Suppose $\vec h : [a, b] \to R^n$ is continuous. Then show that;
$$\bigg|\bigg| \int_a^b \vec h(t) dt \bigg|\bigg| \leq
\int_a^b \| \vec h(t) \| dt$$
Note: This is used as Lemma in one of the lecture videos of Theodore Shifrin and he says this follows simply from Cauchy-Schwarz, ie $|\vec f. \vec g| \leq \|\vec f\|\|\vec g\|$. I was not able to fill the gaps though.
Best Answer
Hint: call the components of the integral $J_i$:
$$ \mathbf{J} := \int_a^b \mathbf{h}(t)dt, \qquad J_i := \langle \mathbf{e_i}, \mathbf{J} \rangle = \int_a^b h_i(t)dt,$$ with $\mathbf{e_i}$ the $i$th standard basis vector and $\langle \cdot, \cdot \rangle$ the default inner product on $\mathbb{R}^n$. Then we have:
$$ \lVert \mathbf{J} \rVert^2 = \sum_{i=1}^n J_i^2 = \sum_{i=1}^n J_i \int_a^b h_i(t)dt = \int_a^b \sum_{i=1}^n J_i h_i(t)dt.$$
Apply Cauchy-Schwarz to $\sum_{i=1}^n J_i f_i(t)$ and simplify.