Suppose we have $u\in C([0,T];L^1(\mathbb{R}^n))$. If we define $f(t)=\widehat{u(t)}$.
Is the map going from $[0,T]\to C([0,T];L^{\infty}(\mathbb{R}^n))$?
How can I prove the continuity of this map? I suppose that we need the norm $\displaystyle \lim_{t\to t_0}\lVert f(t)-f(t_0) \rVert=0$. So passing through the definition of $f$ we have to prove $\displaystyle \lim_{t\to t_0}\lVert \widehat{u(t)}-\widehat{u(t_0)} \rVert=0$ but I'm confused about the last norm. Any help will be appreciated.
The space $C([0,T]; L^1(\mathbb R^n))$ is comprised of continuous maps from $[0,T] \to L^1(\mathbb R^n)$; that is, if $u \in C([0,T]; L^1(\mathbb R^n))$, then $u(t) \in L^1(\mathbb R^n)$ for each $t \in [0,T]$.
The Fourier Transform on this space is then a map $\mathcal F: C([0,T]; L^1(\mathbb R^n)) \to C([0,T]; L^\infty(\mathbb R^n))$ so that $\hat u \in C([0,T]; L^\infty(\mathbb R^n))$ whenever $u \in C([0,T]; L^1(\mathbb R^n))$ [so to answer your first question: $\hat u(t)$ is a continuous map from $[0,T] \to L^\infty(\mathbb R^n)$]. To prove that $\hat u$ is indeed continuous, you simply need the inequality $$\| \hat u(t) \|_{L^\infty(\mathbb R^n)} \le \| u(t) \|_{L^1(\mathbb R^n)}$$ which shows that $$\lim_{t\to t_0}\|\hat u(t) - \hat u(t_0)\|_{L^\infty(\mathbb R^n)} =\lim_{t\to t_0} \| u(t) -u(t_0)\|_{L^1(\mathbb R^n)} = 0$$ where the latter limit is zero since $u \in C([0,T];L^1(\mathbb R^n))$.