Prove the inequalites of complex vectors Let $x(t), y(t)\in \mathbb{C}_{N}$, where
$$   \mathbb{C}_{N}=\{x:\mathbb{Z}\to \mathbb{C}|x(j)=x(j+N)~\text{for any}~j\in \mathbb{Z}\} $$
Define the functions
$$ R_{xy}(j)=\sum_{k=0}^{N-1}x(k)\overline{y(k-j)} $$
and
$$ R_{xx}(j)=\sum_{k=0}^{N-1}x(k)\overline{x(k-j)} $$
$$ R_{yy}(j)=\sum_{k=0}^{N-1}y(k)\overline{y(k-j)} $$
Show that
\begin{equation}
\sum_{j=0}^{N-1}|R_{xy}(j)|^2\geq R_{xx}(0)R_{yy}(0)-\left(\sum_{j=1}^{N-1}|R_{xx}(j)|^{2}  \right)^{\frac{1}{2}}\left(\sum_{j=1}^{N-1}|R_{yy}(j)|^{2}  \right)^{\frac{1}{2}}
\end{equation}
and
\begin{equation}
\sum_{j=0}^{N-1}|R_{xy}(j)|^2\leq R_{xx}(0)R_{yy}(0)+\left(\sum_{j=1}^{N-1}|R_{xx}(j)|^{2}  \right)^{\frac{1}{2}}\left(\sum_{j=1}^{N-1}|R_{yy}(j)|^{2}  \right)^{\frac{1}{2}}.
\end{equation}
When does the equality holds?
I have trying the Cauchy-Schwarz inequality, but I stuck here. Can someone help me with above inequalites? Thank you very much!
 A: Since the functions $R$ are convolutions, we might use the Fourier transform. Let's define it as
$$\hat{x}(k) := \frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}x(j)e^{-2\pi ijk/N}.$$
You may see that
$$\begin{align*}
(i)\quad &x(j) = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\hat{x}(k)e^{2\pi i jk/N}; \\
(ii)\quad &\lVert x\rVert_2 = \lVert \hat{x}\rVert_2; \\
(iii)\quad &\hat{R}_{xy}(k) = \sqrt{N}\hat{x}\bar{\hat{y}}(k).\end{align*}$$
By homogeneity we can assume that $\lVert \hat{x}\rVert_2 = \lVert \hat{y}\rVert_2 = 1$, and using these properties both inequalities get into
$$N\lVert \hat{x}\hat{y} \rVert_2^2 \geq 1-(N\lVert \hat{x}\rVert_4^4-1)^{1/2}(N\lVert \hat{y}\rVert_4^4-1)^{1/2}$$
and
$$N\lVert \hat{x}\hat{y} \rVert_2^2 \leq 1+(N\lVert \hat{x}\rVert_4^4-1)^{1/2}(N\lVert \hat{y}\rVert_4^4-1)^{1/2}.$$
Now we can think in terms of multiplication instead of convolution. We can write these equations more briefly, changing notation also, as
$$\lvert N\lVert fg \rVert_2^2-1\rvert \le (N\lVert f\rVert_4^4-1)^{1/2}(N\lVert g\rVert_4^4-1)^{1/2}. \qquad(*)$$
We don't lose anything if we think of $f$ and $g$ as positive functions.
Now we exploit some probabilistic intuition. Since $\lVert f\rVert_2 = 1$, we have that the mean value of $f$ is $1/\sqrt{N}$, so let's rewrite the left hand side of (*) as
$$N\Big(\sum f^2g^2 \Big) - 1 = N\sum\Big(f^2-\frac{1}{N}\Big)\Big(g^2-\frac{1}{N}\Big).$$
We use Cauchy-Schwarz to see that
$$\begin{align*}
\lvert\sum\Big(f^2-\frac{1}{N}\Big)\Big(g^2-\frac{1}{N}\Big)\rvert &\le \Big[\sum\Big(f^2-\frac{1}{N}\Big)^2\Big]^{1/2}\Big[\sum\Big(g^2-\frac{1}{N}\Big)^2\Big]^{1/2} \\
&= [\lVert f\rVert_4^4-1/N]^{1/2}[\lVert g\rVert_4^4-1/N]^{1/2},
 \end{align*}$$
which implies (*). I hope there is no mistake.
