$A \leqslant B$ for operators implies $\lVert{A\phi}\rVert \leqslant \lVert{B\phi}\rVert$.

Consider two (possibly unbounded) closed operators $A, B$ on a Hilbert space $\mathcal{H}$, such that $\langle \phi, A \phi \rangle \leqslant \langle \phi, B \phi \rangle$ for all $\phi \in \mathcal{D}$, where $\mathcal{D}$ denotes the common domain of both operators.

It this enough to conclude that there exists some constant $C>0$ such that $\lVert{A \phi}\rVert \leqslant C \lVert{B \phi}\rVert$? Both operators are also lower bounded, if that helps. The scenario I am considering is one in which $B = A + A'$, where $A$ and $A'$ are non-negative, and I would like to obtain a bound of the form $\lVert{A'\phi}\rVert \leqslant (1+C)\lVert{B\phi}\rVert$.

Solution 1:

It this enough to conclude that there exists some constant $C > 0$ such that $\| A \phi \| \leq \| B \phi \|$?

No, this is not enough to make this conclusion. Observe that by squaring both sides and expanding the squared norm via the inner product, the inequality $\| A \phi \| \leq \| B \phi \|$ is equivalent to $$ \langle A \phi, A \phi \rangle \leq \langle B \phi, B \phi \rangle \quad \forall \phi \in \mathcal{D} \quad (\dagger) $$ which is different from your assumption that $$ \langle \phi, A \phi \rangle \leq \langle \phi, B \phi \rangle \quad \forall \phi \in \mathcal{D} . \quad (\ddagger)$$ In particular, to see that $(\dagger)$ does not imply $(\ddagger)$, it's enough to consider $D = \mathbb{R}^2$ and $$ A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \quad B = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} $$

Functions that map operators to operators which preserve the Loewner ordering are called operator monotone. See this answer for more information.

You mentioned that you are interested in the case where $B = A + A'$. In this case, we can show that

$$ \| B \phi \| = \| (A + A') \phi \| \leq \| A \phi \| + \| A' \phi \| = 2 \| A \phi \| \enspace , $$ where the inequality follows from the triangle inequality.