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 $x \in [0,1]$? 
 A: No, not in general. For $n$ a natural number greater than or equal to $3$, let
$$f_i(x,y) = \left\{ \begin{array}{l l} (i+1)(xy)^i + (i+1)y^i & \text{for $i \in \{1,...,n-1\}$}\\
\sum_{j=1}^{n-1} (j+1)(xy)^j + (n-1)& \text{for $i=n$}\end{array}\right.$$
Then the the functions $f_1,...,f_n$ are linearly independent. Let us see why this is so. For any linear combination we have:
$$\sum_{i=1}^n a_i f_i(x,y) = \sum_{i=1}^{n-1}(a_i+a_n)(i+1)(xy)^i + \sum_{i=1}^{n-1}a_i(i+1)y^i +a_n(n-1)$$
and since a polynomial is zero if and only if all its coefficients are zero we see that the linear combination is zero if and only if $a_1=a_2=..=a_n=0$.
 However the functions
$$g_i(x) = \left\{\begin{array}{l l} x^i+1 & \text{for $i\in\{1,..,n-1\}$}\\
\sum_{j=1}^{n-1}x^j + (n-1) & \text{for $i=n$}\end{array}\right.$$
are linearly dependent since $$\sum_{i=1}^{n-1} g_i(x) = g_n(x).$$ One can show more that any $n-1$ of the $g_i$'s are linearly independent (which is much stronger that your requirement that none of them are scalar multiples of each other).
From one of your comments "I believe the solution relies on the fact that the mapping from $f_i$ to $g_i$ is an isomorphism" it seems that you perhaps wanted to ask a different question (since definite integration is far from injective). Perhaps you wanted to ask if $f_i:[0,1]\times [0,1]\to \mathbb{R}$ and $g_i : [0,1]\times [0,1] \to \mathbb{R}$ for $i \in \{1,..,n\}$ are functions such that the partial derivative of $g_i$ with respect to $y$ is $f_i$, then $f_1,..,f_n$ being linearly independent implies $g_1,..,g_n$ are linearly independent. This is indeed so since $\sum_{i=1}^n a_ig_i(x,y) = 0$ implies (by taking the partial derivative) that $\sum_{i=1}^{n} a_i f_i(x,y)=0$ which then implies that $a_1=a_2=..=a_n=0$.
A: 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.
