= Solution
Put $I=\int_0^1f(t)\,dt$. If $I\ne0$, take the <inner product> with the <unit vector> $u=I/\|I\|_2$. Then
$$
\|I\|_2=u\mathbin{\cdot}I=\int_0^1u\mathbin{\cdot}f(t)\,dt\leq\int_0^1\|f(t)\|_2\,dt
$$
by the <Cauchy-Schwarz inequality>; the case $I=0$ is immediate. This is the <triangle inequality for a vector-valued integral>.
Equality in the Euclidean argument requires $u\mathbin{\cdot}f(t)=\|f(t)\|_2$ for every $t$: the nonnegative continuous difference has integral zero. Equality in <Cauchy-Schwarz inequality> then says $f(t)=\lambda(t)u$ with $\lambda(t)\geq0$. Conversely, if $f(t)=\lambda(t)v$ for one fixed $v\in\mathbb R^m$ and a continuous $\lambda\geq0$, then for \b[every] <norm>,
$$
\left\|\int_0^1f\right\|=\left(\int_0^1\lambda\right)\|v\|=\int_0^1\|f(t)\|\,dt.
$$
Thus these nonnegative-ray functions, including $f\equiv0$, are exactly the required functions.
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