Conditions for Linear Independence for functions defined by integration Given that the set of strictly positive and continuous functions $$f_i(x,y) &gt;0, \quad i=1,\dots,n$$  are defined on $[0,1]^2$  and
 $\mathbb{R}$-linearly  independent for $(x,y)  \in [0,1]^2$. That is if $c_1, \ldots, c_n \in \mathbb{R}$ and  $\sum_{i=1}^n c_i f_i(x,y)=0$ for all $(x,y) \in [0,1]^2$ then each $c_i$ is zero.
 Let $g_i$ be defined by
$$
g_i(x) = \int_{y\in [0,1] } f_i(x,y) d y, \quad  i=1,\dots,n
$$
what are other conditions are needed on $f_i(x,y)$ so that the set of functions $g_i(x)$ are  $\mathbb{R}$-linearly  independent for $(x,y)  \in [0,1]$.

I'm having difficulties on this problem and any suggestions or references to read would be greatly appreciated. Some thoughts or attempts that I have done are 


If $f_i(x,y)= h_i(x) k_i(y)$ where the set of $h_i(x)$ is  $\mathbb{R}$-linearly  independent for $x  \in [0,1]$ and if the set of $k_i(y)$ is  $\mathbb{R}$-linearly  independent for $y  \in [0,1]$ then both $f_i(x,y)$ and $g_i$ are $\mathbb{R}$-linearly  independent?
If $f_i(x,y)$ is also $\mathbb{R}$-linearly independent when $x \in [0,1]$ for every fixed $y  \in [0,1]$. I'm just not sure how the proof would work in this case.
$\renewcommand{\p}{\partial}$
HINT:
Recall that any set of linearly independent functions has non zero Wronskian. 

\begin{align} 
\forall c_{1}, c_{2} \ldots, c_n \in \mathbb R  \quad  
c_{1} g_{1}\left(x\right) + c_{2} g_{2}\left(x\right) + \ldots + c_{n} g_{n}\left(x\right)  \not\equiv 0
 \impliedby 
W\big( g_{1}, g_{2}, \ldots, g_{n} \big) 
\not \equiv 0
\end{align}

Recall also  Leibniz formula  for differentiation under the integral with variable limits:

$$
\frac{d}{dx}\left( 
\int_{a\left(x\right)}^{b\left(x\right)} f\big(x,t\big)\, dt  \right)
= f\big(x,b\left(x\right)\big) \cdot b'\left(x\right)
- f\big(x,b\left(x\right)\big) \cdot a'\left(x\right)
+ \int_{a\left(x\right)}^{b\left(x\right)} 
\frac{\partial}{\partial x}\,f\left(x,t\right)\, dt 
$$



In your particular case

\begin{align} 
g_{i} \left(x\right) = \int_{0}^{1} f_{i} \left(x,y\right) \,d y
\implies
\frac{d\,g_i}{dx} = \int_{0}^{1} \frac{\partial\,}{\partial x}\Big( f_i\left(x,y\right) \Big)\,d y 
\implies
\frac{d^{k}g_i}{dx} = \int_{0}^{1} \frac{\partial^{k}\,}{\partial x^{k}} \Big( f_i\left(x,y\right) \Big)\,d y, 
\end{align}
for $\,k = 1, \,\ldots,\, n-1\,$ under assumption that functions $\,f_{i}\,$ have $\,n-1\,$ derivatives. 

In order for a set of functions $\,g_{1}, \ldots, g_{n}\,$ we require the non-zero Wronskian

\begin{align} 
W = 
\begin{vmatrix}
g_{1} &amp; g_{2} &amp; \cdots &amp; g_{n} \\
g_{1}^{\left(1\right)} &amp; g_{2}^{\left(1\right)} &amp;\cdots&amp; g_{n}^{\left(1\right)}\\
g_{1}^{\left(2\right)} &amp; g_{2}^{\left(2\right)} &amp;\cdots&amp; g_{n}^{\left(2\right)}\\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots\\
g_{1}^{\left(n-1\right)} &amp; g_{2}^{\left(n-1\right)} &amp; \cdots &amp; g_{n}^{\left(n-1\right)}
\end{vmatrix} =
\begin{vmatrix}
\int_{0}^{1} f_{1} \,dy
&amp; \int_{0}^{1} f_{2}\,dy
&amp; \cdots 
&amp; \int_{0}^{1} f_{n}\,dy 
\\
\int_{0}^{1} \frac{\partial \,f_{1} }{\partial x }\,dy 
&amp; \int_{0}^{1} \frac{\partial \,f_{2} }{\partial x}\,dy
&amp; \cdots 
&amp; \int_{0}^{1} \frac{\partial \,f_{n} }{\partial x}\,dy
\\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots\\
\int_{0}^{1} \frac{\partial^{n-2}\,f_{1} }{\partial x^{n-2}}\,dy 
&amp; \int_{0}^{1} \frac{\partial^{n-2}\,f_{2} }{\partial x^{n-2}}\,dy 
&amp; \cdots 
&amp; \int_{0}^{1} \frac{\partial^{n-2}\,f_{n} }{\partial x^{n-2}}\,dy 
\end{vmatrix}
\not \equiv 0
\end{align}

Nonzero Wronski determinant  is nonzero means that the column vectors of the matrix  are linearly independent.
Hope you can pick it from here.
$\renewcommand{\p}{\partial}$
HINT:
Recall that any set of linearly independent functions has non zero Wronskian. 

\begin{align} 
\forall c_{1}, c_{2} \ldots, c_n \in \mathbb R  \quad  
c_{1} g_{1}\left(x\right) + c_{2} g_{2}\left(x\right) + \ldots + c_{n} g_{n}\left(x\right)  \not\equiv 0
 \impliedby 
W\big( g_{1}, g_{2}, \ldots, g_{n} \big) 
\not \equiv 0
\end{align}

Recall also  Leibniz formula  for differentiation under the integral with variable limits:

$$
\frac{d}{dx}\left( 
\int_{a\left(x\right)}^{b\left(x\right)} f\big(x,t\big)\, dt  \right)
= f\big(x,b\left(x\right)\big) \cdot b'\left(x\right)
- f\big(x,b\left(x\right)\big) \cdot a'\left(x\right)
+ \int_{a\left(x\right)}^{b\left(x\right)} 
\frac{\partial}{\partial x}\,f\left(x,t\right)\, dt 
$$



In your particular case

\begin{align} 
g_{i} \left(x\right) = \int_{0}^{1} f_{i} \left(x,y\right) \,d y
\implies
\frac{d\,g_i}{dx} = \int_{0}^{1} \frac{\partial\,}{\partial x}\Big( f_i\left(x,y\right) \Big)\,d y 
\implies
\frac{d^{k}g_i}{dx} = \int_{0}^{1} \frac{\partial^{k}\,}{\partial x^{k}} \Big( f_i\left(x,y\right) \Big)\,d y, 
\end{align}
for $\,k = 1, \,\ldots,\, n-1\,$ under assumption that functions $\,f_{i}\,$ have $\,n-1\,$ derivatives. 

In order for a set of functions $\,g_{1}, \ldots, g_{n}\,$ we require the non-zero Wronskian

\begin{align} 
W = 
\begin{vmatrix}
g_{1} &amp; g_{2} &amp; \cdots &amp; g_{n} \\
g_{1}^{\left(1\right)} &amp; g_{2}^{\left(1\right)} &amp;\cdots&amp; g_{n}^{\left(1\right)}\\
g_{1}^{\left(2\right)} &amp; g_{2}^{\left(2\right)} &amp;\cdots&amp; g_{n}^{\left(2\right)}\\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots\\
g_{1}^{\left(n-1\right)} &amp; g_{2}^{\left(n-1\right)} &amp; \cdots &amp; g_{n}^{\left(n-1\right)}
\end{vmatrix} =
\begin{vmatrix}
\int_{0}^{1} f_{1} \,dy
&amp; \int_{0}^{1} f_{2}\,dy
&amp; \cdots 
&amp; \int_{0}^{1} f_{n}\,dy 
\\
\int_{0}^{1} \frac{\partial \,f_{1} }{\partial x }\,dy 
&amp; \int_{0}^{1} \frac{\partial \,f_{2} }{\partial x}\,dy
&amp; \cdots 
&amp; \int_{0}^{1} \frac{\partial \,f_{n} }{\partial x}\,dy
\\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots\\
\int_{0}^{1} \frac{\partial^{n-2}\,f_{1} }{\partial x^{n-2}}\,dy 
&amp; \int_{0}^{1} \frac{\partial^{n-2}\,f_{2} }{\partial x^{n-2}}\,dy 
&amp; \cdots 
&amp; \int_{0}^{1} \frac{\partial^{n-2}\,f_{n} }{\partial x^{n-2}}\,dy 
\end{vmatrix}
\not \equiv 0
\end{align}

Nonzero Wronski determinant  is nonzero means that the column vectors of the matrix  are linearly independent.
Hope you can pick it from here.