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Consider the following so-called bi-linear functional equation $$\sum_{i=1}^{n}f_i(x)g_i(y)=f_1(x)g_1(y)+f_2(x)g_2(y)+\cdot\cdot\cdot+f_n(x)g_n(y)=0 \tag{1}$$ where $f_i(x)$ and $g_i(y)$ are arbitrary functions. What conditions are required for such a functional equation to hold?

For example, for $n=2$ we have

$$\begin{align} f_1(x)g_1(y)+f_2(x)g_2(y) &= 0 \\ \frac{f_1(x)}{f_2(x)}+\frac{g_2(y)}{g_1(y)} &= 0 \end{align} \tag{2}$$

where I assumed that $f_2(x) \ne 0$ and $g_1(y) \ne 0$. Consequently, we can conclude that Eq.$(2)$ will hold if and only if

$$\begin{align} f_1(x) &= +\lambda f_2(x) \\ g_2(y) &= -\lambda g_1(y) \end{align} \tag{3}$$

where $\lambda$ is some constant. Considering other cases, I ended up with

$$\boxed{ \begin{array}{ll} 1.\, \text{if} \, f_2(x)=0 \, \text{and} \, g_1(y)=0 \, \text{then} & 0=0 \\ 2.\, \text{if} \, f_2(x)\ne0 \, \text{and} \, g_1(y)=0 \, \text{then} & g_2(y)=0 \\ 3.\, \text{if} \, f_2(x)=0 \, \text{and} \, g_1(y)\ne0 \, \text{then} & f_1(x)=0 \\ 4.\, \text{if} \, f_2(x)\ne0 \, \text{and} \, g_1(y)\ne0 \, \text{then} & f_1(x) = +\lambda f_2(x) \, \text{and} \, g_2(y) = -\lambda g_1(y) \end{array} }\tag{4}$$

How can I generalize the result in $(4)$ for functional equation $(1)$?

Any help is appreciated.

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    $\begingroup$ $F(x)=(f_1(x),f_2(x),\cdots)$ and $G(y)=(g_1(y),g_2(y),\cdots)$ are two orthogonal vectors in an $n$-dimensional Euclidean space. $\endgroup$
    – entrelac
    Jul 20, 2016 at 19:09
  • $\begingroup$ @GeorgSaliba: Does that help? :) $\endgroup$
    – Hosein Rahnama
    Jul 20, 2016 at 19:22
  • $\begingroup$ Well, for $2$D space, the product of the slopes of two perpendicular lines is equal to $-1$ which is equivalent to $(3)$. Reading Robert's answer I realize that my comment is similar, but in much simpler, less detailed terms. $\endgroup$
    – entrelac
    Jul 20, 2016 at 19:27

1 Answer 1

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I'll suppose $f_i$ and $g_i$ are defined on sets $D_f$ and $D_g$ respectively, taking values in a field $\mathbb F$ (you're probably thinking of $\mathbb F = \mathbb R$ or $\mathbb C$).

Suppose the linear span of $(f_1, \ldots, f_n)$ (as functions of $x \in D_f$) has dimension $r$. Thus the vectors $(c_1, \ldots, c_n)$ such that $c_1 f_1 + \ldots + c_n f_n = 0$ form a subspace $V$ of $\mathbb F^n$ of dimension $n-r$. Then (1) is equivalent to: $(g_1(y), \ldots, g_n(y)) \in V$ for all $y \in D_g$.

For example, in the case $n=2$ your solution corresponds to the case $r=1$, where $V$ is spanned by $(1, -\lambda)$.

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  • $\begingroup$ Sorry Robert but I encountered some questions! :) Here is what I got from your answer. We introduce two spaces $V=\text{Span}\{f_1,f_2,..,f_n\}$ and $W=\text{Span}\{g_1,g_2,..,g_n\}$. Afterwards, we claim that if $\text{Dim}(V)=r$ and $\sum_{i=1}^{n}f_i(x)g_i(y)=0$ then $\text{Dim}(W)=n-r$. Can you explain that why the claim is valid? Sorry if this is a rudimentary question! I am confused! :) $\endgroup$
    – Hosein Rahnama
    Jul 23, 2016 at 22:39
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    $\begingroup$ That's not at all what I said. Let me try to clarify. Take any $f_1, \ldots, f_n$. What is the condition on $(g_1, \ldots, g_n)$ to make your equation (1) hold? Let $r$ be the dimension of the span of $f_1, \ldots, f_n$. There is a subspace $V$ of $\mathbb F^n$ of dimension $n-r$ consisting of those $(c_1, \ldots, c_n)$ such that $c_1 f_1 + \ldots + c_n f_n$. Then (1) says $g_1(y) f_1 + \ldots + g_n(y) f_n = 0$ for all $y \in D_g$, i.e. that $(g_1(y), \ldots, g_n(y))$ is a member of $V$ for all $y \in D_g$. $\endgroup$
    – Robert Israel
    Jul 24, 2016 at 6:04
  • $\begingroup$ Ah! Thanks. I just mixed things in my head! :) What can we say about the dimension of $W=\text{Span}\{g_1,g_2,..,g_n\}$? :) $\endgroup$
    – Hosein Rahnama
    Jul 25, 2016 at 8:45
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    $\begingroup$ It should have dimension at most $n-r$. $\endgroup$
    – Robert Israel
    Jul 25, 2016 at 15:55
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    $\begingroup$ Rank-nullity theorem. If $W$ is the linear span of $f_1, \ldots, f_n$, the linear operator $T: c \mapsto c_1 f_1 + \ldots + c_n f_n$ from $\mathbb F^n$ to $W$ is surjective and thus has rank $r$. Its null space $V$ has dimension $n - r$. $\endgroup$
    – Robert Israel
    Oct 17, 2017 at 15:55

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