If $A, B$ are self adjoint operators and $A \le B$, is $imA \subset imB$? is $ker(B) \subset ker(A)$?

Question

If $A, B$ be self adjoint operators and $A \le B$, is $imA \subset imB$? is $ker(A) \subset ker(B)$?

Let $H$ be a Hilbert space is $A,B$ self adjoint operators on $H$, such that $A \le B$.

Is $ImA \subset ImB$ ?

from the definition of operator order I get that for every $x \in H$: $$<Ax, x> \le <Bx,x>$$

but I don't see how that helps me to find the images of $A$ and $B$.

Help would be appreciated.

Now when also $A \ge 0$, is $Ker(B) \subset Ker(A)$ ?

I got to $<Ax,x> = 0$ for every $x \in Ker(B)$, since $<Ax,x> \ge 0$ and $<Ax,x> \le <Bx,x> = <0,x> = 0$

But I'm not sure how to continue.

Answer 1

Regarding ranges:

Simply assuming that $A\leq B$ does not imply that $\text{Ran}(A)\subseteq \text{Ran}(B)$, a counter-example being $$-1\cdot I \leq 0\cdot I.$$

Perhaps assuming, in addition, that $A\geq 0$, might improve the chances of an affirmative answer.

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Regarding kernels:

If $0\leq A\leq B$, and if $x\in\text{Ker}(B)$, then $$ \Vert A^{1/2}x\Vert^2 = \langle A^{1/2}x,A^{1/2}x \rangle = \langle Ax,x \rangle \leq \langle Bx,x \rangle = 0, $$ so $A^{1/2}x=0$, and hence also $$ Ax = A^{1/2}A^{1/2}x = 0, $$ so $x\in\text{Ker}(A)$.