Reducing Subspace and invariant subspace I know that If $A\in\mathbb{B}(\mathbb{H})$,$\mathbb{M}\subset\mathbb{H}$, and $P=P_{\mathbb{M}}$(Projection) , then T.F.A.E (a)~(b)

(a)$\mathbb{M}$ is invariant for A

(b)PAP=AP

Also, T.F.A.E (c)~(d)

(c) $\mathbb{M}$ reduces A

(d) PA=AP

Let $\mu$=Area measure on $D=\{z\in\mathbb{C}:\vert z\vert &lt;1\}$ and define $A:L^2(\mu)\rightarrow L^2(\mu)$ by $(Af)(z)=zf(z)$ for $\vert z \vert&lt;1$ and f in $L^2(\mu)$.Find a nontrivial reducing subspace for A and an invariant subspace that does not reduce A.

How to find Invariant subspace and reducing subspace?
Let $D_{\frac12}=\left\{z\in\mathbb C:|z|&lt;\frac12\right\}$ and $$L^2\left(D_{\frac12}\right)=\left\{f\in L^2(D):f(z)=0, |z|\geqslant \frac12\right\} $$ the set of $L^2$ functions that vanish outside of $D_{\frac12}$. Then if $f\in L^2\left(D_{\frac12}\right)$, we have $zf(z)=0$ for $|z|\geqslant\frac12$ so $Af\in L^2\left(D_{\frac12}\right)$ and hence $L^2\left(D_{\frac12}\right)$ is invariant under $A$. Similarly, $$L^2\left(D_{\frac12}\right)^\perp = \left\{f\in L^2(D):f(z)=0,|z|&lt;\frac12 \right\} $$ is invariant  under $A$, so $L^2\left(D_{\frac12}\right)$ is a reducing subspace for $A$.

For an invariant subspace that is not reducing, I am still trying to find one.
I want to thank to Math1000 for a great idea finding a nontrivial reducing subspace.
For an invariant but not reducing subspace of A, the answer would be Bergmann space $A^{2}(\mathbb{D})$, where $A^2(\mathbb{D})$:={$f \in L^2(\mathbb{D})~|~ f$ is holomorphic on $\mathbb{D}$}. The reason is followed:

Since a closed linear subspace reduces $A$ if and only if it is invariant under $A$ and its adjoint $A^{*}$, the idea of finding such "invariant but not reducing" subspace becomes to find a closed subspace invariant under $A$ but not under $A^{*}$.

For our given linear operator $A$, its adjoint $A^{*}$ is computed by
$$A^{*}(f)=\bar{z}f$$

Then, we can easily check the following claims:


$A^{2}(\mathbb{D})$ is a closed subspace of $L^2(\mathbb{D})$.
$A^{2}(\mathbb{D})$ is invariant under $A$
$A^{2}(\mathbb{D})$ is not invariant under $A^{*}$


All of above claims can be proved by some simple techniques in Complex Analysis. 
(Hint: For 1, use Morera's Theorem and for 2,3 use definition of holomorphic functions)

I think these claims have no problem, but please mail me if it has an logical error.