3
$\begingroup$

I am stuck with an exercise that I found in a textbook by Conway. First, I would like to clarify what is meant by a semi-inner product.

Definition. Suppose that $\mathscr X$ is a vector space over the complex field $\mathbb C$. A semi-inner product on $\mathscr X$ is a function $u:\mathscr X\times\mathscr X\to\mathbb C$ such that for all $\alpha,\beta$ in $\mathbb C$, and $x,y,z$ in $\mathscr X$, the following are satisfied:

  • $u(\alpha x+\beta y,z)=\alpha u(x,z)+\beta u(y,z)$,
  • $u(x,x)\ge 0$,
  • $u(x,y)=\overline{u(y,x)}$,

where $\bar\alpha$ is the complex conjugate of $\alpha$.

The difference between an inner product and a semi-inner product is that an inner product also satisfies the following:

  • if $u(x,x)=0$, then $x=0$.

Now I formulate the exercise from the textbook.

Let $u(\cdot,\cdot)$ be a semi-inner product on $\mathscr X$. Then $$\left|u(x,y)\right|^2=u(x,x)u(y,y)$$ if and only if there are $\alpha$ and $\beta$ in $\mathbb C$, not both $0$, such that $u(\beta x+\alpha y,\beta x+\alpha y)=0$.

How can I show that if there are $\alpha$ and $\beta$ in $\mathbb C$, not both $0$, such that $u(\beta x+\alpha y,\beta x+\alpha y)=0$, then $\left|u(x,y)\right|^2=u(x,x)u(y,y)$?

  • 0
    I rather strongly suspect that the requirement should say “not both $0$”, not “both not $0$”.2012-10-19
  • 0
    Have you tried expanding out $u(\beta x+\alpha y,\beta x+\alpha y)$ and look at the result? Note that the result must always be nonnegative for all choices of $\alpha$, $\beta$, and zero for the given pair. This should be useful.2012-10-19
  • 0
    The second condition in the definition should be $u(x,x) \geq 0$ not $u(x,y) \geq 0$. (If $u(x,y) \geq 0$ for all $x$ and $y$ then $u$ is identically equal to $0$).2012-10-19
  • 0
    Thank you, Yury, for correcting my typo.2012-10-19
  • 0
    Reply to Harald's remark: the requirement in the textbook says "both not $0$". I guess that otherwise the proposition is false. If $\alpha=0$ and $\beta\ne0$, then $u(x,x)=0$. Assume that $x\ne0$ (the equality is valid if $x=0$), but $u(x,x)=0$. Unless we show that $x(x,y)=0$ for all $y\in\mathscr X$ if $u(x,x)=0$, we get a contradiction. Is it possible to construct an example of a semi-inner product that for some $x\ne 0$ $u(x,x)=0$, but there exists $y\in\mathscr X$ such that $u(x,y)\ne0$?2012-10-19
  • 0
    @Harald: Yes, that is exactly what I tried to do. Let us denote $\gamma=\alpha/\beta$. Then we obtain the following: $$u(x+\gamma y,x+\gamma y)=u(x,x)+\bar\gamma u(x,y)+\gamma u(y,x)+\left|\gamma\right|^2u(y,y).$$ I am not sure what to do next... If we restrict our attention to the vector space $\mathscr X$ over the real field $\mathbb R$, then we can get the result in the following way. The quadratic polynomial $$q(t)=u(x,x)+2u(x,y)t+u(y,y)t^2$$ is non-negative and is equal to $0$ for some $t\in\mathbb R$. Hence the discriminant is equal to zero $$4\left|u(x,y)\right|^2-4u(x,x)u(y,y)=0.$$2012-10-19
  • 0
    Well, that was my point; if $u(x,x)=0$ then $u(x,y)=0$ for all $y$ by the C–B–S inequality.2012-10-19
  • 0
    Here is a point to consider: Note that $u(y,x)=\overline{u(x,y)}$ (this follows by polarization). So $\bar\gamma u(x,y)+\gamma u(y,x)$ is real. Now replace $\gamma$ by $\gamma t$ with $\gamma$ fixed and $t$ real and variable, and employ your discriminant argument. You're almost there.2012-10-19
  • 0
    @Harald: Thank you very much for your advice. It seems that I only need one final step. Let $\gamma=\left|\gamma\right|e^{i\theta}$ and $u(x,y)=\left|u(x,y)\right|e^{-i\varphi}$. We have that $$\left(\bar\gamma u(x,y)+\gamma u(y,x)\right)^2=4\left|\gamma\right|^2u(y,y)u(x,x)$$ and $$\left|u(x,y)\right|^2\left(e^{-i\theta}e^{i\varphi} +e^{i\theta}e^{-i\varphi}\right)^2=4u(y,y)u(x,x).$$ How can I show that $e^{-i\theta}e^{i\varphi}+e^{i\theta}e^{-i\varphi}=2$? Or am I making a mistake somewhere?2012-10-19
  • 0
    @Harald: I agree with your remark about the requirement. It should say "not both $0$". My previous comment is wrong.2012-10-19
  • 0
    That quantity *must* be $\pm2$, or else your equation cannot hold (because $\lvert u(x,y)\rvert\le u(x,x)u(y,y)$).2012-10-19

2 Answers 2