Bound on $f_h(t) := \frac 1h \int_t^{t+h}f$

Question

Given $f \in L^2(0,T;L^2(\Omega))$ define $$f_h(t) = \frac 1h \int_t^{t+h}f(s)\;\mathrm{d}s$$ for $t \in (0,T-h)$ and $f_h(t) = 0$ for $t > T-h$.

In this paper (http://dml.cz/bitstream/handle/10338.dmlcz/119003/CommentatMathUnivCarolRetro_39-1998-2_3.pdf), see Section 2, the author states that for any $0 < t_1 < T$, we have the bound $$\int_0^{t_1}\int_\Omega |f_h|^2 \leq \int_0^T\int_\Omega |f|^2$$

My question is why cannot $t_1 \leq T$, i.e., can't we pick $t_1=T$ so the bound holds on $(0,T)$? What goes wrong?

This "restriction" is made also in eg. https://pub.uni-bielefeld.de/download/2624153/2624155 see Lemma 1.14 on page 10.

Answer 1

Here is a proof of the inequality: first Hoelder, then estimate $L^1$ against $L^2$, Fubini, estimate the inner integral, done: $$ \frac1{h^2}\int_0^{t_1}\int_\Omega \left|\int_t^{t+h} f(s) ds\right|^2dx \ dt \le\frac1{h^2}\int_0^{t_1}\int_\Omega \left(\int_t^{t+h} |f(s)| ds\right)^2dx \ dt\\ \le\frac1h\int_0^{t_1}\int_\Omega \int_t^{t+h} |f(s)|^2 ds\ dx \ dt\\ \le\frac1h\int_0^{t_1+h}\int_\Omega \int_{\max(0,s-h)}^{\min(t_1,s)} |f(s)|^2 dt\ dx \ ds\\ \le\int_0^{t_1+h}\int_\Omega |f(s)|^2 \ dx \ ds\\ \le \int_0^T\int_\Omega |f(s)|^2 \ dx \ ds. $$ The only step that seems to need $t_1+h\le T$ is the last bound. However, since $f$ was set to zero outside of $(0,T)$, this estimate is also valid for $t_1+h>T$. I see no reason, why this estimate should not be true for every $h$, $t_1$.

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Another interpretation is the following: let $F(t):=\int_{0}^t f(s) ds$. Then $f_h$ is a finite difference approximation of $F'=f$, whose $L^2(0,T;L^2(\Omega))$ norm can be bounded by $\|F\|_{H^1(0,T;L^2(\Omega))}\approx \|f\|_{L^2(0,T;L^2(\Omega))}$.

Answer 2

There may be other reasons, but simple in order to integrate something which depends on $f_h$, $f_h$ has to be well defined. So if you integrate from $0$ to $t_1$, you need $t_1+h \le T$, or, equivalently, $t_1 \le T-h < T$ (since you have to assume $h>0$ to make sense of this).