# Prove a property using shift theorem and duality

I'm reading Lectures on the Fourier Transform and Its Applications and I'm going to prove shift theorem for the inverse Fourier transform using duality. According to the mentioned source, the duality equations are $$\mathcal{F}f = (\mathcal{F}^{-1}f)^{-} \tag{1}$$ $$\mathcal{F}f^- = \mathcal{F}^{-1}f \tag{2}$$ $$\mathcal{F}\mathcal{F}f = f^{-} \tag{3}$$ Where $$(\mathcal{F}f)(s) = \int_{-\infty}^{+\infty}e^{-2\pi i st}f(t)dt$$ $$(\mathcal{F}^{-1}f)(t) = \int_{-\infty}^{+\infty}e^{+2\pi i st}f(s)ds$$ And $$f^{-}(t) = f(-t)$$ Using shift theorem which says $$f(t) \leftrightarrow F(s) \implies f(t-b) \leftrightarrow e^{-2\pi isb}F(s) \tag{4}$$ And duality equations I want to show $$F(s) \leftrightarrow f(t) \implies F(s-b) \leftrightarrow e^{2\pi itb}f(t) \tag{5}$$ First of all $$\text{(5)}$$ seems ambiguous to me but I guess it's $$\mathcal{F}\{ e^{2\pi itb}f(t)\} = F(s - b) \tag{6}$$ It's straightforward to show that directly $$\mathcal{F}\{ e^{2\pi itb}f(t)\} = \int_{-\infty}^{+\infty}e^{-2\pi i st}e^{2\pi itb}f(t)dt = \int_{-\infty}^{+\infty}e^{-2\pi it(s-b)}f(t)dt = F(s-b)$$ But I don't know how to combine duality equations and shift theorem to prove $$\text{(5)}$$. There is another version of duality which can be found for example here. It says $$x(t) \leftrightarrow X(s) \implies X(t) \leftrightarrow x(-s) \tag{7}$$ Is $$\text{(7)}$$ same as duality equations i.e. equations $$\text{(1)}, \text{(2)}$$ and $$\text{(3)}$$ or they are different from each other?

Edit: Equation $$\text{(6)}$$ is called modulation theorem. Also it seems in $$\text{(5)}$$ by $$F$$ we mean $$\mathcal{F}^{-1} f$$. So it's completely different from $$\text{(6)}$$. I'm really confused by these similar equations.

Edit2: I realized that the other version of duality is same as $$\text{(3)}$$. I mean the equations $$\text{(3)}$$ and $$\text{(7)}$$ are identical.

If $$X(f)=\mathcal{F}\{x(t)\}$$ you have, in analogy with Eq. $$(7)$$,

$$x(f)\Longleftrightarrow X(-t)\tag{1}$$

where we assume that functions with the independent variable $$f$$ are Fourier transforms of the corresponding functions with independent variable $$t$$.

Applying $$(1)$$ to

$$x(t-b)\Longleftrightarrow e^{-2\pi jbf} X(f)\tag{2}$$

gives

$$x(f-b)\Longleftrightarrow e^{2\pi jbt} X(-t)\tag{3}$$

According to $$(1)$$, $$X(-t)$$ is the inverse Fourier transform of $$x(f)$$, so with $$g(t)=X(-t)$$ and $$G(f)=\mathcal{F}\{g(t)\}=x(f)$$ you obtain the desired result:

$$G(f-b)\Longleftrightarrow e^{2\pi jbt} g(t)\tag{4}$$

• Sorry, I'm still confused. If $X(f)=\mathcal{F}\{x(t)\}$ then what do you mean by $x(f)\Longleftrightarrow X(-t)$? Aug 9 '20 at 18:08
• @S.H.W: It means that $x(f)$ is the Fourier transform of $X(-t)$, which is a consequence of $X(f)$ being the Fourier transform of $x(t)$. Aug 9 '20 at 19:20
• I see. Thank you very much. Do you think in addition to $\text{(7)}$ we can consider equations $\text{(1)}$ and $\text{(2)}$, which I've quoted, as duality equations? Aug 9 '20 at 19:51
• @S.H.W: I guess you could say so. Aug 9 '20 at 20:46