Proof that $\left| \int_X \mathbf{f} d\mu \right|_p \leq \int_X \left| \mathbf{f} \right|_p d\mu$ for $\mathbf{f} = (f_1, ..., f_n)$

Question

I am having trouble with the following proof.

Define $\mathbf{f}(x) = (f_1(x), ..., f_n(x))$ where $f_i:X\to\mathbb{R}$ for each positive integer $i \leq n$, and each $f_i$ is integrable.

Prove that $$\left| \int_X \mathbf{f} \ \mathrm{d}\mu \right|_p \leq \int_X \left| \mathbf{f} \right|_p \ \mathrm{d}\mu$$

for $1 \leq p \leq \infty$

Now, we can write the left-hand side explicitly as

$$\left| \int_X \mathbf{f} \ \mathrm{d}\mu \right|_p = \left[\sum_{j=1}^n \left| \int_X f_j \ \mathrm{d}\mu \right|^p \right]^{1/p}$$

Now, the exponent gives me trouble. If we had the case $p=1$, then it is simple since for every integrable function $g: X \to \mathbb{R}$, we have

$$\left| \int_X g \ \mathrm{d}\mu \right| \leq \int_X |g| \ \mathrm{d}\mu$$

Thus,

$$\left| \int_X \mathbf{f} \ \mathrm{d}\mu \right| = \sum_{j=1}^n \left| \int_X f_j \ \mathrm{d}\mu \right| \leq \sum_{j=1}^n \int_X|f_j| \ \mathrm{d}\mu = \int_X \left(\sum_{j=1}^n |f_j| \right) \ \mathrm{d}\mu = \int_X| \mathbf{f}| \ \mathrm{d}\mu$$

But, I can't work around the exponential for $1 < p < \infty$

Thank you!

Answer 1

Let $\frac 1 p +\frac 1 q =1$, $a \in \mathbb R^{n}$ with $\|a\|_q \leq 1$ and consider $\langle a , \int fd\mu \rangle$. This is same as $\int \langle a , f \rangle d\mu$. Since $|\langle a , f \rangle| \leq |f|_p$ we get $|\langle a , \int f d\mu \rangle| \leq \int |f|_p d\mu$. Now taking sup over all $a$ gives you $|\int f d\mu|_p \leq \int |f|_p d\mu$. [ I have taken $p \in (1,\infty)$, I will leave the cases $p=1$ and $p=\infty$ to you].