Vector norm of integral inequality On the Wikipedia page for the mean value theorem there is this Lemma (Lemma 2 under Mean value theorem for vector-valued functions) that says:
Let $v:[a,b]\rightarrow\mathbb{R}^m$ be a continuous function on $[a,b]$, then we have that
$$ \left\Vert\int_a^b v(t)\,dt\right\Vert\leq\int_a^b\Vert v(t)\Vert dt. $$
I found a paper that uses/states the following similar result
$$ \left\Vert\int_0^1 v(t)\,dt\right\Vert^2\leq\int_0^1\Vert v(t)\Vert^2 dt. $$
I can follow the proof from Wikipedia, but I'm not sure how to proof the 'squared' version above. I guess the fact the integral is taken from 0 to 1 is important. I did find this post which asks a similar question, but it seems rely on it being a scalar function there.
 A: Squaring the first inequality gives
$$
\left\Vert\int_0^1 v(t)\,dt\right\Vert^2\leq\left(\int_0^1\Vert v(t)\Vert \, dt \right)^2 \, .
$$
Now apply Inequality releating squared absolute value of an integral to the integral of the squared absolute values of the integrand to the scalar-valued function $f(t) = \Vert v(t)\Vert$ to conclude that the right-hand side is
$$
 \le \int_0^1\Vert v(t)\Vert^2 \, dt \,  .
$$
A: You have
$$\left(||\int_0^1v(t)dt||\right)^2\leq\left(\int_0^1||v(t)||dt\right)^2$$
by the first result and the monotonicity of $x\mapsto x^2$ on $[0,\infty)$. If $t\mapsto||v(t)||$ is summable then by the convexity of $x\mapsto x^2$ and Jensen's inequality it follows
$$\left(\int_0^1||v(t)||dt\right)^2\leq\int_0^1||v(t)||^2dt$$
and thus the desired result.
A: If $\{e_1, \dots, e_m\}$ is the canonical basis of $\mathbb R^m$ and $v : t \mapsto v_1(t) e_1 + \dots + v_m(t) e_m$ then by definition of the vector integral
$$\int_0^1 v(t) \ dt = \sum_{i=1}^m \left(\int_0^1 v_i(t) \ dt \right) e_i$$ and therefore if $\Vert \cdot \Vert$ stands for the Euclidean norm
$$\left\Vert \int_0^1 v(t) \ dt \right\Vert^2 = \sum_{i=1}^m\left(\int_0^1 v_i(t) \ dt \right)^2 \le \int_0^1 \left(\sum_{i=1}^m v_i^2(t) \right) \ dt = \int_0^1 \Vert v(t) \Vert^2 \ dt$$
