Vector norm of integral inequality

Question

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.

Answer 1

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 $$ \le \int_0^1\Vert v(t)\Vert^2 \, dt \, . $$