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Suppose we know $$f(x+y) + f(x-y) = 2f(x) + 2f(y)$$ with $f: G \rightarrow K$ and $G$ an abelian group and $K$ a field.

How can we prove that for $$\langle x,y\rangle := \frac{1}{2}(f(x+y)-f(x)-f(y))$$ the following is satisfied: $$\langle a+b,c\rangle = \langle a,c\rangle+\langle b,c\rangle$$ for all $a,b,c$.

We can assume $\langle 0,0\rangle = 0$ and $\langle x,y\rangle = \langle y,x\rangle$.

I have tried calculating it and come to the conclusion that we can equivalently prove the following equation: $$f(a+b+c) = f(a+b) + f(b+c) + f(a+c) - f(a) - f(b) - f(c),$$ but I couldn't find a way to prove this.

The question is in the category of elliptic curves because in my case $f = \deg$, but it should not be important in this case.

egreg
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NightRain23
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3 Answers3

5

$$\langle a+b,c\rangle = \frac {1} {2} (f(a+b+c)-f(a+b)-f(c)).$$ On the other hand $$\langle a, c \rangle + \langle b,c \rangle = \frac 1 2 [f(a+c)-f(a)-f(c)+f(b+c) -f(b) - f(c)].$$

Need to show that $$\langle a+b,c \rangle = \langle a,c \rangle + \langle b,c \rangle.$$

Putting $a=b$ in the equation $$f(a+b)+f(a-b)=2[f(a)+f(b)]$$ we get $$f(2a) = 4f(a)$$ since $f(0) = \langle 0, 0 \rangle = 0.$

Now $f(a+c)-f(a)-f(c) = \frac 1 2 (f(a+c)-f(a-c))$ and $f(b+c)-f(b)-f(c) = \frac 1 2 (f(b+c)-f(b-c)).$ Therefore we have $$\begin{align*} 2[\langle a,c \rangle + \langle b,c \rangle] & = \frac 1 2 [(f(a+c)+f(b+c))-(f(a-c)+f(b-c))]. \\ & = \frac 1 4 [(f(a+b+2c)+f(a-b)) - (f(a+b-2c)+f(a-b))]. \\ & = \frac 1 4 [f(a+b+2c) - f(a+b-2c)].\ {(*)} \end{align*}$$

Now $f(a+b-2c) = 2[f(a+b)+f(2c)]-f(a+b+2c)]$ and $f(a+b+2c) = 2 [f(a+b+c)+f(c)]-f(a+b).$

Plugging all these in $(*)$ we have

$$\begin{align*} 2[\langle a,c \rangle + \langle b,c \rangle] & = \frac 1 2 [2f(a+b+c)+2f(c)-2f(a+b) - f(2c)]. \end{align*}$$

Now using the fact that $f(2c)=4f(c)$ we have

$$\begin{align*} 2[\langle a,c \rangle + \langle b,c \rangle] & = f(a+b+c)-f(a+b)-f(c). \\ & = 2 \langle a+b,c \rangle. \end{align*}$$

So we have $$\langle a+b,c \rangle = \langle a,c \rangle + \langle b,c \rangle.$$

little o
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1

This is very closely related to the well known result that if a norm satisfies the parallelogram identity then it induces an inner product:

Norms Induced by Inner Products and the Parallelogram Law

You, of course, aren't proving linearity since there's no vector space structure, but for $\langle a + b, c \rangle = \langle a, c \rangle + \langle b, c \rangle$ the same algebra they go through in Step 2 of the second answer (by t.b.) applies (replacing $\lVert x \rVert^2$ with $f(x)$).

bitesizebo
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1

With $x = y$ we have

$$ f(2x)-4f(x) = f(0)\Rightarrow f(0) = 0 $$

so we follow with $f(2x)-4f(x) = 0$ and with $u = \log_2 x$ we have the recurrence

$$ F(u+1)-4F(u) = 0 $$

with solution

$$ F(u) = \Theta(u) 4^u $$

here $\Theta(u) = \Theta(u+1)$ is an arbitrary periodic function with unit period. Assuming $\Theta(u) = C_0$ constant we have

$$ f(x) = C_0 2^{\log_2 x^2} = C_0 x^2 $$

So with

$$ <x,y> := f(x+y)-f(x)-f(y) $$

we have

$$ <x,y> = C_0 x y\Rightarrow <a+b,c> = <a,c> + <b,c> $$

Cesareo
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