Let $x[n]$ be a discrete time signal with DFT given by $X(f)=\sum_n x[n]e^{-2\pi inf}$ supported on $[-1/2M,1/2M]$ with $f\in[-1/2,1/2]$.
I can then down-sample to get $y[n]:=x[nM]$. Then, let
$$\widetilde{x}[n]=\begin{cases}My[n/M],& M|n,\\0,&\text{otherwise}.\end{cases}$$
Then its DFT is given by
$$ \begin{aligned}\widetilde{X}(f)&=\sum_{n\in\mathbb{Z}}\widetilde{x}[n]e^{-2\pi inf}\\&=\begin{cases}M\sum_{n\in\mathbb{Z}}y[n/M]e^{-2\pi inf},& M|n,\\0,&\text{otherwise}\end{cases}\\&=\begin{cases}M\sum_{n}x[n]e^{-2\pi inf},&M|n,\\0,&\text{otherwise}.\end{cases}\end{aligned} $$
Now, let $\hat{x}[n]=(\tilde{x}\ast h)[n]$ be the discrete Hilbert transform of $\tilde{x}$, with $h$ an "almost" ideal low-pass filter with cut-off frequency $f_c=1/M$.
My question is, how do I then apply the Shannon interpolation formula to reconstruct $x(t)$?
Intuitively, I would guess that it would be something along the lines of
$$x(t)=\left(\sum_{n\in\mathbb{Z}}x[n]\cdot\delta(t-n\Delta t)\right)\ast H(f),$$
with
$$H(f)=\begin{cases}\frac{DTFT\{\hat{x}[\cdot]\}(f)}{M\cdot X(f)},&\text{if }M|n, \\ 0,&\text{otherwise}. \end{cases}$$
Am I correct?