Fourier tranform's inverse is it's adjoint

${\mathcal{F}}^{†}={\mathcal{F}}^{-1}$, where the Fourier Transform $\mathcal{F}$ has an Adjoint Operator ${\mathcal{F}}^{†}$ and Inverse Operator ${\mathcal{F}}^{-1}$.

Proof

We will show that $⟨{\mathcal{F}}^{-1}f,g⟩=⟨f,\mathcal{F}g⟩\forall f,g$, where $⟨\cdot ,\cdot ⟩$ is the Inner Product

\begin{array}{rll} \braket{\mathcal{F}^{-1}f,g} &= \braket{\int_{\mathbb{R}}f(x)e^{2 \pi i t x}dx, g} & \text{By definition of the Inner Product in $L_2$} \\ &= \int_{\mathbb{R}} \overline{ \int_{\mathbb{R}}f(x)e^{2 \pi i t x}dx } g(t)dt & \text{Absorbing the complex conjugate} \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} \overline{ f(x) }e^{- 2 \pi i t x}dx g(t)dt$ & \text{Rearranging the expression} \\ &=\int_{\mathbb{R}} \overline{ f(x)} \int_{\mathbb{R} }e^{- 2 \pi i t x} g(t) dt dx$ & \text{By definition of the Fourier Transform} \\ &=\int_{\mathbb{R}} \overline{ f(x)} \mathcal{F} g(x) dx & \text{By definition of the Inner Product in $L_2$} \\ &=\braket{f,\mathcal{F} g} \end{array}

Since the $f,g$ were arbitrary, the above together with Riesz Representation Theorem implies that ${\mathcal{F}}^{†}={\mathcal{F}}^{-1}$.