A trivial fact about factorisations of functions of two variables: how to prove it?

Question

Let $s<t \in [0,T]$. Let $g_1,g_2$ and $f_i, \tilde f_i, \, i=0,\ldots, n$, be real continuous functions on $[0,T]$ that are never $0$ such that

$$g_1(s) g_2(t) = \sum_{i=0}^n f_i(s) \tilde f_i(t).$$

I want to prove the trivial fact that either all $f_i(s)$ are multiples of $g_1(s)$, i.e $g_1(s)=\lambda_i f_i(s)$ where $\lambda_i \in \mathbb R_0$, or all $\tilde f_i(t)$ are multiples of $g_2(t)$.

It seems so simple, yet I cannot produce a rigorous proof. It seems a proof by contradiction would be suited for this, yet I do not see the contradiction.

Answer 1

This is not true.

We will find a counter-example when $g_1$ and $g_2$ are both constant functions equal to $1,$ and $n=1, T=2.$

Let $$f(s)=\begin{cases}(2,-1)&0\leq s<1\\\left(1+s,-s\right)&1\leq s\leq 2\end{cases}$$

$$\tilde f(t)=\begin{cases}(1,1)&1\leq t\leq 2\\\left(\frac32-\frac{t}2,2-t\right)&0\leq t<1\end{cases}$$

Now if $s<t$ then:

• Case: $0\leq s<t<\frac{1}{2},$ then $f(s)\cdot \tilde f(t)=(2,-1)\cdot \left(\frac32-\frac{t}2,2-t\right) =1.$
• Case: $\frac12\leq s<t\leq 1,$ then $f(s)\cdot \tilde f(t)=(1+s,-s)\cdot (1,1)=1.$
• Case: $0\leq s<\frac12\leq t\leq1.$ Then $f(s)\cdot \tilde f(t)=(2,-1)\cdot(1,1)=1.$

You can write out the four functions $f_i,\tilde f_i$ as case statements to ensure they are never zero, and none are constant functions.

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It is not even true if you make the equality true for all $s,t.$

If $U_t$ is a continuous set of planes, all containing the same common line in $\mathbb R^3,$ and none containing $0,$ you can define $f$ to be any continuous function restricted to the line, and define $\tilde f(t)$ to be a vector $u$ such that $U_t=\{v\mid u\cdot v=1\}.$

For example, the common line being $(1+3\alpha,1+\alpha,\alpha),$ the planes are $ax+by+cz=1$ with $a+b=1$ and $2a+2+c=1.$ So for $t\in [0,1],$ we get the equation for $U_t$ is $(3-t)x+(t-2)y+(2t-7)z=1$ and $$f(s)=(4+3s,2+s,1+s)\\\tilde f(t)=(3-t,t-2,2t-7)$$

Or $$1=(4+3s)(3-t)+(2+s)(t-2)+(1+s)(2t-7)$$ For all $s,t.$ With $s,t\in[0,1],$ all of $4+3s,2+s,1+s,3-t,t-2,2t-7$ are non-zero.

So, give any $g_1,g_2$ and $T=1,$ you can define:

$$\begin{align} f_0(s)&=(3s+4)g_1(s),\\f_1(s)&=(2+s)g_1(s),\\f_2(s)&=(1+s)g_1(s),\\\tilde f_0(t)&=(3-t) g_2(t)\\ \tilde f_1(t)&=(t-2) g_2(t)\\ \tilde f_2(t)&= (2t-7)g_2(t) \end{align} $$

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A little more work gives us strictly positive example:

$$36=(10+s)(1+t)+(2-s)(7-2t)+(2+s)(6-3t)$$ All the terms are positive when $s,t\in[0,1].$