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Given $f(\mathbf{x},t)\in L^2\big((t_1,t_2);\mathbf{L}^2(\Omega)\big)$, how to prove the following inequality?

$$ \Bigg\|\int_{t_1}^{t_2}f(\mathbf{x},t)dt\Bigg\|_{\mathbf{L}^2(\Omega)} \le \int_{t_1}^{t_2}\Bigg\|f(\mathbf{x},t)\Bigg\|_{\mathbf{L}^2(\Omega)}dt. $$

The above result is simlar to $$ \Bigg|\int_a^b f(x) dx\Bigg| \le \int_a^b\Bigg|f(x) \Bigg|dx. $$

chentaocuc
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  • For which functions can you prove the inequality quiet easily? Or how is (strong-)measurabilty defined in such spaces? – Quickbeam2k1 Jul 13 '13 at 08:31
  • I have seen someone used this result in the paper, however, I don't kown how to prove it. – chentaocuc Jul 14 '13 at 04:31
  • Inner product and norm in $L^2\big((t_1,t_2);\mathbf{L}^2(\Omega)\big)$ are define as follows. For any $\mathbf{u},\mathbf{v} \in L^2\big((t_1,t_2);\mathbf{L}^2(\Omega)\big)$, $$(\mathbf{u},\mathbf{v}):=\int_{t_1}^{t_2}\int_{\Omega}\mathbf{u}\cdot\mathbf{v}~d\mathbf{x}dt,$$ $$|\mathbf{u}|:=\sqrt{(\mathbf{u},\mathbf{u})}.$$ – chentaocuc Jul 14 '13 at 04:38

1 Answers1

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Recall that the norm of vector $x$ in a Hilbert space is equal to $\sup \langle x,y\rangle$ where the supremum is taken over all unit vectors $y$. Let $g\in L^2(\Omega)$ be a unit norm function (we can even choose it smooth, with compact support). Then $$\left\langle \int_{t_1}^{t_2}f(\cdot,t)\,dt , g \right\rangle_{L^2(\Omega)} = \int_{t_1}^{t_2}\,dt \int_\Omega f(\mathbf x,t)g(\mathbf x)\,d\mathbf x $$ The latter does not exceed $$\begin{split} \int_{t_1}^{t_2}\,dt \int_\Omega |f(\mathbf x,t)|\,|g(\mathbf x)|\,d\mathbf x &\le \int_{t_1}^{t_2} \|f(\cdot, t)\|_{L^2(\Omega)} \|g\|_{L^2(\Omega)} \,dt \\&= \int_{t_1}^{t_2} \|f(\cdot, t)\|_{L^2(\Omega)} \,dt \end{split}$$ as desired.

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