$\Omega\subset\mathbb{R}^3$ is a bounded, and $u(\mathbf{x},t) \in C\big(0,T,L^2(\Omega)\big)$. We divide the interval $[0,T]$ in $N$ equal subintervals with the time step $\tau$. With the notaion $$ u_i=u(t_i),~ \delta u_i=\frac{u_i-u_{i-1}}{\tau}, $$ define linear and piecewise constant interpolation functions of u as follows: \begin{gather} u_\tau = u_i+(t-t_{i-1})\delta u_i, & t\in(t_{i-1},t_i], \\ \bar{u}_\tau = u_i, & t\in(t_{i-1},t_i]. \end{gather}
How to prove the following result: $$ C_1\int_0^T\|u_{\tau}\|_{L^2(\Omega)}^2 \le \int_0^T\|\bar{u}_{\tau}\|_{L^2(\Omega)}^2 \le C_2\int_0^T\|u_{\tau}\|_{L^2(\Omega)}^2. $$
The choice of normed space does not really matter: you could simply say it's $X$ instead of $L^2(\Omega)$. The second part, with $C_2=27$, follows from the lemma below (applied on each subinterval with $x$ being the value of piecewise constant interpolation).
Lemma. For every element $x$ of a normed space $X$, and for every $v\in X$ we have $$\int_0^1 \|x+tv\|^2\,dt \ge \frac{1}{27}\|x\|^2 \tag{1}$$
Proof. The function $\phi(t)=\|x+tv\|^2$ is convex, which implies that for any $[a,b]\subset [0,1]$ we have $$\int_a^b \phi(t)\,dt\ge \frac{b-a}{2} f((a+b)/2)\tag{2}$$ Applying (2) to $[0,2/3]$ and $[1/3,1]$, and picking the better estimate, we obtain $$ \int_0^1 \phi(t)\,dt \ge \frac13 \max(\phi(1/3),\phi(2/3)) \tag{3} $$ Since $x=2(x+v/3)-(x+2v/3)$, it follows that $\|x\|\le 2\|x+v/3\|+\|x+2v/3\|$, hence $$\|x\|^2 \le 9 \max(\phi(1/3),\phi(2/3)) \tag{4}$$ Comparing (3) and (4) yields (1). $\quad \Box$
However, there is a problem with the left side of the inequality. For every $N$, there is a function $u$ that is zero at all partition points except the leftmost one. In this case $\bar u$ is identically zero, but $u_\tau$ is not. Thus, one has to make $N$ dependent on $u$, which I'm not sure is acceptable in this context. Of course, as $N\to \infty$, both integrals converge to $\int_0^T \|u\|^2$, which implies that any choice of $C_1<1$ and $C_2>1$ will work when $N$ is large enough; but I don't think this is what you are after.