Prove a vector integral inequality

Question

Given continuous $f:\mathbb{R}\rightarrow\mathbb{R}^n$, $x_1<x_2$, how to prove $||\int_{x_1}^{x_2}f(x)dx|| \leq \int_{x_1}^{x_2}||f(x)||dx$?

I think it needs to use Cauchy-Schwarz inequality, and so far I got this: replace $x_2$ with $x$, then to prove $$\sqrt{\sum_i(\int_{x_1}^{x}f_i(t)dt)^2}\leq \int_{x_1}^{x}\sqrt{\sum_if_i(t)^2}dt$$ let $\int_{x_1}^{x}f_i(t)dt=F_i(x)$, differentiate both sides by $x$, then I'd like to prove the left side's derivative is $\leq$ the right side's derivative, that is: $$\frac{\sum_iF_i(x)f_i(x)}{\sqrt{\sum_iF_i(x)^2}}\leq\sqrt{\sum_if_i(x)^2}$$ which is true by Cauchy-Schwarz, as long as $\sum_iF_i(x)^2\neq0$. Then if $\sum_iF_i(x)^2\neq0$ for all $x\in[x_1,x_2]$, we can integrate both sides to get the original inequality.

So that's where I've got to so far. How do deal with the cases where for some $x\in[x_1,x_2]$, $\sum_iF_i(x)^2=0$?

Answer 1

Exchangeability of inner product with the integral

So for any vector $V\in\mathbb R^n$ we have by applying Cauchy-Schwarz :

$\displaystyle \bigg|\langle\int_a^b f(x)dx,V\rangle\bigg|=\bigg|\int_a^b\langle f(x),V\rangle dx\bigg|\le\int_a^b ||f(x)||\,||V||dx=||V||\int_a^b ||f(x)||dx$

Now this is true for $\displaystyle V=\int_a^b f(x)dx\quad$ giving $\quad\displaystyle \bigg|\bigg|\int_a^b f(x)dx\bigg|\bigg|^2\le\bigg|\bigg|\int_a^b f(x)dx\bigg|\bigg|\int_a^b ||f(x)||dx$

And we get the result by dividing by $||V||\neq 0$ and for $||V||=0$, the inequality is trivially true.

Answer 2

Because just $|a|+|b|\geq|a+b|$ and use the definition of an integral.