inequality about linear and piecewise constant interpolation? $\Omega\subset\mathbb{R}^3$ 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:
$$
C_1\int_0^T\|u_{\tau}\|_{L^2(\Omega)}^2 \le \int_0^T\|\bar{u}_{\tau}\|_{L^2(\Omega)}^2 \le C_2\int_0^T\|u_{\tau}\|_{L^2(\Omega)}^2.
$$ 
 A: 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. 
