Continuity of the Fourier Transform I came across this and in Lemma 2.1, it is proving the continuity of the Fourier Transform. From what I can understand, it seems that the continuity is based the difference of two Fourier Transforms of 2 sequences of Schwartz functions converges to zero. I understand the lemma explaining this but I still cant get a hold of the physical representation of the continuity of the Fourier Transform. Any explanation is appreciated thanks
 A: The Fourier Transform is a beast that takes some getting used to.  We begin by having a nice 'physical' intuition about the Fourier Transform on the space of square integrable periodic functions; i.e. $L^2(\mathbb{T^n})$ functions on the torus.  This gives us the Fourier Series which is a new function defined on $\mathbb{Z^n}$ that is square summable i.e. a member of $\mathscr{l}^2(\mathbb{Z^n})$.  The new function gives us a recipe for building the initial periodic function from complex exponentials.  Now the generalization of this process to functions that aren't necessarily periodic is what we usually refer to as the Fourier Transform.  i.e. given $f \in L^p(\mathbb{R}^n)$ can we build this function from complex sinusoids?  The answer is sometimes yes and sometimes no; we can always do this if  $p \in [1,2]$.
So where do you begin?  Well it turns out that the convenient place to start is $L^1(\mathbb{R}^n)$ and the FT will be consistent with the computation of the Fourier Coefficients and the formula is what most people call the FT. 
For $f\in L^1(\mathbb{R}^n)$:
$$\hat{f}(\xi)=\int_{\mathbb{R}^n}f(t)e^{-2\pi it \xi}dt$$
We can show that the map defined above will give: 
$$\mathscr{F}:L^1(\mathbb{R}^n)\rightarrow BC(\mathbb{R}^n)$$
with $BC(\mathbb{R}^n)$ the space of bounded continuous functions on $\mathbb{R}^n$.
Now, why Schwartz functions? Well for one they're easy to handle when we're dealing with integration. Additionally, for $p \in [1,\infty)$ the Schwartz Space $\mathscr{S}$ is dense in $L^p$. 
Also per your question, when we look at the effect of $\mathscr{F}$ on $\mathscr{S}$ it turns out to be a continuous operator.
So what about continuity? 
What does it mean to be continuous? 
Well when we have a metric space, the metric will define which sequences will converge and then we say a functions is continuous if it preserves the convergence of sequences; i.e. $x_n\to x \implies f(x_n) \to f(x)$.  Informally, close points get mapped to close values.  We don't even need a metric to define this, just a topology. 
When you investigate $\mathscr{S}$ you discover that it's a Fréchet Space and it comes with a topology (not a norm though), the topology is born from a family of seminorms. So to understand continuity of $\mathscr{F}$ on $\mathscr{S}$ you really just need to understand how continuity works for topological spaces. I probably said too much. I hope this helps.  Take Care 
