Linear Independence for functions defined by integration I came across this problem while doing some work. I'm been unable to make any progression on it. Any suggestions would be greatly appreciated. 
Given that the set of strictly positive and continuous functions $$f_i(x,y) >0, \quad i=1,\dots,n$$
are linearly independent for $(x,y)  \in [0,1]^2$. Let $g_i$ be defined by
$$
g_i(x) = \int_{y\in [0,1] } f_i(x,y) d y, \quad  i=1,\dots,n
$$
where $g_i$ is unique up to some positive multiple (There is a better way to say this?, say $g_i(x) \neq c g_j(x)$ for some $c >0$ and $i$ &  $j \in \left\{1,\ldots,n\right\}$ and $x \in(0,1)$ )
Is the set of functions, $g_1,\ldots,g_n$, also linearly independent for $x \in [0,1]$? 

No, it is not (always) linear independent.
Take $f_1(x,y) := 1+y^2$ and $f_2 = 1+(y-1)^2$ (on $[0,1]^2$).
As far as I understood it, $S:=\{f_1,f_2\}$ is good to go.  
$$g_1(x) = 1.5 = g_2(x)\quad \forall x \in [0,1].$$
So this proposition can't be true.
Possible explanation (no proof)
You lose information during integration, so they lose their distinction.
The $f_1(x,1-y)=f_2(x,y)\quad \forall x \in[0,1]^2$ is the problem here. The symmetry of the integral kills the difference.
