The span of the orthorgonal projections is norm dense in $B(H)$

Question

This is a question in my functional analysis book.

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How to use the spectral theorem to prove that the span of the orthogonal projections is norm dense in $B(H)$?

Answer 1

Let $M$ be a von Neumann algebra and $P(M)=\{p\in M\;;\; p^2=p=p^*\}$ the set of its projections.

Claim: $\mbox{Span }P(M)$ is norm dense in $M$.

In particular, this is true for the von Neumann algebra $B(H)$. And it would not make it easier to restrict to this case.

Proof: Take $x$ in $M$. We can write $$ x=\frac{x+x^*}{2}+i\frac{x-x^*}{2i}=h+ik $$ with $h,k$ self-adjoint elements. So, without loss of generality, we can assume that $x$ is self-adjoint.

Then we have $L^\infty$ functional calculus for $x$.

Let $\epsilon>0$.

Construct a simple function $\phi=\sum_{1}^n\lambda_j1_{[a_j,b_j]}$ on $[-\|x\|;\|x\|]$ such that $$ \sup_{[-\|x\|;\|x\|]} |t-\phi(t)|\leq \epsilon. $$ To do that, draw the graph of the identity and approximate it by constant steps.

Note the above inequality is a fortiori true for the sup over the spectrum of $x$.

Then set $$ x_\epsilon:=\phi(x)=\sum_{1}^n\lambda_j1_{[a_j,b_j]}(x)=\sum_{1}^n\lambda_jp_j $$ and note that each $p_j$ is a projection in $M$, so $x_\epsilon$ belongs to the span of $P(M)$.

Finally, we have $$ \|x-x_\epsilon\|=\sup_{t\in \sigma(x)}|t-\phi(t)|\leq \epsilon. $$ So the span of $P(M)$ is indeed norm dense in $M$.