Let $f \in L^\infty(0,T;L^2(\Omega))$ be non-negative ($\Omega$ bounded domain) and define $$f_n = \frac 1h \int_{t_{n-1}}^{t_n} f(t) dt$$ where $\{t_n\}$ partittions $[0,T]$ into subintervals of size $h$, i.e., $t_{n+1}-t_n = h$.
I have read that $$\sum_{n=1}^m h\lVert f_n \rVert_{L^2(\Omega)} \leq T^{1/2}\lVert f \rVert_{L^2(0,T;L^2(\Omega))}$$
but I cannot prove this. I can only show that $$\lVert{f_i}\rVert_{L^2(\Omega)}^2 \leq \frac 1h \lVert{f}\rVert_{L^2(t_{n-1},t_n;H)}^2$$ and I can't prove it unless we say something like the sum of squares is equal to square of the sum.
Actually, it's pretty straightforward. All you need is the triangle inequality for integrals and the Cauchy-Schwarz inequality. $$ \sum_{n=1}^m h\lVert f_n\rVert_{L^2(\Omega)}=\sum_{n=1}^m h\left\lVert \frac 1 h\int_{t_{n-1}}^{t_n} f(t)\,dt\right\rVert_{L^2(\Omega)}\leq\sum_{n=1}^m \int_{t_{n-1}}^{t_n}\lVert f(t)\rVert_{L^2(\Omega)}\,dt= \int_{0}^{T}\lVert f(t)\rVert_{L^2(\Omega)}\,dt\leq T^{1/2}\left(\int_0^T \lVert f(t)\rVert_{L^2(\Omega)}^2\,dt\right)^{1/2}. $$