Let's say I have a continuous function $f \in {L^2}\left( { - \infty ,\infty } \right)$ for which it is known that: $$\forall \omega \notin \left[ {0,L} \right] \cup \left[ {3L,4L} \right]{\rm{ }} \to {\rm{ }}\widehat f\left( \omega \right) = 0$$ How can I reconstruct $f$ only based on the values $f\left( {{{n\pi } \over L}} \right)$?
Meaning I wish to find a series representation such that $$f\left( x \right) \sim \sum\limits_{n = - \infty }^\infty {f\left( {{{n\pi } \over L}} \right){h_n}\left( x \right)} $$ for some ${{h_n}\left( x \right)}$, where the convergence is in the ${L^2}\left( { - \infty ,\infty } \right)$ sense.
I am familiar with the Shannon Sampling Theorem which states that if a continuous function $$g \in {L^2}\left( { - \infty ,\infty } \right)$$ satisfies $$\forall \left| \omega \right| > L > 0{\rm{ }} \to {\rm{ }}\widehat g\left( \omega \right) = 0$$ Then $$g\left( x \right) = \sum\limits_{n = - \infty }^\infty {g\left( {{{n\pi } \over L}} \right){{\sin \left( {Lx - \pi n} \right)} \over {Lx - \pi n}}} $$ Where the equality is in the ${L^2}\left( { - \infty ,\infty } \right)$ sense.
I tried to follow the same proof our professor gave for Shannon's Theorem and get a similar result. Note: I'm using the following definition of the Fourier transform: $$\widehat f\left( \omega \right) = \int\limits_{ - \infty }^\infty {f(x){e^{ - i\omega x}}dx} $$
My attempt:
I tried to split $\widehat f\left( \omega \right)$ into two functions: $$\widehat {{f_1}}\left( \omega \right) = \widehat f\left( \omega \right) \cdot {1_{\left[ {0,L} \right]}}\left( \omega \right)$$ $$\widehat {{f_2}}\left( \omega \right) = \widehat f\left( \omega \right) \cdot {1_{\left[ {3L,4L} \right]}}\left( \omega \right)$$ and it's clear that $$\widehat f\left( \omega \right) = \widehat {{f_1}}\left( \omega \right) + \widehat {{f_2}}\left( \omega \right)$$
I then proceeded to compute the complex fourier series of $\widehat {{f_1}}\left( \omega \right)$ on $\left[ { - L,L} \right]$ and I got: $$\widehat {{f_1}}\left( \omega \right) \sim \sum\limits_{n = - \infty }^\infty {{\pi \over L}{f_1}\left( { - {{n\pi } \over L}} \right){e^{{{i\pi n\omega } \over L}}}} $$ Then the same for $\widehat {{f_2}}\left( \omega \right)$ on $\left[ {2L,4L} \right]$: $$\widehat {{f_2}}\left( \omega \right) \sim \sum\limits_{n = - \infty }^\infty {{\pi \over L}{f_2}\left( { - {{n\pi } \over L}} \right){e^{{{i\pi n\omega } \over L}}}} $$ By Parseval's formula: $${1 \over {2L}}\int\limits_{ - L}^L {{{\left| {\widehat {{f_1}}\left( \omega \right)} \right|}^2}d\omega } = \sum\limits_{n = - \infty }^\infty {{{\left| {{\pi \over L}{f_1}\left( { - {{n\pi } \over L}} \right)} \right|}^2}} $$ $${1 \over {2L}}\int\limits_{2L}^{4L} {{{\left| {\widehat {{f_2}}\left( \omega \right)} \right|}^2}d\omega } = \sum\limits_{n = - \infty }^\infty {{{\left| {{\pi \over L}{f_2}\left( { - {{n\pi } \over L}} \right)} \right|}^2}} $$ meaning: $${1 \over {2L}}\int\limits_{ - \infty }^\infty {{{\left| {\widehat f\left( \omega \right)} \right|}^2}d\omega } = {\left( {{\pi \over L}} \right)^2}\sum\limits_{n = - \infty }^\infty {{{\left| {{f_1}\left( {{{n\pi } \over L}} \right)} \right|}^2} + {{\left| {{f_2}\left( {{{n\pi } \over L}} \right)} \right|}^2}} $$ By Plancherel's theorem: $$\int\limits_{ - \infty }^\infty {{{\left| {f\left( x \right)} \right|}^2}dx} = {1 \over {2\pi }}\int\limits_{ - \infty }^\infty {{{\left| {\widehat f\left( \omega \right)} \right|}^2}d\omega } = {\pi \over L}\sum\limits_{n = - \infty }^\infty {{{\left| {{f_1}\left( {{{n\pi } \over L}} \right)} \right|}^2} + {{\left| {{f_2}\left( {{{n\pi } \over L}} \right)} \right|}^2}} $$
Here is where I got stuck.
a) I have no idea how to link $${{{\left| {{f_1}\left( {{{n\pi } \over L}} \right)} \right|}^2} + {{\left| {{f_2}\left( {{{n\pi } \over L}} \right)} \right|}^2}}$$ with $${{{\left| {f\left( {{{n\pi } \over L}} \right)} \right|}^2}}$$
I could use a convolution of $f$ with a sinc function to get ${{f_1}}$ but it seems it would only make things more complicated.
b) I don't really know how the series representation of $f$ should look like.
c) I had another approach in mind: I thought about defining $$\widehat h\left( \omega \right) = \left\{ \matrix{ \widehat f\left( \omega \right){\rm{ ,}}0 \le \omega \le L \hfill \cr \widehat f\left( {\omega + 4L} \right){\rm{ ,}} - L \le \omega \le 0 \hfill \cr} \right.$$ which appears to satisfy the conditions of the original Shannon sampling theorem but I wasn't able to procceed.
Sorry for the lengthy post, I thought that if I explained in as much detail as possible what I've tried to do in order to solve this problem it would increase the chances of response.
Any idea, hints, references would be welcome!