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I'm having trouble with the following problem in John Conway's book on Functional Analysis:

Let $u$ be a semi-inner product on $\mathcal{X}$ and put $\mathcal{N} = \{x \in \mathcal{X} \, : \, u(x,x) = 0 \}$.

(a) Show that $\mathcal{N}$ is a linear subspace of $\mathcal{X}$.

(b) Show that if

\begin{align*} \langle x + \mathcal{N}, y + \mathcal{N} \rangle = u(x,y) \end{align*} for all $x + \mathcal{N}$ and $y + \mathcal{N}$ in the quotient space $\mathcal{X}/\mathcal{N}$, then $\langle \cdot, \cdot \rangle$ is a well-defined inner product on $\mathcal{X}/\mathcal{N}$.

I was able to prove part (a), but I'm having some trouble showing that $\langle \cdot, \cdot \rangle$ is a well-defined on $\mathcal{X}/\mathcal{N}$. I would like to prove that if $x \in \mathcal{N}$, then for any $y \in \mathcal{X}$ we have $u(x,y) = 0$. This would imply that if $x + \mathcal{N} = y + \mathcal{N}$, then for any $z \in \mathcal{X}$

\begin{align*} \langle x + \mathcal{N}, z + \mathcal{N} \rangle - \langle y + \mathcal{N}, z + \mathcal{N} \rangle & = \langle (x-y) + \mathcal{N}, z + \mathcal{N} \rangle \\ & = u(x-y, z) \\ & = 0 \end{align*}

since $x-y \in \mathcal{N}$. A similar argument should work for $\langle z + \mathcal{N}, x + \mathcal{N} \rangle = \langle z + \mathcal{N}, y + \mathcal{N} \rangle$. I just can't see why $u(x,y) = 0$ for all $x \in \mathcal{N}$ and $y \in \mathcal{X}$.

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2 Answers 2

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What you require is Cauchy -Schwarz inequality : $|u(x,y)| \leq \sqrt {|u(x,x)|} \sqrt {|u(y,y)|}$. To prove this inequality start with $u(x+ay,x+ay) \geq 0$, expand the inner product and choose $a$ suitably. Since this is a standard argument used in all versions of Cauchy -Schwarz inequality I am omitting the details, but feel free to aks me for details if needed.

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You need to show that if $x+N = x'+N, y+N=y'+N$ then $u(x,y) = u(x',y')$.

Recall that $|u(a,b)|^2 \le u(a,a) u(b,b)$, hence if either $a$ or $b$ is in $N$ then $u(a,b) = 0$.

Then \begin{eqnarray} u(x',y') &=& u(x+x'-x),y+y'-y) \\ &=& u(x,y+y'-y) + u(x'-x,y+y'-y) \\ &=& u(x,y+y'-y) \\ &=& u(x,y)+u(x,y'-y) \\ &=& u(x,y) \end{eqnarray}

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