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Let $X_t(\omega)$ be a continuous function $t\rightarrow L^p(\omega)$ (i.e., if we fixed the variable $t$ we obtain a function which belongs to $L^p$), with $t\in[0,T]$ and $\omega\in\mathbb{R}$.

I would like to know if this property of Lebesgue integrals:

$|\int_0^TX_t(\omega)dt|\leq\int_0^T|X_t(\omega)|dt$

is also valid for $L^p$-norm, i.e. if this is true:

$||\int_0^TX_t(\omega)dt||_{L^p(\omega)}\leq\int_0^T||X_t(\omega)||_{L^p(\omega)}dt$

Indeed, I have just proved that:

$||\int_0^TX_t(\omega)dt||_{L^p(\omega)}=(\int_\mathbb{R}(\int_0^TX_t(\omega)dt)^pd\omega)^\frac{1}{p}\leq(\int_0^T||X_t(\omega)||_{L^p(\omega)}^pdt)^\frac{1}{p}\cdot T^\frac{p-1}{p}$,

where I applied Hölder's inequality and Fubini's theorem.

Thank you for your attention.

  • 0
    I didn't know it existed this inequality... Sorry!2012-12-04

1 Answers 1

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We have for each integer $N$ that $\left\lVert\frac TN\sum_{j=1}^NX_{TkN^{-1}}(\cdot)\right\rVert_{L^p(\Omega)}\leqslant\frac TN\sum_{j=1}^N\left\lVert X_{TkN^{-1}}(\cdot)\right\rVert_{L^p(\Omega)}.$ As $t\mapsto X_t(\cdot)$ and $t\mapsto \lVert X_t(\cdot)\rVert_{L^p(\Omega)}$, we conclude by a Riemann sum argument.