In my introductory complex analysis course I am given some particularly phrased theorems. One of these is apparently the equivalent of the Dominated Convergence Theorem. The way it is phrased is:
$$\textrm{Let} \ f:]a,b[\times\mathbb{R}\rightarrow\mathbb{R}, \ f=f(x,\sigma) \ \textrm{for} \ a<x<b \ \textrm{and}\ \sigma \in \mathbb{R}$$
$$ \textrm{If} \ f(x,\cdot)\in C^0(]a,b[)\ \textrm{and}\ \exists h(x)\in L^1 \ \textrm{such that}: $$
$$sup\left|f(x,\sigma)\right|\le h(x)$$Then $$ F(\sigma)=\int_{-\infty}^{\infty} f(x,\sigma) dx\in C^0(\mathbb{R})$$
I am trying to apply this theorem to prove that the Fourier transform of an $L^1$ function is continuous. What I have done so far is to affirm that
$$f(x,\sigma)=f(x)e^{-i\sigma x}$$ where $\sigma$ is the frequency in the traditional definition of the Fourier transform. It is clear to me that:
$$sup\left|f(x,\sigma)\right|\le \left|f(x)\right|$$ and $\left|f(x)\right|$ is $L^1$ by hypothesis. However the part missing is that $f(x,\cdot)\in C^0(\mathbb{R})\ $. With this I would be able to conclude according to the theorem given to me.
Can anyone clarify this?
ANSWER: continuity is required with respect to the variable not being integrated over. Therefore $f(x,\sigma)$ must be continuous with respect to $\sigma$ and $L^1$ with respect to $x$.
Let $\sigma_n \longrightarrow s_0$.Use the continuity of $e^{i \sigma x}=\cos({\sigma x})+i \sin({\sigma x})$ and then aplly Lebesgue's dominated convergence theorem for the sequence $g_n=f(x)e^{i\sigma_nx}$