[proofplan]
We verify both identities by direct computation from the definition of the [Fourier transform](/page/Fourier%20Transform). For the translation identity, we substitute $z = x - a$ in the Fourier integral and factor the resulting exponential. For the modulation identity, we combine the exponentials $e^{ia \cdot x}$ and $e^{-ix \cdot \xi}$ into a single phase $e^{-ix \cdot (\xi - a)}$, recognizing the result as $\hat{f}(\xi - a)$.
[/proofplan]
[step:Compute the Fourier transform of the translated function $T_a f$]
By definition of the [Fourier transform](/page/Fourier%20Transform), with $T_a f(x) = f(x - a)$,
\begin{align*}
\widehat{T_a f}(\xi) &= \int_{\mathbb{R}^n} f(x - a) \, e^{-ix \cdot \xi} \, d\mathcal{L}^n(x).
\end{align*}
Apply the substitution $z = x - a$, so $x = z + a$ and $d\mathcal{L}^n(x) = d\mathcal{L}^n(z)$. The domain of integration remains $\mathbb{R}^n$ since translation is a bijection on $\mathbb{R}^n$. Substituting:
\begin{align*}
\widehat{T_a f}(\xi) &= \int_{\mathbb{R}^n} f(z) \, e^{-i(z + a) \cdot \xi} \, d\mathcal{L}^n(z).
\end{align*}
Factor the exponential using $e^{-i(z+a) \cdot \xi} = e^{-iz \cdot \xi} \, e^{-ia \cdot \xi}$, which holds by the multiplicativity of the complex exponential. The factor $e^{-ia \cdot \xi}$ is independent of $z$ and may be extracted from the [integral](/page/Integral):
\begin{align*}
\widehat{T_a f}(\xi) &= e^{-ia \cdot \xi} \int_{\mathbb{R}^n} f(z) \, e^{-iz \cdot \xi} \, d\mathcal{L}^n(z) = e^{-ia \cdot \xi} \, \hat{f}(\xi).
\end{align*}
[guided]
We want to compute $\widehat{T_a f}(\xi)$ where $T_a f(x) = f(x - a)$. By definition of the [Fourier transform](/page/Fourier%20Transform),
\begin{align*}
\widehat{T_a f}(\xi) &= \int_{\mathbb{R}^n} f(x - a) \, e^{-ix \cdot \xi} \, d\mathcal{L}^n(x).
\end{align*}
The presence of the shift $x - a$ inside $f$ suggests a change of variables to remove it. Apply the substitution $z = x - a$, so $x = z + a$ and $d\mathcal{L}^n(x) = d\mathcal{L}^n(z)$ (Lebesgue measure is translation-invariant). The integration domain remains $\mathbb{R}^n$ since the map $x \mapsto x - a$ is a bijection on $\mathbb{R}^n$. After substitution:
\begin{align*}
\widehat{T_a f}(\xi) &= \int_{\mathbb{R}^n} f(z) \, e^{-i(z + a) \cdot \xi} \, d\mathcal{L}^n(z).
\end{align*}
Now factor the exponential: $e^{-i(z+a) \cdot \xi} = e^{-iz \cdot \xi} \, e^{-ia \cdot \xi}$. The factor $e^{-ia \cdot \xi}$ does not depend on $z$, so we extract it from the integral:
\begin{align*}
\widehat{T_a f}(\xi) &= e^{-ia \cdot \xi} \int_{\mathbb{R}^n} f(z) \, e^{-iz \cdot \xi} \, d\mathcal{L}^n(z) = e^{-ia \cdot \xi} \, \hat{f}(\xi).
\end{align*}
The integral on the right is exactly $\hat{f}(\xi)$ by definition. The identity says that translating in space produces a phase shift $e^{-ia \cdot \xi}$ in the frequency domain.
[/guided]
[/step]
[step:Compute the Fourier transform of the modulated function $M_a f$]
By definition, with $M_a f(x) = e^{ia \cdot x} f(x)$,
\begin{align*}
\widehat{M_a f}(\xi) &= \int_{\mathbb{R}^n} e^{ia \cdot x} \, f(x) \, e^{-ix \cdot \xi} \, d\mathcal{L}^n(x).
\end{align*}
Combine the two exponentials: $e^{ia \cdot x} \, e^{-ix \cdot \xi} = e^{ix \cdot (a - \xi)} = e^{-ix \cdot (\xi - a)}$. Substituting:
\begin{align*}
\widehat{M_a f}(\xi) &= \int_{\mathbb{R}^n} f(x) \, e^{-ix \cdot (\xi - a)} \, d\mathcal{L}^n(x) = \hat{f}(\xi - a),
\end{align*}
where the final equality recognizes the integral as the [Fourier transform](/page/Fourier%20Transform) of $f$ evaluated at $\xi - a$.
[guided]
We want to compute $\widehat{M_a f}(\xi)$ where $M_a f(x) = e^{ia \cdot x} f(x)$. By definition of the [Fourier transform](/page/Fourier%20Transform),
\begin{align*}
\widehat{M_a f}(\xi) &= \int_{\mathbb{R}^n} e^{ia \cdot x} \, f(x) \, e^{-ix \cdot \xi} \, d\mathcal{L}^n(x).
\end{align*}
The key observation is that the two exponentials can be combined. Since both are functions of $x$, we write $e^{ia \cdot x} \, e^{-ix \cdot \xi} = e^{ix \cdot a - ix \cdot \xi} = e^{-ix \cdot (\xi - a)}$. No change of variables is needed here -- the exponentials simply merge:
\begin{align*}
\widehat{M_a f}(\xi) &= \int_{\mathbb{R}^n} f(x) \, e^{-ix \cdot (\xi - a)} \, d\mathcal{L}^n(x).
\end{align*}
The right-hand side is the Fourier transform of $f$ evaluated at the shifted frequency $\xi - a$:
\begin{align*}
\widehat{M_a f}(\xi) &= \hat{f}(\xi - a).
\end{align*}
This identity is dual to the translation identity: modulating by a frequency $a$ in space translates the Fourier transform by $a$ in the frequency domain.
[/guided]
[/step]
