How to show that the integral inequality holds for vector-valued functions. If I define the integral for $\int_a^b\textbf{F}(t)dt$ as $(\int_a^b F_1(t)dt, \int_a^bF_2(t)dt,\ldots, \int_a^bF_n(t)dt )$. How do I then show that
$$\left|\int_a^b\textbf{F}(t)dt \right|\le\int_a^b\left|\textbf{F}(t) \right|dt?$$
If we write out the inequality we have
$$\sqrt{\left(\int_a^b F_1(t)dt\right)^2+\left(\int_a^bF_2(t)dt\right)^2+\cdots+ \left(\int_a^bF_n(t)dt\right)^2 }\le\int_a^b\sqrt{F_1(t)^2+F_2(t)^2+\cdots+F_n(t)^2}dt.$$
attempt:
If I use Jenssen's inequality I get for the left side:
$$\sqrt{\left(\int_a^b F_1(t)dt\right)^2+\left(\int_a^bF_2(t)dt\right)^2+\cdots+ \left(\int_a^bF_n(t)dt\right)^2 }\le\sqrt{\int_a^b F_1^2(t)dt+\int_a^bF_2^2(t)dt+\cdots+ \int_a^bF_n^2(t)dt }\\=\sqrt{\sum\limits_{i=1}^n\int_a^bF_i^2(t)dt}=\sqrt{\int_a^b\sum\limits_{i=1}^nF_i^2(t)dt}.$$
The problem I have now is that the square root function is concave, so if I use Jenssen again, I get the opposite inequality as the one I need. Any idea on how to solve this?
 A: The linearity of the integral, that of the inner product and the Schwarz inequality imply, for any $\vec v\in \mathbb R^n,$
$\tag 1\left \langle \int_a^bF(t)dt,\vec v\right \rangle=\int_a^b\langle F(t),\vec v\rangle dt\le \int_a^b|\langle F(t),\vec v\rangle| dt\le \int_a^b|F(t)|\vec v| |dt\le |\vec v|\int_a^b |F(t)|dt .$
$\tag 2\text{Taking}\quad \vec v=\int_a^bF(t)dt,\quad \text{we get}\quad |\vec v|^2\le |\vec v|\int_a^b |F(t)|dt$
$\tag3  \text{which is}\quad  \left |\int_a^bF(t)dt\right |\le \int_a^b |F(t)|dt.$
A: The exact proof depends a little on what definition of integral you are using, but the following observation is the key...
$$\vert \sum_i \vec{F}_i \Delta t \vert \leq \sum_i \vert \vec{F}_i \vert \Delta t $$
This holds for any choice of vectors $F_i$ by triangle inequality. Now make them either vectors appropriate for a Riemann sum or for a simple function approximating your given function...
A: This is slightly overkill, but this theorem is a consequence of the Minkowski integral inequality.
