Let $\omega > 0$ and $f \in L_1(\mathbb{R}) \cap C(\mathbb{R})$ be a function that is bandlimited on $[-\omega,\omega]$, i. e. $f^\wedge(x) = 0$ for all $|x|>\omega$. Here, $f^\wedge$ denotes the Fourier transform
$$f^\wedge(x) = \int_{-\infty}^{\infty} f(t) \exp(-ixt) dt$$
of $f$.
As homework, we need to show that the function $f \operatorname{sinc}(\omega (x- \cdot))$ is bandlimited on $[-2\omega, 2\omega]$ for all $x \in \mathbb{R}$, i. e.
$$(f \operatorname{sinc}(\omega (x- \cdot)))^\wedge(2 \omega y) = 0$$
for all $|y|>1$. Here, $\operatorname{sinc}(x) = sin(x)/x$ for $x \neq 0$ and $\operatorname{sinc}(0)=1$.
We were told to make use of the fact that
$$(\operatorname{sinc}(a \cdot + b))^\wedge (x) = \begin{cases} \frac{\pi}{a} \exp(i \frac{b}{a} x) & , |x|< a, \\ \frac{\pi}{2a} \exp(i \frac{b}{a} x) & , |x| = a, \\ 0 & , |x|>a. \end{cases}$$
where $a,b \in \mathbb{R}$ with $a > 0$ (cf. here).
I know that if $f$ is continuous and bandlimited, then there is a result in our lecture that says $f(t) = \frac{a}{\pi} (f * \operatorname{sinc}(a \cdot))(t).$ Here,
$$(f * g)(t) = \int_{-\infty}^\infty f(t-\tau) g(\tau) d\tau$$
denotes the convolution of $f$ and $g$.
Another consequence of the properties of $f$ is $f^\wedge \in L_1(\mathbb{R})$, which means that $$f = (f^\wedge)^\vee,$$ which we have already shown in our lecture. Here, $f^\vee(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(t) \exp(ixt) dt$ denotes the inverse Fourier transform.
However, I am stuck here and I have no idea about how to make use of these results.
Could anyone please help me here?