Product of $L^2$-norms Let $H(e^{j\omega})$ represent the frequency domain response of a minimum phase filter such that $H(e^{j\omega})^{-1}$ exists. Is there a closed-form expression or a bound for the following expression?
$$
\left\lVert H \right\rVert_2^2 \left\lVert \frac{1}{H} \right\rVert_2 ^2
$$
EDIT:
$L^2$-norm is defined as a summation over all frequencies. Obviously, the lower bound will be 1 (if averaging), and equality holds if and only if $H(e^{j\omega})=\text{constant}$. But is there a better bound for a general filter response?
Thanks.
But is there a better bound for a general filter response?

No. The argument about it $H$ being a minimum phase filter doesn't make a difference. In essence you are asking a math question on some bound for
$$L = \sum X^2 \cdot \sum \frac{1}{X^2}$$
There is none. Let's assume a very simple signal $x = [a,1]$ . The product of the norms is
$$L = (a^2+1)(1/a^2+1) = a^2 + 2 + 1/a^2.$$
It's clear that $L &gt;=4$ and the equation is easily solvable for $a$ for any given value of $L$. In other words: by choosing $a$ you can make $L$ as large as you want it to be, so there is upper bound.