Is a sum of a set of linearly independent functions and their time-reversed versions also linearly independent? Let $\{f_i (t) : [0,T] \mapsto \mathbb{R}\}_{i\in I}$ be a set of linearly independent functions, i.e.:

\begin{equation}
\sum_{i\in I} c_i \, f_i (t) = 0   \Rightarrow   c_i = 0,
\end{equation}

for all $i\in I$ and for all $t$ in the support of $f_i$.

Now let the 'time-reversed' versions of $f_i$, namely $\{f_i (T-t)\}_{i\in I}$ also
be linearly independent, i.e.

\begin{equation}
\sum_{i\in I} c_i \, f_i (T-t) = 0  \Rightarrow c_i = 0,         
\end{equation}

for all ${i\in I}$ and for all $(T-t)$ in the support of $f_i$.

Question: do these linear independence assumptions (and any other assumptions that you might need to make $f_i$ 'well-behaved') imply that the set of functions:

\begin{equation}
\{ f_i (t) + f_i (T-t)\}_{i \in I}
\end{equation}

are linearly independent? I.e. does it follow that: 

\begin{equation}
\sum_{i\in I} c_i \, \big(f_i (t) + f_i (T-t)\big) = 0   \Rightarrow  c_i = 0,       
\end{equation}

for all $i\in I$ and for all $t$ in the support of $f_i (t) + f_i (T-t)$.

(You can make any assumptions you need on the smoothness or
differentiability etc of $f_i$ to make this work!)

Thanks for any help! :)
No, consider $f:[0,1]\rightarrow R$ defined by $f(x)=2x-1$, $f(1-x)=2(1-x)-1=-2x+1$, thus $f(1-x)+f(x)=-2x+1+2x-1=0$.
OK, well, clearly a set of one function is linearly independent and if it is also an odd function (about $T/2$) then $f(t)+f(T-t)=0$ everywhere but this is a very pathological case. How about for functions that are not odd about $T/2$ (and for sets, $I$, which have cardinality greater than one--- i.e. where there is a set of linearly independent functions $\{f_i(t), i \in I\}$)?