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Given that the set of strictly positive and continuous functions $$f_i(x,y) >0, \quad i=1,\dots,n$$ 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.
confused_dragon
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  • Your conditions may not be true. For example, $f_i(x,y) = x^i(y-1/2)^{2i+1}$ then $g_i(x)$ are all zero. – Quang Hoang Sep 18 '15 at 14:17
  • @QuangHoang- thanks for your input but the set of $f_i(x,y)$ need to be strictly positive and continuous functions – confused_dragon Sep 18 '15 at 18:15
  • Well, a real example is not hard to construct, e.g. $f_n(x,y) = 1+ \sin (2\pi n(x+y))$. No idea what feasible assumptions can be imposed... The problem is that it the linear independence is obviously true for functions of the form $h(x) k(y)$. But already for functions like $h(x) k(y) + l(x)m(y)$ this is false... – zhoraster Sep 28 '15 at 18:33

1 Answers1

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$\renewcommand{\p}{\partial}$ 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} & g_{2} & \cdots & g_{n} \\ g_{1}^{\left(1\right)} & g_{2}^{\left(1\right)} &\cdots& g_{n}^{\left(1\right)}\\ g_{1}^{\left(2\right)} & g_{2}^{\left(2\right)} &\cdots& g_{n}^{\left(2\right)}\\ \vdots & \vdots & \ddots & \vdots\\ g_{1}^{\left(n-1\right)} & g_{2}^{\left(n-1\right)} & \cdots & g_{n}^{\left(n-1\right)} \end{vmatrix} = \begin{vmatrix} \int_{0}^{1} f_{1} \,dy & \int_{0}^{1} f_{2}\,dy & \cdots & \int_{0}^{1} f_{n}\,dy \\ \int_{0}^{1} \frac{\partial \,f_{1} }{\partial x }\,dy & \int_{0}^{1} \frac{\partial \,f_{2} }{\partial x}\,dy & \cdots & \int_{0}^{1} \frac{\partial \,f_{n} }{\partial x}\,dy \\ \vdots & \vdots & \ddots & \vdots\\ \int_{0}^{1} \frac{\partial^{n-2}\,f_{1} }{\partial x^{n-2}}\,dy & \int_{0}^{1} \frac{\partial^{n-2}\,f_{2} }{\partial x^{n-2}}\,dy & \cdots & \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.

Vlad
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  • I've heard of the Wronskian but I have only seen it in action on particular sets like $x^2$, $sin(x)$ and $cos(x)$. I'm sorry but I'm not sure how to pick it up from here. Also should the last row have $$ \frac{\partial^{n-1} }{\partial x^{n-1}} $$ ? – confused_dragon Sep 19 '15 at 12:47