Suppose $f\in L^{1}(\mathbb{R}^{n}\rightarrow \mathbb{C})$. What do we know about
$$F(y_{1},\cdots,y_{n-1},x):=\int_{\mathbb{R}} e^{\dot{\imath} xy_{n}} f(y_{1},\cdots,y_{n-1},y_{n})\,dy_{n} $$
Can we still say $x\mapsto F(y_{1},\cdots,y_{n-1},x)$ is uniformly continuous and $$\|F(y_{1},\cdots,y_{n-1},x)\|_{L_{x}^{\infty}(\mathbb{R})}\leq \|f(y_{1},\cdots,y_{n-1},y_{n})\|_{L_{y_{n}}^{1}(\mathbb{R})}.$$
More importantly, does the one-dimensional Fourier transform of $y_{i}\mapsto F(y_{1},\cdots,y_{n-1},x)$ for some $i\in \{1,\cdots,n-1\}$ still make sense in $L^{1}(\mathbb{R})$.
This is trivially true if $f(y_{1},\cdots,y_{n})$ is a tensor product $f_{1}(y_{1})\cdots f_{n}(y_{n})$. Is it true in general and why ?
Of course. By Fubini, $f(y_1, \ldots, y_{n-1}, \cdot)$ is in $L^1(\mathbb R)$ for almost every $y_1, \ldots, y_{n-1}$, with $$\| f\|_{L^1} = \int_{\mathbb R^{n-1}} \|f(y_1, \ldots, y_{n-1}, \cdot)\|_{L^1(\mathbb R)} $$ $F(y_1, \ldots, y_{n-1}, \cdot)$ is the Fourier transform of $f(y_1, \ldots, y_{n-1}, \cdot)$ and so the usual results for Fourier transform of functions of one variable apply, in particular $ \|F(y_1, \ldots, y_{n-1}, \cdot)\|_\infty \le \|f(y_1, \ldots, y_{n-1},\cdot)\|_1$.