What you write is true for band-limited periodic functions. If you know that a function is periodic with period $T$ and its Fourier representation contains no frequencies higher than $\nu$, then the function is completely determined either by the Fourier coefficients for frequencies up to $\nu$ or by as many function values at points dividing any interval of length $T$ into equal intervals (where "as many" can be taken literally if you're dealing with a complex-valued function, and has to be spelled out appropriately if you're dealing with real-valued functions). This is related to the Nyquist–Shannon sampling theorem. You can transform between these two representations of the function using the discrete Fourier transform.
I've never seen this specifically referred to as a "dual problem"; with regard to viewing the relationship between a function and its Fourier transform as a duality, this section in Wikipedia and this one may be of interest to you.