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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:


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


Indeed, I have just proved that:

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

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

Thank you for your attention.

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up vote 2 down vote accepted

It holds more generally replacing $L^p(\Omega)$ by a normed space $(B,\lVert \cdot \rVert_B)$, assuming $M\colon t\mapsto X_t(\cdot)$ goes from $[0,T]$ to $B$ is continuous.

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

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Or: more generally, if $\phi$ is a continuous convex function on a Banach space $E$, and $f : [0,T] \to E$ is continuous, then we have $$ \phi\left(\frac{1}{T}\int_0^T f(t)\;dt\right) \le \frac{1}{T}\int_0^T \phi(f(t))\;dt $$ And $[0,T]$ can be replaced by a more abstract finite measure space, continuity of $f$ by appropriate measurability.

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