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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?

user394334
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    It is trivially true for simple functions (triangle inequality). Use a limiting argument to conclude for all integrable ones. – peek-a-boo Dec 29 '21 at 02:28
  • @peek-a-boo Could you please give a little more information on how to do that? I get that the square-roots and squares messes that up. – user394334 Dec 29 '21 at 02:40
  • For simple/step functions, nothing should be messing you up. It is literally by triangle inequality of any norm. – peek-a-boo Dec 29 '21 at 02:41

3 Answers3

1

The linearity of the integral, that of the inner product and the Schwarz inequality imply, for any $\vec v\in \mathbb R^n,$

$\tag 1\left \langle \int_a^bF(t)dt,\vec v\right \rangle=\int_a^b\langle F(t),\vec v\rangle dt\le \int_a^b|\langle F(t),\vec v\rangle| dt\le \int_a^b|F(t)|\vec v| |dt\le |\vec v|\int_a^b |F(t)|dt .$

$\tag 2\text{Taking}\quad \vec v=\int_a^bF(t)dt,\quad \text{we get}\quad |\vec v|^2\le |\vec v|\int_a^b |F(t)|dt$

$\tag3 \text{which is}\quad \left |\int_a^bF(t)dt\right |\le \int_a^b |F(t)|dt.$

Matematleta
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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...

STW
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This is slightly overkill, but this theorem is a consequence of the Minkowski integral inequality.

Mason
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