Hi I am having trouble verifying the equivalence of the following two inequalities.
\begin{align*}\tag{1} \|u(t)\|_{L^2(\Omega)}&\le \|u_0\|_{L^2(\Omega)} \end{align*}
\begin{align*}\tag{2} \sup_{t\ge 0}\int_{\Omega}|u(x,t)|^2\,dx&\le \int_{\Omega}|u_0(x)|^2\,dx \end{align*} can see the right hand side of both equations are equivalent and independent to $t$. However for the left hand side, I cannot see how they could be equivalent. For (2), the lhs is $$\sup_{t\ge 0}\int_{\Omega}|u(x,t)|^2 dx=\sup_{t\ge 0}\|u(t)\|_{L^2(\Omega)}^2.$$ And I cannot see how this could be equivalent to $\|u(t)\|_{L^2(\Omega)}$. I mean, $\sup_{t\ge 0}\|u(t)\|_{L^2(\Omega)}^2$ is really not quite $\|u(t)\|_{L^2(\Omega)}$.
Appreciate for any helps.
(I saw someone asked about $u(t)$. I do not think that is needed to answer the basic questions I asked. Nevertheless, $u(t)\in H_0^1(\Omega)\cap H^2(\Omega)$, $u(0)=u_0$.
Note that (1) is a statement about all $t$, that is, we have to read (1) as $$ \def\norm#1{\left\|#1\right\|_{L^2(\Omega)}} \norm{u(t)} \le \norm{u_0}, \qquad \text{for all $t \ge 0$} $$ as $t$ is not specified any further. This is of course equivalent - by the very definition of the supremum - to (2), that is $$ \sup_{t\ge 0} \norm{u(t)} \le \norm{u_0} $$ as (1) states that $\norm{u_0}$ is an upper bound for the set $\{\norm{u(t)} : t \ge 0\}$ and the supremum is the least upper bound.