Fourier tranform supported on interval yields continuity? Let $f\in L^2(\mathbb{R})$ be a function such that the Fourier transform $\hat{f}$ is supported on $[-\pi,\pi]$. That is, $$\hat{f}(y)=\int_{-\infty}^\infty f(x)e^{-ixy}dx$$ is supported on $[-\pi,\pi]$. 
I wonder if it's true that $f$ is always continuous?
 A: Caveat: Two functions differing only on a set of measure zero have the same Fourier transform, and the $L^2$-Fourier transform is only defined up to equality almost everywhere. So in the strictest sense, the answer is: No, $f$ need not be continuous.
However, the more interesting question is whether $f$ is equal almost everywhere to a continuous function, or whether the most regular representant of the class of $f$ is continuous.
The answer in that case is yes, and we have much much more. Since a compactly supported $L^2$ function is also in $L^1$, we have (for the most regular representant), the integral representation
$$f(x) = N\cdot \int_{-\pi}^\pi \hat{f}(y)e^{ixy}\,dy,$$
where $N$ is a normalisation factor depending on the used definition of the Fourier transform, with $\hat{f}(y) = \int_{-\infty}^\infty f(x)e^{-ixy}\,dx$, we would have $N = (2\pi)^{-1}$.
By the dominated convergence theorem, the integral depends continuously on $x$.
More, if we differentiate the integrand, we get
$$\left(\frac{d}{dx}\right)^k \left(g(y)e^{ixy}\right) = i^ky^k g(y)e^{ixy},$$
and the differentiated integrand is dominated by $\pi^k \lvert g(y)\rvert$ for all $x$, so by the dominated convergence theorem, $f$ is infinitely often differentiable, with
$$f^{(k)}(x) = i^kN\int_{-\pi}^\pi y^kg(y)e^{ixy}\,dy,$$
and since $y^kg(y) \in L^2(\mathbb{R})$, all derivatives of $f$ also belong to $L^2(\mathbb{R})$.
Yet more, for an $L^2$ function $g$ with support in the bounded interval $[a,b]$, consider the function
$$G(z) = \int_a^b g(y)e^{izy}\,dy,\quad z \in \mathbb{C}.$$
The integrand is measurable ($g$ is measurable, and $y \mapsto e^{izy}$ is continuous for every $z$), and
$$\int_a^b \left\lvert g(y)e^{izy}\right\rvert\,dy \leqslant \int_a^b \lvert g(y)\rvert e^{-y\operatorname{Im}z}\,dy \leqslant e^{\lvert z\rvert\max \{\lvert a\rvert,\lvert b\rvert\}}\int_a^b \lvert g(y)\rvert\,dy,$$
so $G$ is well-defined, and the dominated convergence theorem tells us it is continuous. Since the integrand is holomorphic in $z$, $G$ is holomorphic too (most easily seen by Morera's theorem, but one can also differentiate under the integral by basically the same argument as above, and verify the Cauchy-Riemann equations).
Thus $G$ is an entire function of exponential type,
$$\lvert G(z)\rvert \leqslant C\cdot e^{K\lvert z\rvert}$$
for some constant $C$ (can be chosen as $\int_a^b \lvert g(y)\rvert\,dy$) and some $K$, here $K = \max \{\lvert a\rvert,\lvert b\rvert\}$ is a possible choice.
The restriction of any derivative of $G$ to any line parallel to the real axis belongs to $L^2$, since for fixed $\eta$ we have
$$G(\xi + i\eta) = \int_a^b g(y) e^{i(\xi+i\eta)y}\,dy = \int_{-\infty}^\infty \left(g(y)e^{-\eta y}\right)e^{i\xi y}\,dy,$$
so $\xi \mapsto G(\xi+i\eta)$ is - modulo a normalisation factor - the inverse Fourier transform of the $L^2$-function $y\mapsto g(y)e^{-\eta y}$, hence belongs to $L^2(\mathbb{R})$.
Conversely, by one of the classical Paley-Wiener theorems, any entire function $F$ of exponential type $K$ such that the restrictions of $F$ to lines parallel to the real axis are square integrable, is the holomorphic Fourier transform of an $L^2$-function with support in $[-K,K]$.
A: The Fourier operator acting on $L^2(\mathbb R)$ is an invertible isometry. 
So $\mathcal{F}^{-1}(\mathcal{F}(f))=f$ and since the map $\mathcal{F}(f)$ is supported on $[-\pi,\pi]$, it belongs to $L^1(\mathbb R)$.
Indeed, using Cauchy-Schwarz in $L^2([-\pi,\pi])$, we get: 
$$\int_{\mathbb R}  |\mathcal{F}(f)(y)|dy = \int_{-\pi}^\pi |\mathcal{F}(f)(y)|dy \leq \sqrt{2\pi}\left(\int_{-\pi}^\pi |\mathcal{F}(f)(y)|^2dy \right)^{1/2} \leq \sqrt{2\pi}\|\mathcal{F}(f)\|_2$$ 
Now, you should know that the Fourier operator (or its inverse) sends $L^1(\mathbb R)$ to $\mathcal{C}_0(\mathbb R)$, the space of continuous functions vanishing at $\pm\infty$.
Hence $\mathcal{F}^{-1}(\mathcal{F}(f))$ has to be continuous i.e. $f$ has to be continuous.
