Easy question that I can't do: Find all continuous functions satisfying $\left\|\int_0^1 f\right\|= \int_0^1 \|f\|$ I am trying to find all continuous functions $f: [0,1] \to \mathbb R^n$ satisfying:
$$\left\|\int_0^1 f\right\|= \int_0^1 \|f\|$$
Any hints? 
I have tried lots of things, none of which have worked. I have found plenty of functions that work, but have no idea how to describe the entire set. I have tried writing the integral as Riemann sums and using the analogy of the vector triangle inequality, which wasn't helpful. I have tried looking at specific norms to get some necessary conditions, but didn't get far with that either. For $\mathbb R^1$ a necessary condition would be that $f$ is always positive or always negative.
 A: The answer depends on the used norm.
If we use $\lVert v\rVert_1 = \sum_{k = 1}^n \lvert v_k\rvert$, the equality becomes
$$\sum_{k = 1}^n \biggl\lvert \int_0^1 f_k(x)\,dx\biggr\rvert = \int_0^1 \sum_{k = 1}^n \lvert f_k(x)\rvert\,dx = \sum_{k = 1}^n \int_0^1 \lvert f_k(x)\rvert\,dx.\tag{1}$$
Since for every $k$ the inequality
$$\biggl\lvert \int_0^1 f_k(x)\,dx\biggr\rvert \leqslant \int_0^1 \lvert f_k(x)\rvert\,dx\tag{2}$$
holds, we have equality in $(1)$ if and only if we have equality in $(2)$ for every $k$, that is, if none of the components of $f$ changes sign.
If we use $\lVert v\rVert_{\infty} = \max \{ \lvert v_k\rvert : 1 \leqslant k \leqslant n\}$, the equality becomes
$$\max \left\{ \biggl\lvert \int_0^1 f_k(x)\,dx\biggr\rvert : 1 \leqslant k \leqslant n\right\} = \int_0^1 \max \left\{ \lvert f_k(x)\rvert : 1 \leqslant k \leqslant n\right\}\,dx. \tag{3}$$
If we then pick an $m$ such that
$$\biggl\lvert \int_0^1 f_m(x)\,dx\biggr\rvert = \max \left\{ \biggl\lvert \int_0^1 f_k(x)\,dx\biggr\rvert : 1 \leqslant k \leqslant n\right\},$$
we see that we must have
$$\biggl\lvert \int_0^1 f_m(x)\,dx\biggr\rvert = \int_0^1 \lvert f_m(x)\rvert\,dx = \int_0^1 \max \left\{ \lvert f_k(x)\rvert : 1 \leqslant k \leqslant n\right\}\,dx,$$
which implies that $f_m$ doesn't change sign, and $\lvert f_k(x)\rvert \leqslant \lvert f_m(x)\rvert$ for all $x\in [0,1]$ and all $k$ (let's call that a dominating component of $f$). So there must be at least one dominating component of $f$ that doesn't change sign. Conversely, one checks that if $f$ has a dominating component that doesn't change sign, then
$$\biggl\lVert\int_0^1 f(x)\,dx\biggr\rVert_{\infty} = \int_0^1 \lVert f(x)\rVert_{\infty}\,dx.$$
Despite the answer depending on the used norm, there is a unified approach to finding the set of functions with the desired property.
First, we treat the trivial case. If $\int_0^1 f(x)\,dx = 0$, then the equality holds if and only if $f\equiv 0$. That follows easily from the continuity and positive definiteness of the norm.
So we may assume that
$$I := \int_0^1 f(x)\,dx$$
is not $0$ and the equality holds. Now we invoke a big theorem (Hahn-Banach) to note that there is a $\lambda \in (\mathbb{R}^n,\lVert\,\cdot\,\rVert)'$ with $\lVert \lambda\rVert = 1$ and $\lambda(I) = \lVert I\rVert$.
Then we have
\begin{align}
\int_0^1 \lVert f(x)\rVert\,dx &= \biggl\lVert \int_0^1 f(x)\,dx\biggr\rVert \tag{assumption}\\
&= \lambda\Biggl(\int_0^1 f(x)\,dx\Biggr)\\
&= \int_0^1 \lambda\bigl(f(x)\bigr)\,dx \tag{linearity of integral}\\
&\leqslant \int_0^1 \bigl\lvert \lambda\bigl(f(x)\bigr)\bigr\rvert\,dx\\
&\leqslant \int_0^1\lVert\lambda\rVert\cdot \lVert f(x)\rVert\,dx\\
&= \int_0^1 \lVert f(x)\rVert\,dx.
\end{align}
Since the outer terms of the chain of inequalities are the same, we must have equality everywhere, and hence
$$\lambda\bigl(f(x)\bigr) = \lVert f(x)\rVert\tag{4}$$
for all $x\in [0,1]$. Conversely, if $4$ holds for some $\lambda$ with $\lVert \lambda\rVert = 1$, then we have the equality $\lVert I\rVert = \int_0^1 \lVert f(x)\rVert\,dx$.
This also works for $f\equiv 0$, and we can summarise:
$$\biggl\lVert\int_0^1 f(x)\,dx\biggr\rVert = \int_0^1 \lVert f(x)\rVert\,dx$$
if and only if there is a $\lambda \in (\mathbb{R}^n,\lVert\,\cdot\,\rVert)'$ with $\lVert\lambda\rVert = 1$ such that for all $x\in [0,1]$ we have $f(x) \in C(\lambda)$, where $C(\lambda)$ is the cone
$$C(\lambda) = \{ v\in \mathbb{R}^n : \lambda(v) = \lVert v\rVert\}.$$
If the norm is an $\ell^p$-norm, $\lVert v\rVert = \left(\sum_{k = 1}^n \lvert v_k\rvert^p\right)^{1/p}$ for some $1 < p < \infty$, the cones $C(\lambda)$ are easily described.
In particular the Euclidean case - $p = 2$ - should now not pose a big problem.
A: So there is a way to do this without using any big theorems (like Hahn-Banach).
First we will get a necessary condition. Suppose that $$\left\| \int_0^1 f\right\|= \int_0^1 \|f\|$$
If we take $\| \cdot \|$ to be the Euclidean norm, then we can write this as 
$$\left(\int_0^1 f\right)\cdot \left( \int_0^1 f\right) = \left\| \int_0^1 f\right\|\int_0^1 \|f\|$$
Or equivalently
$$\int_0^1 \left( f \cdot \int_0^1 f \right)= \int_0^1 \left(\left\| \int_0^1 f\right\| \|f\|\right) \tag 1$$
Now by Cauchy-Schwarz we have $\displaystyle  f \cdot \int_0^1 f  \le \left\| \int_0^1 f\right\| \|f\|$
So for $(1)$ to hold we require that $\displaystyle  f \cdot \int_0^1 f  = \left\| \int_0^1 f\right\| \|f\|$, since we are given that $f$ is continuous.
This occurs if and only if $$f(t) = \lambda(t) \int_0^1 f(t) \, \mathrm d t$$ where $\lambda$ is a scalar.
This implies that $f(t) = \lambda(t) v$ where $v$ is a constant vector. This turns out to be sufficient for any norm on $\mathbb R^n$ if $\lambda$ does not change sign for all $t \in [0,1]$, since 
\begin{align}\left \| \int_0^1 \lambda(t) v \, \mathrm d t \right \| &= \left \| v \int_0^1 \lambda(t) \, \mathrm d t \right \| \\ &=  \| v \| \left| \int_0^1 \lambda(t) \, \mathrm d t \right|\\ &= \|v\|\int_0^1 |\lambda(t)| \, \mathrm d t \\ &= \int_0^1 \| \lambda(t) v \| \, \mathrm d t
\end{align} 
A: Let $g(x)=\int_0^x f(t) dt$. So $g'(x)=f(x)$, and $\vert\vert g'(x)\vert\vert =\vert\vert f(x)\vert\vert$. The hypothesis is $\vert \vert g(1)-g(0)\vert\vert=\int_0^1 \vert\vert g'(t)\vert\vert dt$. So your ppt holds iff the curve $g : [0,1]\to \bf R^n$ realize the distance between its extreme points (a geodesic). For a strictly convex normed space, the geodesic between two points is unique and is a line segment, and $f$ also describe injectively a line segment. For other spaces such as $d_\infty$ on $\bf R^2$ the "sup" norm, you can have plenty of geodesics between two points : the geodesic between $(-1,1)$ and $(1,1)$ are of length 2, and $[(-1,1)(0,x)\cup [(0,x),(1,1)]$ is a one parameter family of geodesics of the same length.
