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. $$ \sum_{n=1}^m h\lVert f_n\rVert_{L^2(\Omega)}=\sum_{n=1}^m h\left\lVert \frac 1 h\int_{t_{n-1}}^{t_n} f(t)\,dt\right\rVert_{L^2(\Omega)}\leq\sum_{n=1}^m \int_{t_{n-1}}^{t_n}\lVert f(t)\rVert_{L^2(\Omega)}\,dt= \int_{0}^{T}\lVert f(t)\rVert_{L^2(\Omega)}\,dt\leq T^{1/2}\left(\int_0^T \lVert f(t)\rVert_{L^2(\Omega)}^2\,dt\right)^{1/2}. $$

  • $\begingroup$ But how the first step? I get $\lVert f_n \rVert_{L^2}^2 = (1/h^2)\int_\Omega\left(\int_{t_{n-1}}^{t_n} f(t)\right)^2 \leq (1/h)\int_\Omega \int_{t_{n-1}}^{t_n} |f(t)|^2 = (1/h)\lVert f \rVert_{L^2(t_{n-1}, t_n; L^2(\Omega))}^2$ which is different. I used Cauchy Schwarz for the last inequality.. $\endgroup$ – Upin Feb 28 '18 at 10:38
  • $\begingroup$ @Upin, I just canceled the factor $h$ and applied the triangle inequality for integrals. $\endgroup$ – MaoWao Feb 28 '18 at 10:41
  • $\begingroup$ please resee the comment it compiles now $\endgroup$ – Upin Feb 28 '18 at 10:41
  • $\begingroup$ I added one more step. I hope it's clear now. $\endgroup$ – MaoWao Feb 28 '18 at 10:44
  • $\begingroup$ Oh so you use $\lVert \int_0^T f\rVert_{L^2(\Omega)} \leq \int_0^T \lVert f \rVert_{L^2(\Omega)}$, ok I didn't know this existed. I thought it was only true for the absolute value and not the norm function. Thanks $\endgroup$ – Upin Feb 28 '18 at 10:58

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