An example $A$ such that $R(A)\not\subset R(A^2)$ for positive self-adjoint $A$

Question

Can you provide an example such that $R(A)\not\subset R(A^2)$ for positive self-adjoint $A$ where $R()$ denotes the range?

Thanks.

Math.

Answer 1

We have to distinguish positive and positive definite operators. If $A$ is self-adjoint and positive definite i.e. $(Ax,x) \ge \gamma (x,x)$, $x \in D(A)$ for some constant $\gamma > 0$ then it is invertible. And therefore $R(A) = H$ and in this case $R(A^2) = R(A) = H$. Here an example doesn't exist.

For positive operators $A$, i.e. $(Ax,x) > 0$ for all $x \ne 0$ we can provide an example. You can take arbitrary self-adjoint bounded positive operator that is not surjective. For example consider operator $A = M_f:L_2(\mathbb{R}) \rightarrow L_2(\mathbb{R})$ that is multiplication by $f(x) = \frac{1}{1 + x^2}$. Obviuosly $R(A) = \hat{H^2}(\mathbb{R})$ and $R(A^2) = \hat{H^4}(\mathbb{R})$, where $\hat{H^s}$ are Fourier images of Sobolev spaces $H^s$ (they are not equal).

In general if $A$ is an injective bounded operator ($D(A) = H$) and not surjective then $R(A) \ne R(A^2)$. If $R(A) = R(A^2)$ then for all $x \in H$ there exists $x' \in H$ s.t. $Ax = A^2x'$. But $A$ is injective and therefore $x = Ax'$. It follows that $A$ is surjective. That's a contradiction. Now you can observe that positive operators are injective. So, we can conclude that bounded self-adjoint positive operator that is not surjective satisfies your conditions.