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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.

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Actually, it's pretty straightforward. All you need is the triangle inequality for integrals and the Cauchy-Schwarz inequality. $$ \sum_{n=1}^m h\lVert f_n\rVert_{L^2(\Omega)}=\sum_{n=1}^m h\left\lVert \frac 1 h\int_{t_{n-1}}^{t_n} f(t)\,dt\right\rVert_{L^2(\Omega)}\leq\sum_{n=1}^m \int_{t_{n-1}}^{t_n}\lVert f(t)\rVert_{L^2(\Omega)}\,dt= \int_{0}^{T}\lVert f(t)\rVert_{L^2(\Omega)}\,dt\leq T^{1/2}\left(\int_0^T \lVert f(t)\rVert_{L^2(\Omega)}^2\,dt\right)^{1/2}. $$

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  • $\begingroup$ 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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