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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 $\left|u(x,y)\right|^2=u(x,x)u(y,y)$?

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I rather strongly suspect that the requirement should say “not both $0$”, not “both not $0$”. –  Harald Hanche-Olsen Oct 19 '12 at 11:17
    
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. –  Harald Hanche-Olsen Oct 19 '12 at 11:20
    
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 $0$). –  Yury Oct 19 '12 at 12:00
    
Thank you, Yury, for correcting my typo. –  V. C. Oct 19 '12 at 12:03
    
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 $u(x,y)\ne0$? –  V. C. Oct 19 '12 at 12:27

2 Answers 2

I don't like to introduce $\lambda = \beta/\alpha$ because this destroys the symmetry between $x$ and $y$. I'd rather begin with an algebraic observation: the polynomial (aka quadratic form) $P(t,s)=At^2+2Bts+Cs^2 $ is:

  • is strictly positive on $\mathbb R^2\setminus \{(0,0)\}$ iff $AC>B^2$ and $A>0$
  • is nonnegative on $\mathbb R^2\setminus \{(0,0)\}$ iff $AC\ge B^2$

Both follow from the quadratic formula.

Expand $u(\beta x+\alpha y, \beta x+\alpha y)$ by linearity: it is equal to $$ |\beta|^2 u(x,x) + |\alpha|^2 u(y,y) + 2 \operatorname{Re}(\beta \bar \alpha u(x,y)) \\ \le |\beta|^2 u(x,x) + |\alpha|^2 u(y,y) + 2 |\beta \alpha| \,|u(x,y)| $$ The right hand side is nonnegative. Apply the above fact about quadratic forms with $|\beta|=t$ and $|\alpha|=s$. This gives the CBS inequality and relates the equality case to the vanishing of $u(\beta x+\alpha y, \beta x+\alpha y)$.

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Instead of $\mu <x, y> $ I have used $ <x, y>.$

Let $ <\mathcal{X}, .>$ be semi inner product. Let $x $, $y $ be fixed vectors in $\mathcal{X}$ and $\gamma$ be scalar. Consider $$ <x - \gamma y, x - \gamma y> = <x, x> - \gamma<y, x> - \bar\gamma<x, y> + |\gamma|^2<y,y> $$ Put $<y,x> = b \mathrm{e}^{i\lambda} (b \ge 0) $ , $ \gamma = t\mathrm{e}^{-i\lambda}$ (t is real), $ a = <y, y>, c = <x, x> $ Note here that $\lambda, a, c $ are constants whereas $ t $ is real variable. With this, we have \begin{equation} <x - \gamma y, x - \gamma y> = c - 2bt + at^2 ....(1) \end{equation} Now, $$ |<x,y>|^2 = <x, x><y,y> \iff b^2 - ac = 0 \iff 4b^2 - 4ac = 0 $$ $\iff c - 2bt + at^2 = 0 $ have equal roots. This is true if and only if $ c - 2bt + at^2 = 0$ has unique real root, say $ t_0 $. So by taking $\gamma_0 = t_0\mathrm{e}^{-i\lambda}$, from the equation (1), we obtain that $$ <x - \gamma_0 y, x - \gamma_0 y>= c - 2bt_0 + at_0^2 = 0 $$ Thus, the required scalars in your problem are $\beta = 1 $ and $\alpha = -\gamma_0 $

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