# Equality in the Cauchy-Bunyakowsky-Schwarz inequality for a semi-inner product

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. – Cm7F7Bb 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 $u(x,y)\ne0$? – Cm7F7Bb Oct 19 '12 at 12:27

Instead of $\mu \langle x, y\rangle$ I have used $\langle x, y\rangle.$
Let $\langle\mathcal{X}, .\rangle$ be semi inner product. Let $x$, $y$ be fixed vectors in $\mathcal{X}$ and $\gamma$ be scalar. Consider $$\langle x - \gamma y, x - \gamma y\rangle = \langle x, y\rangle - \gamma\langle y,x\rangle - \bar\gamma\langle x, y\rangle + |\gamma|^2\langle y, y\rangle$$ Put $\langle y,x\rangle = b \mathrm{e}^{i\lambda} (b \ge 0)$ , $\gamma = t\mathrm{e}^{-i\lambda}$ (t is real), $a = \langle y, y\rangle, c = \langle x, x\rangle$ Note here that $\lambda, a, c$ are constants whereas $t$ is real variable. With this, we have $$\langle x - \gamma y, x - \gamma y\rangle = c - 2bt + at^2 \tag{1}$$ Now, $$|\langle x, y\rangle|^2 = \langle x, x\rangle\langle y, y\rangle \iff b^2 - ac = 0 \iff 4b^2 - 4ac = 0$$ $\iff c - 2bt + at^2 = 0$ 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 $$\langle x - \gamma_0 y, x - \gamma_0 y\rangle = c - 2bt_0 + at_0^2 = 0$$ Thus, the required scalars in your problem are $\beta = 1$ and $\alpha = -\gamma_0$
If $\langle y,x\rangle=be^{i\lambda}$, how do we know that $\gamma=t e^{-i\lambda}$? We only know that $\langle x-\gamma y,x-\gamma y\rangle=0$, where $\gamma\in\mathbb C$. We can express any complex number as $\gamma=te^{i\varphi}$, but how do we know that $\varphi=-\lambda$? This trick is used in the proof of the CBS inequality, but there we can choose $\gamma\in\mathbb C$ freely and here $\gamma\in\mathbb C$ is fixed. – Cm7F7Bb Feb 3 '15 at 14:30