$x(n)={1,2,3,4}$

If $x(n) (↔)^{DFT} X(k)$ & $y(n)(↔)^{DFT} Y(k)$then

$∑_{n=0}^{N-1}x(n).y^* (n)=\frac{1}{N} ∑_{n=0}^{N-1}X(k).Y^* (k)$

Proof:-We have

$r_{xy} (m)=∑_{n=0}^{N-1}x(n).y^* (n-m)_N$

At

$m=0 ,r_{xy} (m)=∑_{n=0}^{N-1}x(n).y^* (n)$ ………(1)

By DFT, we have

$DFT{r_{xy} (m) }=X(K).Y^* (k)$

$r_{xy} (m)=IDFT{X(k).Y^* (k)}$

By IDFT eqn

$r_{xy} (m)=\frac{1}{N} ∑_{k=0}^{N-1}X(k).Y^* (K).e^{\frac{j2πkm}{N}}$

at m=0

$r_{xy} (0)=\frac{1}{N} ∑_{k=0}^{N-1}X(k).Y^* (k)$ ……………(2)

By comparing …