Expression for bilinear function

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Theorem

Let $f$ be a real function of two independent variables, $f \in \R \times \R \to \R$.

Then:

$\map f {x, y}$ is a linear function of $x$ when $y$ is equal to a real constant

$\map f {x, y}$ is a linear function of $y$ when $x$ is equal to a real constant

if and only if $f$ has the form:

$\exists a, b, c, d \in \R: \forall {x, y} \in \R: \map f {x, y} = a x y + b x + c y + d$

Proof

Sufficient Condition

Let:

$\map f {x, y}$ be a linear function of $x$ when $y$ is equal to a real constant

$\map f {x, y}$ be a linear function of $y$ when $x$ is equal to a real constant

We need to show that:

$\exists a, b, c, d \in \R: \forall {x, y} \in \R: \map f {x, y} = a x y + b x + c y + d$

Expressions for $\map f {x, y}$ when either $x$ or $y$ is constant

We have:

$\map f {x, y}$ is a linear function of $x$ when $y$ is equal to a real constant.

This means that:

$\exists \alpha, \beta, y \in \R: \forall x \in \R: \map f {x, y} = \alpha x + \beta$

The real constants $\alpha$ and $\beta$ may depend on $y$ as $y$ is constant.

In other words:

$\forall {x, y} \in \R: \map f {x, y} = \map {a_1} y x + \map {a_2} y$

where $a_1$ and $a_2$ are real functions, $a_1, a_2 \in \R \to \R$.

By symmetry, we find from:

$\map f {x, y}$ is a linear function of $y$ when $x$ is equal to a real constant

that:

$\forall {x, y} \in \R: \map f {x, y} = \map {b_1} x y + \map {b_2} x$

where $b_1$ and $b_2$ are real functions, $b_1, b2 \in \R \to \R$.

$\Box$

Expression for $\map f {x, y}$ for every value of $x$ and $y$

We have:

\(\ds \forall {x, y} \in \R: \map f {x, y}\)

\(=\)

\(\ds \map {a_1} y x + \map {a_2} y\)

\(\, \ds \land \, \)

\(\ds \forall {x, y} \in \R: \map f {x, y}\)

\(=\)

\(\ds \map {b_1} x y + \map {b_2} x\)

\(\ds \leadsto \ \ \)

\(\ds \forall {x, y} \in \R: \map {a_1} y x + \map {a_2} y\)

\(=\)

\(\ds \map {b_1} x y + \map {b_2} x\)

\(\ds \leadsto \ \ \)

\(\ds \forall x \in \R: \map {a_1} 0 x + \map {a_2} 0\)

\(=\)

\(\ds \map {b_2} x\)

$y = 0$

\(\ds \leadsto \ \ \)

\(\ds \forall x \in \R: \map {b_2} x\)

\(=\)

\(\ds \map {a_1} 0 x + \map {a_2} 0\)

In other words, $b_2$ is a linear function of $x$.

We have:

\(\ds \forall {x, y} \in \R: \map {a_1} y x + \map {a_2} y\)

\(=\)

\(\ds \map {b_1} x y + \map {b_2} x\)

\(\ds \leadsto \ \ \)

\(\ds \forall {x, y} \in \R: \map {a_1} y x + \map {a_2} y\)

\(=\)

\(\ds \map {b_1} x y + \map {a_1} 0 x + \map {a_2} 0\)

$\map {b_2} x = \map {a_1} 0 x + \map {a_2} 0$

\(\ds \leadsto \ \ \)

\(\ds \forall x \in \R: \map {a_1} 1 x + \map {a_2} 1\)

\(=\)

\(\ds \map {b_1} x + \map {a_1} 0 x + \map {a_2} 0\)

$y = 1$

\(\ds \leadsto \ \ \)

\(\ds \forall x \in \R: \map {b_1} x\)

\(=\)

\(\ds \map {a_1} 1 x + \map {a_2} 1 - \paren {\map {a_1} 0 x + \map {a_2} 0}\)

\(\ds \leadsto \ \ \)

\(\ds \forall x \in \R: \map {b_1} x\)

\(=\)

\(\ds \paren {\map {a_1} 1 - \map {a_1} 0} x + \map {a_2} 1 - \map {a_2} 0\)

In other words, $b_1$ is a linear function of $x$.

In conclusion, we have:

\(\ds \forall {x, y} \in \R: \map f {x, y}\)

\(=\)

\(\ds \map {b_1} x y + \map {b_2} x\)

\(\ds \leadsto \ \ \)

\(\ds \forall {x, y} \in \R: \map f {x, y}\)

\(=\)

\(\ds \map {b_1} x y + \map {a_1} 0 x + \map {a_2} 0\)

$\map {b_2} x = \map {a_1} 0 x + \map {a_2} 0$

\(\ds \leadsto \ \ \)

\(\ds \forall {x, y} \in \R: \map f {x, y}\)

\(=\)

\(\ds \paren {\paren {\map {a_1} 1 - \map {a_1} 0} x + \map {a_2} 1 - \map {a_2} 0} y + \map {a_1} 0 x + \map {a_2} 0\)

$\map {b_1} x = \paren {\map {a_1} 1 - \map {a_1} 0} x + \map {a_2} 1 - \map {a_2} 0$

\(\ds \leadsto \ \ \)

\(\ds \forall {x, y} \in \R: \map f {x, y}\)

\(=\)

\(\ds \paren {\map {a_1} 1 - \map {a_1} 0} x y + \paren {\map {a_2} 1 - \map {a_2} 0} y + \map {a_1} 0 x + \map {a_2} 0\)

\(\ds \leadsto \ \ \)

\(\ds \forall {x, y} \in \R: \map f {x, y}\)

\(=\)

\(\ds \paren {\map {a_1} 1 - \map {a_1} 0} x y + \map {a_1} 0 x + \paren { \map {a_2} 1 - \map {a_2} 0} y + \map {a_2} 0\)

This expression for $\map f {x, y}$ is of the sought for form:

$\exists {a, b, c, d} \in \R: \forall {x, y} \in \R: \map f {x, y} = a x y + b x + c y + d$

$\Box$

Necessary Condition

Let:

$\exists {a, b, c, d} \in \R: \forall {x, y} \in \R: \map f {x, y} = a x y + b x + c y + d$

We need to show that:

$\map f {x, y}$ is a linear function of $x$ when $y$ is equal to a real constant

$\map f {x, y}$ is a linear function of $y$ when $x$ is equal to a real constant

Let:

$y$ be equal to a real constant.

We have:

\(\ds \exists {a, b, c, d} \in \R: \forall {x, y} \in \R: \map f {x, y}\)

\(=\)

\(\ds a x y + b x + c y + d\)

\(\, \ds \land \, \)

\(\ds y\)

\(=\)

\(\ds \text {a real constant}\)

\(\ds \leadsto \ \ \)

\(\ds \exists {a, b, c, d, y} \in \R: \forall x \in \R: \map f {x, y}\)

\(=\)

\(\ds a x y + b x + c y + d\)

\(\ds \leadsto \ \ \)

\(\ds \exists {a, b, c, d, y} \in \R: \forall x \in \R: \map f {x, y}\)

\(=\)

\(\ds \paren {a y + b} x + c y + d\)

So, $\map f {x, y}$ is a linear function of $x$ as $a y + b$ and $c y + d$ are constant.

Now, let:

$x$ be equal to a real constant.

We have:

\(\ds \exists a, b, c, d \in \R: \forall {x, y} \in \R: \map f {x, y}\)

\(=\)

\(\ds a x y + b x + c y + d\)

\(\, \ds \land \, \)

\(\ds x\)

\(=\)

\(\ds \text {a real constant}\)

\(\ds \leadsto \ \ \)

\(\ds \exists {a, b, c, d, x} \in \R: \forall y \in \R: \map f {x, y}\)

\(=\)

\(\ds a x y + b x + c y + d\)

\(\ds \leadsto \ \ \)

\(\ds \exists {a, b, c, d, x} \in \R: \forall y \in \R: \map f {x, y}\)

\(=\)

\(\ds \paren {a x + c} y + b x + d\)

So, $\map f {x, y}$ is a linear function of $y$ as $a x + c$ and $b x + d$ are constant.

Hence the result.

$\blacksquare$