Let $G(j\omega)=C(j\omega I-A)^{-1}B+D \in \mathbb{C}^{m\times n}$ be the transfer function (multi-input multi-output) matrix of state space representation.
We know if $G(j\omega)$ is called bounded-real if $G(j\omega)^*G(j\omega)\leq \gamma I$ with $\gamma>0$.
Also, we know that the above condition is equivalent to the following time-domain condition:
$$\int_0^{\infty}y(t)^Ty(t)dt<\gamma \int_0^{\infty}u(t)^Tu(t)dt$$
where $\forall u\in \mathcal{L}_2$, i.e., square integrable. $y(t)=G(t)u(t)$ and of course $Y(j\omega) = G(j\omega)U(\omega)$
How to show this equivalence? I try to write down the matrix form and have no idea how to use Parseval's theorem to explicitly obtain the result.
I use $*$ to refer to the "conjugate transpose". With Parseval's theorem, we have $$ \int_{0}^\infty y(t)^*y(t)\,dt = \int_{-\infty}^\infty Y(j\omega)^*Y(j\omega)\,d\omega \\ = \int_{-\infty}^\infty [G(j \omega)U(j \omega)]^*[G(j \omega)U(j \omega)]\,d\omega\\ = \int_{-\infty}^\infty U(j \omega)^*G(j \omega)^*G(j \omega)U(j \omega)\,d\omega $$ Now, $G(j\omega)^*G(j\omega)\leq \gamma I$ means that for all vectors $v \in \Bbb C^n$, we have $$ v^*[G(j\omega)^*G(j\omega)]v \leq v^*[\gamma I]v = \gamma \,v^*v $$ Thus, we have $$ \int_{-\infty}^\infty U(j \omega)^*G(j \omega)^*G(j \omega)U(j \omega)\,d\omega\\ \leq \int_{-\infty}^\infty \gamma U(j \omega)^*U(j \omega)\,d\omega \\ = \gamma\int_{-\infty}^\infty U(j \omega)^*U(j \omega)\,d\omega \\ = \gamma \int_{0}^\infty u(t)^*u(t)\,dt $$ As desired.