Continuous linear combination of continuously varying vectors? Let ${\bf{e}}_1, {\bf{e}}_2, {\bf{e}}_3:[0,1]\rightarrow \mathbb{R}^3$ be continuous, $\mathbf{0}\neq \mathbf{v}\in \mathbb{R}^3$. 
Suppose that the following condition (C) holds: 
$$
\exists d>0: \;\forall \;{\mathbf{u}}\in \mathbb{R}^3, \;\forall \;t\in [0,1],\;\langle {\bf{e}}_1(t),{\bf{u}}\rangle^2+\langle {\bf{e}}_2(t),{\bf{u}}\rangle^2+\langle {\bf{e}}_3(t),{\bf{u}}\rangle^2\geq d^2 \langle {\bf{v}},{\bf{u}}\rangle^2.
$$
Conjecture: There exist continuous $c_1,c_2,c_3:[0,1]\rightarrow \mathbb{R}$ such that 
$$
\forall t\in [0,1], \;c_1(t){\bf{e}}_1(t)+c_2(t){\bf{e}}_2(t)+c_3(t){\bf{e}}_3(t)={\bf{v}}.\quad \quad (E)
$$
Remarks: 


*

*The condition (C) guarantees that for each $t\in [0,1]$,  ${\bf{v}}$ belongs to the span $S$ of ${\bf{e}}_1(t),{\bf{e}}_2(t),{\bf{e}}_3(t)$. (If not, we can decompose ${\bf{v}}={\bf{v}}_{\perp}+{\bf{v}}_{\parallel}$, with ${\bf {0}}\neq {\bf{v}}_{\perp}\perp S$, and then taking ${\bf{u}}= {\bf{v}}_{\perp}$, we get the contradiction that $0\geq d^2 \langle {\bf{v}}_{\perp},{\bf{v}}_{\perp}\rangle^2>0$.) So there exist $c_1,c_2,c_3:[0,1]\rightarrow \mathbb{R}$ such that ($E$) holds. The question is if we can find continuous such functions $c_1,c_2,c_3$.

*The condition (C) is necessary: As ${\bf{v}}$ is nonzero, $f:=c_1^2+c_2^2+c_3^2$ is everywhere nonzero. So $f$ has a maximum value $M>0$. With $d=M^{-1}$ and Cauchy-Schwarz, 
\begin{eqnarray*}
d^2 \langle {\bf{v}},{\bf{u}}\rangle^2
&=&d^2 (c_1(t)\langle {\bf{e}}_1(t),{\bf{u}}\rangle+c_2(t)\langle {\bf{e}}_2(t),{\bf{u}}\rangle+c_3(t)\langle {\bf{e}}_3(t),{\bf{u}}\rangle)^2
\\
&\leq& \langle {\bf{e}}_1(t),{\bf{u}}\rangle^2+\langle {\bf{e}}_2(t),{\bf{u}}\rangle^2+\langle {\bf{e}}_3(t),{\bf{u}}\rangle^2,
\end{eqnarray*}
for all ${\bf{u}}\in \mathbf{R}^3$ and all $t\in [0,1]$.

 A: In case the vectors $e_1(t),e_2(t),e_3(t)$ form a basis for every $t$, you can solve Equation (E) and get $c_i(t)$ as rational functions in the coefficients of $e_i(t)$ with respect to some fixed basis with non-vanishing denominator. This implies that the functions $c_i(t)$ are continuous. 
The general case is different:
Let $f,g:[0,1]\to \mathbb{R}$ be two continuous functions which are equal to one another only in $t=0$, where $f(t)=g(t)=0$.
Take $e_1(t)=(1,f(t),0)$, $e_2(t)=(1,g(t),0)$, $e_3=(0,0,0)$ and $v=(1,0,0)$. For a vector $u=(x,y,z)$ the condition (C) above reads: $$(x+f(t)y)^2 + (x+g(t)y)^2 \geq d^2x^2.$$
For $d=1$ we get the inequality 
$$x^2 +2(f(t)+g(t))xy + (f(t)^2+g(t)^2)y^2\geq 0$$
The discriminant of the above equation is 
$$4[f^2+g^2+2fg-f^2-g^2]=8fg.$$
Therefore, if $f(t)g(t)\leq 0$ for every $t\in[0,1]$ we get that Condition (C) is satisfied.
By solving the above system of equations, we see that for $t\neq 0$ we have that 
$e_1(t)$ and $e_2(t)$ are linearly independent, and
$$v=\frac{-g(t)}{f(t)-g(t)}e_1(t) + \frac{f(t)}{f(t)-g(t)}e_2(t)$$ (the coefficient of $e_3(t)$ plays no role here). We also see that since $e_1(t)$ and $e_2(t)$ are linearly independent, $c_1(t)$ and $c_2(t)$ are uniquely defined for $t\neq 0$.
We choose now $f(t) = -t$, and $g(t) = |tsin(1/t)|$ and $g(0)=0$.
Notice that: $f(t)g(t)\leq 0$ for every $t\in [0,1]$ $g(t)=f(t)$ only for $t=0$, and $\lim_{t\to 0}\frac{f(t)}{f(t)-g(t)}=\lim_{t\to 0}\frac{1}{1+|sin(1/t)|}$ does not exist, because $\lim_{t\to 0}sin(1/t)$ does not exist.
This shows that the functions $c_2(t)$ and similarly $c_1(t)$ cannot be chosen continuously.
