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Let $E$ be a resolution of the identity on a $\sigma$-algebra $\mathcal{A}$. Then by the Cauchy-Schwarz inequality we have

$$|E_{x,y}(\omega)|^{2}=|\langle E(\omega)x,y\rangle|^{2}\le \langle E(\omega)x,E(\omega)x\rangle\cdot\langle y,y\rangle$$.

and since $E(\omega)=E^{\ast}(\omega)$ and $E^{2}(\omega)=E(\omega)$, we get that the above is equal to

$$\langle E(\omega) x,x\rangle\cdot\langle y,y\rangle=E_{x,x}(\omega)\cdot\langle y,y\rangle,$$

which isn't quite what I wanted. Where did I go wrong?

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    What is a resolution? – 2017-01-06
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    $E(\omega)$ is a projection. So you get that your original expression is equal to $\lvert \langle E(\omega) x , E(\omega) y \rangle \rvert^2$. – 2017-01-06
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    @Vim It's a spectral projection. – 2017-01-06

1 Answers 1

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Method 1: The simplest way is to observe that $[x,y]=\langle E(w)x,y\rangle$ is a pseudo inner product (positive, but may not be positive-definite.) That's enough for Cauchy-Schwartz to hold: $$ |[x,y]|^2 \le [x,x][y,y] \\ |\langle E(w)x,y\rangle|^2 \le \langle E(w)x,x\rangle\langle E(w)y,y\rangle $$ Method 2: Or, you can appeal to Cauchy-Schwarz directly instead, using the fact that $$ \langle E(w)x,y\rangle = \langle E(w)x,E(w)y\rangle, $$ which follows because $E(w)=E(w)^*=E(w)^2$ is an orthogonal projection. To see this way, \begin{align} |\langle E(w)x,y\rangle|^2 &= |\langle E(w)x,E(w)y\rangle|^2 \\ &\le \langle E(w)x,E(w)x\rangle\langle E(w)y,E(w)y\rangle \\ &= \langle E(w)x,x\rangle\langle E(w)y,y\rangle. \end{align}