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I’m studying Conway’s Functional Calculus, and meet this question.

If $E_i, E$ are projections then $E_i \to E$ in WOT is equivalent to $E_i\to E$ in SOT.

I approached to this problem that If it converge in SOT sense, then it trivially converge in WOT.

So we need to care about if WOT converge, then SOT converge.

I tried this formula.

$\langle E_i h, h \rangle \to \langle Eh, h \rangle$ And we know that $E_i $ is adjoint and $E_i^2 = E_i$. So, $\langle E_i h, E_i h \rangle \to \langle Eh, Eh \rangle$

But it does not prove the relation $\|E_i h — Eh \| \to 0$

Can you help me?

s.harp
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probafds123
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1 Answers1

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Note that: $$\|(E-E_i)h\|^2 = \|Eh\|^2+\|E_ih\|^2 - \langle Eh, E_ih\rangle - \langle E_ih,Eh\rangle$$ As you have seen, $E_i\to E$ in WOT implies $\|E_ih\|^2\to \|Eh\|^2$, this is the only step that uses $E_i$ and $E$ being projections. From the definition WOT convergence it follows that $\langle Eh, E_ih\rangle\to\langle Eh,Eh\rangle = \|Eh\|^2$ and that $\langle E_ih,E\rangle \to \langle Eh, Eh,\rangle = \|Eh\|^2$. So in total you get:

$$\|(E-E_i)h\|^2 = \|Eh\|^2+\|E_ih\|^2 - \langle Eh, E_ih\rangle - \langle E_ih,Eh\rangle \to \|Eh\|^2+\|Eh\|^2-\|Eh\|^2-\|Eh\|^2=0$$

for any $h$, which gives SOT convergence.

s.harp
  • 22,576