2(0,T;V^*)\}$ by $S(u_0) = u$, i.e. it takes the initial condition and gives the solution.
GHow do I show that this map is continuous?
The map is not linear so boundedness does not help.
Let $u^n_0 \to u_0$ in $H$ where the $u^n_0$ are initial data such that $$u^n_t - \Delta u^n = f$$ $$u^n(0) = u^n_0$$
i.e. $S(u^n_0) = u^n$. Likewise, let $S(u_0) = u$. We need to show that $S(u^n_0) \to S(u_0)$. Consider the difference of the PDEs $u^n$ and $u$ satisfy: $$(u^n-u)_t - \Delta (u^n - u) = 0$$ $$(u^n-u)(0) = u^n_0 - u_0$$ From standard theory, this PDE has unique solution that depends continuously on data: $$|u^n-u| \leq C|u_0^n-u_0|$$ but the RHS tends to $0$ as $n\to \infty$ by assumption. So we have shown that $S$ is (Lipschitz and) continuous.
Is this a correct proof?