Q:

Does $(1+|\cdot|^2)^{k/2} f\in L^2(\mathbb R^n)$ imply $f^\vee\in C^k(\mathbb R^n)$ and $\partial^\alpha f^\vee\in L^2(\mathbb R^n)$?

Let $f$ be a measurable function such that $(1+|\cdot|^2)^{k/2}f\in L^2(\mathbb R^n)$ where $k\in\mathbb N_0=\mathbb N\cup\{0\}$. This implies $f\in L^2(\mathbb R^n)$ for 
\begin{align*}
\displaystyle \int_{\mathbb R^n} |f(x)|^2\ dx&=\int_{\mathbb R^n} (1+|x|^2)^{-k} (1+|x|^2)^k|f(x)|^2\ dx\\
&\leq \int_{\mathbb R^n} (1+|x|^2)^k|f(x)|^2\ dx\\
&<\infty,
\end{align*}
for $(1+|x|^2)^{-k}\leq 1$. Is this enough to ensure $f^\vee\in C^k(\mathbb R^n)$ and $\partial^\alpha f^\vee\in L^2(\mathbb R^n)$ for all $|\alpha|\leq k$?
Notation: $f^\vee$ denotes the inverse Fourier transform of $f$, that is, $$f^\vee(x):=\int_{\mathbb R^n} e^{2\pi ix\cdot \xi} f(\xi)\ d\xi.$$

A:

The $k$-th derivatives are only guaranteed to be in $L^2$, not continuous. 
To get continuity, you need integration parallel to each coordinate axis. Thus you get $\check f\in C^{k-n}$. 
Using Cauchy-Schwarz there is a more general result in that $H^{r+n/2}$ embeds into $C^r$. For instance, functions in $H^1(\mathbb R)$ are continuous and even Hölder continuous with exponent $1/2$.

