# $A \leq B$ implies $\mathcal{R}(A) \subseteq \mathcal{R}(B)$ for self-adjoint, positive operators

Let $$A$$ and $$B$$ be two compact, self-adjoint, non-negative operators on a separable Hilbert space $$H$$. If would like to show the following:

$$A \leq B \implies \mathcal{R}(A) \subseteq \mathcal{R}(B),$$ where $$\leq$$ denotes the partial ordering on the set of self-adjoint operators and $$\mathcal{R}$$ denotes the range.

So far I can only show the following property: $$A \leq B \implies \overline{\mathcal{R}(A)} \subseteq \overline{\mathcal{R}(B)},$$ which is proven by taking the orthogonal complement and proving the inclusion of the null spaces.

To prove the first property my idea would be to exploit the following characterisation of $$\mathcal{R}(A)$$ (and respectively $$\mathcal{R}(B)$$). Let us write the eigenvalue decomposition of $$A$$: $$A = \sum_{i \in I} \lambda_{A,i} e_{A,i} \langle e_{A,i}, \cdot \rangle_{H},$$ where $$\lambda_{A,1} \geq \lambda_{A,2} \geq \ldots > 0$$ are the positive eigenvalues of $$A$$ and $$e_{A,i}$$ are orthonormal eigenvectors forming an orthonormal basis of $$\overline{\mathcal{R}(A)}$$. Then it is not difficult to show that $$\mathcal{R}(A) = \bigg( x \in \overline{\mathcal{R}(A)} \mid \sum_{i \in I} \frac{\langle e_{A,i}, x \rangle_{H}^2}{\lambda_{A,i}^2} < +\infty \bigg),$$ However, I cannot find a way to use this characterisation to prove that $$A \leq B \implies \mathcal{R}(A) \subseteq \mathcal{R}(B)$$.

• It suffices to consider the case where the null space of $B$ is trivial. If you are comfortable using unbounded operators, I suspect it would be helpful to use the fact that $B^{-1/2}AB^{-1/2} \leq I$ (with both operators restricted to the domain $\mathcal R(A)$) Commented Jul 31 at 17:00
• Could you elaborate on why it "suffices" to consider the case where B is injective? If I can prove the property for B injective, how does this allow to tackle the general case where B is not injective? Commented Jul 31 at 18:48

The conclusion does not hold. First we provide a counterexample for positive operators, and then modify the counterexample in order to cover the case of positive compact operators.

Consider the operator $$A$$ on $$L^2(0,1)$$ defined by $$Af=xf$$ and the one dimensional operator $$C$$ acting as $$Cf=\langle f,1\rangle \, 1=\int\limits_0^1f(t)\,dt.$$ Let $$B=A+C.$$ Then $$0\le A\le B.$$ We have $$x\in \mathcal{R}(A).$$ However $$x\not\in \mathcal{R}(B).$$ Indeed, assume for a contradiction that $$x\in \mathcal{R}(B).$$ Then $$x=xf(x)+c,\quad c=\int\limits_0^1f(t)\,dt$$ Hence $$f(x)=1-{c\over x}.$$ Thus $$f\in L^2(0,1)$$ iff $$c=0,$$ i.e. $$f(x)=1.$$ But $$\int\limits_0^11\,dx \neq 0,$$ a contradiction.

The operator $$A$$ above is not compact. We will modify this example to get compact positive operators. To this end let $$K$$ be any selfadjoint injective compact operator on $$L^2(0,1),$$ so that $$K1=1.$$ Let $$\widetilde{A}=KAK$$ and $$\widetilde{B}=KBK.$$ Then $$0\le \widetilde{A}\le \widetilde{B}.$$ Moreover $$Kx\in \mathcal{R}(\widetilde{A}),$$ as $$Kx=KAK1.$$ Assume $$Kx\in \mathcal{R}(\widetilde{B}).$$ Thus $$Kx=KBKf$$ for some $$f.$$ As $$K$$ is injective we get $$x=BKf\in \mathcal{R}(B),$$ a contradiction.

A weaker conclusion holds. Namely $$\mathcal{R}(A)\subset \mathcal{R}(A^{1/2}) \subset \mathcal{R}(B^{1/2}).$$ Indeed, we have $$\|A^{1/2}x\|^2=\langle Ax,x\rangle \le\langle Bx,x\rangle= \|B^{1/2}x\|^2\quad (*)$$ Consider the operator $$T_0$$ defined on $$\mathcal{R}(B^{1/2})$$ by $$T_0(B^{1/2}x)=A^{1/2}x.$$ The operator $$T_0$$ is well defined since by $$(*)$$ the condition $$B^{1/2}x=0$$ implies $$A^{1/2}x=0.$$ Moreover by $$(*)$$, $$T_0$$ is bounded and $$\|T_0\|\le 1.$$ Hence $$T_0$$ extends uniquely to a bounded operator $$T_1:\overline{\mathcal{R} (B^{1/2})}\to \overline{\mathcal{R} (A^{1/2})}.$$ Let $$T$$ be the operator defined by $$Ty=T_1y$$ if $$y\in \overline{\mathcal{R} (B^{1/2})}$$ and $$Ty=0$$ if $$y\perp \overline{\mathcal{R} (B^{1/2})}.$$ Then $$T$$ is a bounded operator satisfying $$TB^{1/2}=T_0B^{1/2}=A^{1/2}.$$ Hence $$A^{1/2}=B^{1/2}T^*.$$ Thus $$\mathcal{R}(A)\subset\mathcal{R}(A^{1/2})\subset \mathcal{R}(B^{1/2})$$

Remark If $$A$$ and $$B$$ commute then $$0\le A\le B$$ implies $$A^2\le B^2.$$ Applying the above gives $$\mathcal{R}(A) \subset \mathcal{R}(B).$$

• Nice example!  Commented Jul 31 at 20:28
• This is great (both the counter-example and the proof of the weaker result), thank you! Commented Aug 1 at 10:18
• @Ryszard This $A$ isn’t compact, though Commented Aug 1 at 12:23
• I have extended the answer to capture the compact operator case, just by playing around with the example given in my first answer. You are right: the operator $T$ can be represented as $T=A^{1/2}B^{-1/2}$ defined on $\mathcal{R}(B^{1/2})$ and then extended to the closure by the boundedness. Further extended to the entire space by setting $0$ on the orthogonal complement of $\overline{\mathcal{R}(B^{1/2})}$ which is equal $\ker B.$ Commented Aug 1 at 15:27
• @BenGrossmann I have extended the answer to cover the compact case operators. Commented Aug 1 at 16:35

Ryszard has already written an answer which provides a great counterexample. Let me also suggest an alternative approach that may be somewhat more illuminating as to what exactly went wrong with the claim that $$A \leq B$$ implies $$\mathcal{R}(A) \subseteq \mathcal{R}(B)$$.

As already suggested by Ryszard’s answer, the issue is that, without assuming $$A$$ and $$B$$ commute, one cannot get $$A \leq B$$ implies $$A^2 \leq B^2$$ for positive $$A, B$$. In fact, this is already false in dimension $$2$$. For example,

$$A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$$

satisfies $$B - A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$, but,

$$A^2 = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}, B^2 = \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix}$$

So $$B^2 - A^2 = \begin{pmatrix} 3 & 1 \\ 1 & 0 \end{pmatrix} \not\geq 0$$.

Heuristically, this presents an obstacle to the claim due to the following reasons: Let’s assume, for the sake of argument, that $$A, B$$ are injective. It is generally true, for invertible positive bounded operators $$A, B$$, that $$A \leq B$$ implies $$B^{-1} \leq A^{-1}$$. In our case, $$A, B$$ are, of course, not invertible. But they are positive injective, so they have unbounded inverses $$A^{-1}, B^{-1}$$. I’m not sure whether this can be formalized, but let’s assume it still makes sense to say $$B^{-1} \leq A^{-1}$$. Now,

$$\mathcal{R}(A) = \text{dom}(A^{-1}) = \{h \in H: \|A^{-1}h\|^2 < \infty\} = \{h \in H: \langle h, A^{-2}h \rangle < \infty\}$$

and similarly $$\mathcal{R}(B) = \{h \in H: \langle h, B^{-2}h \rangle < \infty\}$$. But this means, in order for us to have $$\mathcal{R}(A) \subseteq \mathcal{R}(B)$$, we want something like $$\langle h, B^{-2}h \rangle \leq \langle h, A^{-2}h \rangle$$ - or, to put it another way, something like $$B^{-2} \leq A^{-2}$$. But, as we already observed, we only have $$B^{-1} \leq A^{-1}$$, which does not imply $$B^{-2} \leq A^{-2}$$.

This suggests that one way to look for such counterexamples would be to find finite-dimensional examples in which $$0 \leq B \leq A$$, $$A, B$$ both invertible, but $$B^2 \not\leq A^2$$, then taking the direct sums of the inverses of such $$A$$ and $$B$$. Indeed, we have,

Claim: Suppose there is a sequence of positive invertible matrices $$A_n, B_n \in \mathbb{M}_{k_n}(\mathbb{C})$$ with $$B_n \leq A_n$$ but $$B_n^2 \leq A_n^2$$ failing “badly”, in the sense that there exists a sequence of nonzero vectors $$x_n$$ s.t.,

$$\frac{\|B_nx_n\|^2}{\|A_nx_n\|^2} = \frac{\langle x_n, B_n^2x_n \rangle}{\langle x_n, A_n^2x_n \rangle} \to \infty$$

as $$n \to \infty$$. Then there exists a sequence of positive numbers $$\beta_n$$ s.t., if $$A = \bigoplus_n \beta_n A_n^{-1}$$, and $$B = \bigoplus_n \beta_n B_n^{-1}$$. Then $$A \leq B$$ and both are compact positive operators, but $$\mathcal{R}(A) \not\subseteq \mathcal{R}(B)$$.

Proof: By replacing $$x_n$$ with $$\frac{x_n}{\|x_n\|}$$, we may assume all $$x_n$$ are unit vectors. For each $$n$$, by re-scaling $$A_n$$ and $$B_n$$ by the same positive constant (which may depend on $$n$$), we may assume $$\|A_nx_n\| \to \infty$$.

Now, we claim any $$\beta_n > 0$$ with $$\beta_n \to 0$$ and $$\beta_n \cdot \max\{\|A_n^{-1}\|, \|B_n^{-1}\|\} \to 0$$ satisfies the desired conclusion. Indeed, as $$\|\beta_n A_n^{-1}\| \to 0$$ and all $$A_n^{-1}$$ are (finite-dimensional) matrices, clearly $$A$$ is compact. Similarly, $$B$$ is compact. $$A_n$$ is positive invertible implies $$A_n^{-1}$$ is positive, so $$A$$ is positive. That $$B$$ is positive is similar. $$B_n \leq A_n$$ implies $$A_n^{-1} \leq B_n^{-1}$$, from which it easily follows that $$A \leq B$$.

Onto the main claim that $$\mathcal{R}(A) \not\subseteq \mathcal{R}(B)$$. Note that $$\|A_nx_n\| \to \infty$$ and $$\beta_n \to 0$$, so $$\beta_n^{-1}\|A_nx_n\| \to \infty$$. Note also that $$\beta_n^{-1}\|A_nx_n\| > 0$$ for all $$n$$, so,

$$(\frac{1}{\beta_n^{-1}\|A_nx_n\|})_n \in \ell^\infty$$

As $$\frac{\|B_nx_n\|}{\|A_nx_n\|} \to \infty$$, there must exists a sequence $$\kappa_n > 0$$ s.t.,

$$(\kappa_n)_n \in \ell^2 \text{ but } (\kappa_n\frac{\|B_nx_n\|}{\|A_nx_n\|})_n \notin \ell^2$$

Let $$\alpha_n = \frac{\kappa_n}{\beta_n^{-1}\|A_nx_n\|}$$, then $$(\alpha_n)_n \in \ell^2$$, so $$\xi := \sum_n \alpha_nx_n \in H = \bigoplus_n \mathbb{C}^{k_n}$$. It is in $$\mathcal{R}(A)$$. Indeed, since $$(\alpha_n\beta_n^{-1}\|A_nx_n\|)_n = (\kappa_n)_n \in \ell^2$$, we have $$\eta := \sum_n \alpha_n\beta_n^{-1}A_nx_n \in H$$. Clearly, we have $$A\eta = \xi$$, so $$\xi \in \mathcal{R}(A)$$. But $$\xi \notin \mathcal{R}(B)$$. Indeed, if $$B\eta’ = \xi$$ for some $$\eta’ = \sum_n \eta’_n \in H$$ where $$\eta’_n \in \mathbb{C}^{k_n}$$. Then examining the definition of $$B$$, we must have,

$$\eta’_n = \alpha_n\beta_n^{-1}B_nx_n = \kappa_n\frac{B_nx_n}{\|A_nx_n\|}$$

But then $$(\|\eta’_n\|)_n = (\kappa_n\frac{\|B_nx_n\|}{\|A_nx_n\|})_n \notin \ell^2$$, so $$\eta’ \notin H$$. $$\square$$

Now, to be specific, here is an example of such $$A_n$$, $$B_n$$, and $$x_n$$: For $$n \geq 3$$, let $$A_n$$ and $$B_n$$ be the $$n \times n$$ matrices,

$$A_n = \begin{pmatrix} n + 2 & 1 & 1 & \cdots & 1 & 1 \\ 1 & n + 2 & 1 & \cdots & 1 & 1 \\ 1 & 1 & n + 2 & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & 1 & \cdots & n + 2 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 2 \end{pmatrix}$$

$$B_n = \begin{pmatrix} 2 & 1 & 1 & \cdots & 1 & 1 \\ 1 & 2 & 1 & \cdots & 1 & 1 \\ 1 & 1 & 2 & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & 1 & \cdots & 2 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 2 \end{pmatrix}$$

That is $$A_n$$ is the matrix with $$n + 2$$ on all diagonal entries, expect the final diagonal entry, which is $$2$$; and with $$1$$ on all off-diagonal entries. And $$B_n$$ is the matrix with $$2$$ on all diagonal entries and with $$1$$ on all off-diagonal entries. We easily see that $$B_n \leq A_n$$. And it is not hard to check both are positive invertible. (Hint for checking this: The matrix with all entries $$1$$ is positive. Both $$A_n$$ and $$B_n$$ are the all $$1$$ matrix adding a diagonal positive invertible matrix, so they are positive invertible themselves.) Let $$x_n \in \mathbb{C}^n$$ be,

$$x_n = \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \\ -2n \end{pmatrix}$$

Then,

$$A_nx_n = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ -3n - 1 \end{pmatrix}, B_nx_n = \begin{pmatrix} -n \\ -n \\ \vdots \\ -n \\ -3n - 1 \end{pmatrix}$$

So $$\|A_nx_n\|^2 = (3n + 1)^2$$ while $$\|B_nx_n\|^2 = n^2(n - 1) + (3n + 1)^2$$. Then $$\frac{\|B_nx_n\|^2}{\|A_nx_n\|^2} \to \infty$$, as required.