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I am trying to prove $\| \int^{b}_{a} \vec{r}(t) dt\| \leq \int^{b}_{a} \| \vec{r}(t) \|dt$.


I am fairly certain that this can be derived from the Cauchy-Schwarz inequality, but I can't quite remember how to make it come out. I tried squaring as such:

$\| \int^{b}_{a} \vec{r}(t) dt\| ^2 = (\int^{b}_{a} \vec{r}(t)dt) \cdot (\int^{b}_{a} \vec{r}(t)dt) $

I can't quite recall the rest of the proof, but there should be a fairly straightforward proof of this. Any help would be appreciated!

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3 Answers 3

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Yes, you can get it out of Cauchy-Schwarz: Let $\vec v = \displaystyle\int_a^b \vec r(t)dt$. If $\vec v=\vec 0$, there's nothing to prove. If not, we have $$\|\vec v\|^2 = \vec v \cdot \int_a^b \vec r(t)dt = \int_a^b \big(\vec v\cdot \vec r(t)\big)dt \le \int_a^b \|\vec v\|\|\vec r(t)\|dt = \|\vec v\| \int_a^b \|\vec r(t)\|dt.$$ Canceling a $\|\vec v\|$, we get what you want.

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We know that

$$\lVert \vec{x}_1+\vec{x}_2+\cdots+\vec{x}_n\rVert \le \lVert\vec{x}_1\rVert+\lVert\vec{x}_2\rVert+\cdots+\lVert\vec{x}_n\rVert.$$

From here, think about the definition of the Riemann integral (in terms of Riemann sums).. You will see how the norm appears inside the integral.

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This is essentially the same result as Ted's but it works in any Banach space (with a suitable integral defined, of course).

Let $f$ be a linear functional with $\|f\| \le 1$. Then $f(\int r(t) dt) = \int f(r(t)) dt \le \int \|r(t)\|dt$.

Now define $f$ on the line $\operatorname{sp} \{ \int r(t) dt \} $ as $f(\lambda \int r(t) dt) = \lambda \|\int r(t)dt \|$ and note that $\|f\| = 1$. Now extend to the rest of the space with Hahn Banach, then we have $f(\int r(t) dt) = \|\int r(t)dt \| \le \int \|r(t)\|dt$.

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