WHY?The Squared Euclidean Imbalance are equal to Fourier coefficients

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I'm reading the classical paper about distinguishing attack, How Far Can We Go Beyond Linear Cryptanalysis ,Thomas Baign`eres, Pascal Junod, and Serge Vaudenay. The only proposition I don't understand is Proposition 9. I post the Proposition 9 first, and then explain the notion.

Proposition 9. In the case where $D_{1}$ is the uniform distribution over $\mathcal{Z}=\left \{0,1 \right \}^{l}$, the SEI(Squared Euclidean Imbalance) and the Fourier coefficients are related by:$ \Delta \left ( D_{0} \right )=\sum_{u\in Z}^{}\hat{\epsilon _{u}}^{2} $.

Here $D_{0}\left [ z \right ]=\frac{1}{\left | \mathcal{Z} \right |}+\epsilon _{z} , \Delta \left ( D_{0} \right )=\left| \mathcal{Z} \right |\sum_{z\in\mathcal{Z}}^{}\epsilon _{z}^{2} $.

And We define the Fourier transform of $D_0$ at point $u∈\mathcal{Z}$ as $\hat{\epsilon }_{u}=\sum_{z\in\mathcal{Z}}^{}\left ( -1 \right )^{u\cdot z}\epsilon _{z}$.

The involution property of the Fourier transform leads to $\epsilon _{z}=\frac{1}{2^{l}}\sum_{u\in\mathcal{Z}}^{}\left ( -1 \right )^{u\cdot z}\hat{\epsilon }_{u}$.

So the Proposition 9 means $\left| \mathcal{Z} \right |\sum_{z\in\mathcal{Z}}^{}\epsilon _{z}^{2}=\sum_{u\in Z}^{}\hat{\epsilon _{u}}^{2}$,BUT WHY?

Here is the paper.