How to prove $\| \int_X f \| \leq \int_X \| f \|$ in higher dimension? Let $E$ be a finite dimensional real vector space with a norm $\|.\|$. Define integral of mesurable functions with value in $E$ by choosing a basis and integrate componentwise. How do we prove the triangle inequality :
$$ \left\| \int_X f  \right\| \leq \int_X \| f \|, $$
($f : X \rightarrow E $) ?
I know a proof when $E=\mathbb{C}$ or $\mathbb{H}$, i.e. when $\|.\|=\|.\|_2$ in dimension $2$ or $4$ (the key is to use the multiplicative law). (I think) I also came up to proof of this inequality by using approximation of $f$ by simple functions, but my proof is very messy...
 A: Since Mark already gave, in the comments, a proof using duals, let me sketch here a proof using convexity. 
We make the following simplifying assumptions: 


*

*You are integrating over a space $X$ with finite total volume. (If not, approximate $f$ by cut-offs of $f$ on subsets of finite volume. That $f$ is integrable guarantees that you can do so (Chebychev's inequality).)

*$X$ has total volume 1. This you can do by rescaling, since the norm scales linearly by definition. 


Observe that the norm is a convex function. We shall prove here Jensen's inequality for a probability space, which will then imply the desired triangle inequality. 
Theorem (Jensen's inequality)
Let $(X,\Sigma,\mu)$ be a probability space (that is, it is a measure space with total volume 1). Let $f:X\mapsto V$ be an integrable function taking values in some (real) topological vector space $V$. Let $\Psi:V\to\mathbb{R}$ a convex function, then we have
$$ \Psi(\int f d\mu) \leq \int \Psi(f) d\mu $$
Sketch of Proof:
Let $g = \int f d\mu \in V$. By convexity, there exists a subdifferential of $\Psi$ at $g$, in the sense that there exists a linear functional $k\in V^*$ such that $\Psi(g) + k(h-g) \leq \Psi(h) $ for any $h\in V$. (This is the generalisation of the supporting hyperplane theorem; in the finite dimensional case you can just use the supporting hyperplane theorem.) Integrate the expression we get
$$ \int \Psi(g) d\mu + \int k(f-g) d\mu \leq \int \Psi(f) d\mu $$
Since the space has total mass 1, and $g$ is independent of the position $x\in X$, the first integral on the LHS is just $\Psi(g) = \Psi(\int f d\mu)$. Now $k$ is a linear functional, so it commutes with integration, but
$$ \int (f-g)d\mu = \int f d\mu - \int f d\mu = 0 $$
so the second term on the LHS is 0. Q.E.D.
