$|\int f(x)\,dx| \le \int|f(x)| \, dx$ for vector valued function

Question

Let $f : \mathbb{R} \to \mathbb{R}^{n}$ be a vector valued function, $f_i \ge 0$. Can we show that

$$\bigg \lvert \int f(x) \,dx \bigg \rvert \le \int \lvert f(x) \rvert \, dx$$

The assertion above is obvious right when $n=1$ . But is it right when $n \gt 1$? For $n=2$ , we need to show

$$\int f\,dx^2+\int g\,dx^2 \le \int \sqrt{f^2+g^2}\,dx^2$$

For each nonnegative function $f, g : \mathbb{R} \to \mathbb{R}$, it is obvious that

$$\int f \,dx^2 + \int g\,dx^2 \le 2 \int \sqrt{f^2+g^2}\,dx^2$$

I think the assertion above is wrong, but I can not find the example of it .

Answer 1

It is true. Let $v=\int f(x)dx$. Then $\langle v , v \rangle =\langle v , \int f(x)dx \rangle =\int \langle v, f(x) \rangle dx\leq \int |v| |f(x)|dx =|v| \int |f(x)|dx.$ Can you finish?

I have used linearity of the integral and Cauchy -Schwraz inequality $\langle x , y \rangle \leq |x||y|$