Is this function surjective; infinite linear combination. 
Let $A(t):[0,T] \rightarrow \mathbb{R^{n\times m}}$ be continuous function. Let $$U = \text{span}\left( \bigcup_{t\in [0,T]} \text{span}(A(t)) \right)$$
  Define function $f(u): L^\infty ([0,T],\mathbb{R}^m) \rightarrow U$ by:
  $$f(u) = \int_0^T A(t)u(t) dt$$
  Is this function surjective?

first idea:
If $u$ would be allowed to be delta function, than for given $v\in U$ I would take finite number of times $t_1,\dots,t_n$ and define $u$ as $u(t) = \sum_i \delta_{t_i}(t) u_i$, that $v = \sum_i A(t_i) u_i$. 
Since I'm not allowed to take delta function, I took only approximations of them. Than I can do only $f(u) =  v \pm \epsilon$. Next I find $u_1$ that $f(u_1) = v-f(u) \pm \epsilon_1$. And I can continue like this. So I get $f(u)+f(u_1)+\dots + f(u_n) = v \pm \epsilon_n$.
The problem is that I cant show that $u + \sum_n u_n$ converge.
second idea:
Approximate $A(t)$ by sequence of simple functions $A_n(t)$ that still $U =\text{span}\left( \bigcup_{t\in [0,T]} \text{span}(A_n(t)) \right)$. Than one can find $u_n(t)$ that $v=\int_0^T A_n(t)u_n(t) dt$. But Now I would like to take convergent subsequence of $u_n$ but I have no idea how. I guess one would have to choose $u_n$ in "right" way so such a convergent subsequence would even exist.
 A: Yes, $f$ is surjective. $U$ is a subspace of a finite-dimensional space, hence finite-dimensional itself. Let $d = \dim U$. Find $d$ points $0 \leqslant t_1 < t_2 < \dotsc < t_d \leqslant T$ and indices $k(i)$ such that $\{ A(t_i)\cdot e_{k(i)} : 1 \leqslant i \leqslant d\}$ is a basis of $U$. (Note that if columns of $A(t)$ are independent, the same columns will be independent for all $s$ in a small neighbourhood of $t$, hence we can choose distinct points for the distinct columns.)
For $\varepsilon > 0$, let $u_i^\varepsilon(t) = \frac{1}{2\varepsilon}\chi_{[t_i-\varepsilon, t_i+\varepsilon]}(t)\cdot e_{k(i)}$ (if $t_1 = 0$ or $t_d = T$, let $u_1^\varepsilon$ resp $u_d^\varepsilon$ the one-sided analogue). For small enough $\varepsilon$, the intervals $[t_i-\varepsilon, t_i+\varepsilon]$ are disjoint, and we have
$$f(u_i^\varepsilon) \xrightarrow{\varepsilon \searrow 0} A(t_i)\cdot e_{k(i)}$$
for all $i$. Hence, choosing $\varepsilon_0 > 0$ small enough, the family $\{b_i : 1 \leqslant i \leqslant d\}$ where $b_i = f(u_i^{\varepsilon_0})$ is linearly independent, hence a basis of $U$. Then
$$\sum_{i=1}^d c_i \cdot b_i = f\left(\sum_{i=1}^d c_i\cdot u_i^{\varepsilon_0}\right).$$
