I have a signal
$$x[0, \ldots , 2^n-1].$$ I would like to compute its DFT$$f[0, \ldots , 2^n-1],$$ take the first half $$f[0, \ldots , 2^{n-1}-1],$$ pad it with zeros,
$$g[i] = \left\{\begin{array}{cc}f[i] & \textrm{ if } i \in [0, \ldots , 2^{n-1}-1]\\ 0 & \textrm{otherwise} \end{array}\right., $$
and convolve that with some window $w$,
$$h = w \ast g.$$
Is there an easy way to use the DFT step to do the convolution for free? Of course I could $0$-pad $f$ and then do the convolution with two FFTs, for a total of three FFTs, but I'm wondering if I can get away with just one.