Non trivial condition for continuity of multivariable Fourier transform Assume $f: \mathbb{R}^2 \to \mathbb{C}$ is a function so that for every $x$ the function $\mathbb{R} \to \mathbb{C}: y \mapsto f(x,y)$ is integrable. 

What is a non-trivial sufficient condition on $f$ so that the function
  $$F: \mathbb{R}^2 \to \mathbb{C}\;\text{ with }\;F(x,t) = \int_{\mathbb{R}}f(x,y) e^{-2\pi iyt}dy$$ 
  is contiuous?

So I was thinking about constructing a sequence $F_n(x,t) = F(x_n,t_n)$ where $x_n \to x$ and $t_n \to t$. To make $F$ continuous, it needs to converge like $F_n \to F$. I was thinking about using the dominated convergence theorem to find a non-trivial condition for $f$ to make $F$ continuous. Because $f$ is integrable for every $x$ so also for every $x_n$, then only condition I can find to fulfill the dominated convergence theorem is that there needs to be a positive itegrable function $g: \mathbb{R}^n \to [0,\infty]$ so that:
$$|F_n(x,t)| \leq g(x,t)$$ 
So that
$$ \int_{\mathbb{R}}f(x,y)| \leq g(x,t)$$ 
But it seems to me that there needs to be a better or different condition.

If $\Phi : x \longmapsto f(x,\cdot) \in L^1(\mathbb{R})$ is continuous, then so is $F$. 
Indeed, for any $x,x’,t$, $|F(x,t)-F(x’,t)| \leq \|\Phi(x)-\Phi(x’)\|_1$.
So if $(x_n,t_n) \rightarrow (x,t)$, $|F(x_n,t_n)-F(x,t)| \leq \|\Phi(x_n)-\Phi(x)\| + |\hat{\Phi(x)}(t_n)-\hat{\Phi(x)}(t)|$ where both terms go to zero.
