Finding some constant of a Fourier transform.

Question

Let $\lambda, \Omega>0$ and $\phi \in PW[-\Omega,\Omega]=\{\phi \in L^2(\mathbb{R}): \hat{\phi}=0 \;\;\text{outside}\;\; [-\Omega,\Omega]\}.$ Define $$f(x)=\frac{\phi(x)\sin(\lambda x)}{x}.$$ For all $x\in \mathbb{R\backslash\{0\}}$.

Q: Determine $\alpha>0$ such that $\hat{f}=0$ outside $[-\alpha,\alpha].$ Here $\hat{f} $ and $\hat{\phi}$ is the fourier transform of $f,\phi$.

I already shown that $f \in L^2(\mathbb{R}$). From this i don't know how to proceed it also hinted that use the fact that $g(z)=z^{-1}\sin(\lambda z)$ is entire. Any help would be much appreciated!!!!

Answer 1

Set $g_\lambda = 1_{[-\lambda,\lambda]}$ where $1_{[-\lambda,\lambda]}$ is the characteristic function on $[-\lambda,\lambda]$. If you define the Fourier transform of a function $f$ via $$ \hat f(k) = \int_\mathbb{R} f(x)e^{-2\pi i x k} dx $$ then by an easy calculation you can get the identity $$ \hat{g_\lambda}(k) = \frac{\sin(2\pi \lambda k)}{\pi k} =:h_\lambda(k) $$ ("Fourier transform of a characteristic function is the sinc function"). By symmetry we have $\hat{h_\lambda} = g_\lambda.$ It follow that the Fourier transform of the function $\frac{\sin(\lambda x)}{x}$ equals $$\pi 1_{[-\frac{\lambda}{2\pi},\frac{\lambda}{2\pi}]}.$$ Now simply use the fact that the Fourier transform of a product of functions is the convolution of the single Fourier transforms and conclude that $\Omega + \frac{\lambda}{\pi}$.