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!
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.