# Showing that $0\leq A\leq B$ and $B \in \mathcal{L}_c(H)$ implies that $A \in \mathcal{L}_c(H)$.

Exercise :

Let $$H$$ be a Hilbert space and $$A,B \in \mathcal{L}(H)$$ be self-adjoint operators with $$0 \leq A \leq B$$ and $$B \in \mathcal{L}_c(H)$$. Show that $$A \in \mathcal{L}_c(H)$$.

Thoughts :

Relying only on the definition of a compact operator, we essentialy need to conclude that $$A$$ transfers bounded sets to relatively compact sets (compact closure).

Now, since $$B$$ is compact and self adjoint, I know that also $$B^*B$$ is compact. This may be of use since the property of $$A$$ and $$B$$ being self adjoint is noted in the exercise.

I think that $$A \leq B \implies \|A\| \leq \|B\|$$ since they are both bounded and we could take $$\mathbf{1} \in H$$ which yields that $$\|A(\mathbf{1})\| \leq \|A\|\|1\| \equiv \|A\| \quad \text{and} \quad \|B(\mathbf{1})\| \leq \|B\|\|1\| \equiv \|B\|$$ and since $$0 \leq A \leq B$$ implies that their values follow the inequality for any $$x \in H$$ thus the implied result.

Request : Beyond these points, I sadly do not have an intuition for a head-start, so I would really appreciate any hints or elaborations.

• I think that this can be proven by using the eigenvalue decomposition of the compact, self-adjoint operator $B$. However, I feel that there is a simpler solution. – gerw May 11 at 11:09

I can think of two arguments. None of them is entirely elementary, though, in the sense that they depend on having square roots available.

• Let $$\{x_n\}$$ be a sequence in the unit ball of $$H$$. Since $$B$$ is compact, there exists a subsequence $$\{x_{n_j}\}$$ such that $$\{Bx_{n_j}\}$$ converges. Now (using twice that $$A$$ is positive) \begin{align} \|A(x_{n_j}-x_{n_k})\|^2&=\langle A^2(x_{n_j}-x_{n_k}),(x_{n_j}-x_{n_k})\rangle \leq \|A\|\,\langle A(x_{n_j}-x_{n_k}),(x_{n_j}-x_{n_k})\rangle\\ &\leq \|A\|\,\langle B(x_{n_j}-x_{n_k}),(x_{n_j}-x_{n_k})\rangle\to0. \end{align} So $$\{Ax_{n_j}\}$$ is Cauchy and thus convergent. So $$A$$ is compact.

• Since $$0\leq A\leq B$$, there exists $$C$$ with $$A^{1/2}=CB^{1/2}$$. Since $$B$$ is compact, so is $$B^{1/2}$$; then $$A^{1/2}$$ is compact and so is $$A$$.

The first argument depends on the inequality $$A^2\leq \|A\|\,A$$ for $$A$$ positive. This can be seen for instance by $$\langle A^2x,x\rangle=\langle Ax,Ax\rangle=\|Ax\|^2=\|A^{1/2}A^{1/2}x\|^2\leq\|A^{1/2}\|^2\|A^{1/2}x\|^2 =\|A^{1/2}\|^2\langle Ax,x\rangle.$$ And $$\|A^{1/2}\|^2=\|A\|$$.

For the second argument, we first note that if $$B^{1/2}x=0$$, then $$A^{1/2}x=0$$. So we have $$H=H_1\oplus H_1^\perp$$, where $$H_1=\overline{\operatorname{ran}B^{1/2}}$$. We have $$\tag1 \|A^{1/2}x\|^2=\langle Ax,x\rangle\leq \langle Bx,x\rangle=\|B^{1/2}x\|^2,$$ so $$A^{1/2}=0$$ on $$H_1^\perp$$. On $$\operatorname{ran}B^{1/2}$$, we define $$CB^{1/2}x=A^{1/2}x.$$ This is well defined by $$(1)$$. We have $$\|CB^{1/2}x\|^2=\|A^{1/2}x\|^2=\langle Ax,x\rangle\leq\langle Bx,x\rangle=\|B^{1/2}x\|^2,$$ so $$C$$ is bounded on $$\operatorname{ran}B^{1/2}$$, so it extends by density to $$H_1$$. We put $$C=0$$ on $$H_1^\perp$$. So $$C$$ is bounded and $$CB^{1/2}=A^{1/2}$$.

• Great answer as well, I really like the second one as well, since in a previous exercise I had proven that $B$ is compact if and only if $B^{1/2}$ is compact. My question is this though : We say that if $0\leq A \leq B$ then there exists C such that $A^{1/2} = CB^{1/2}$. Why use the square root operator ? Couldn't we imply that there exists a $C$ such that $A = CB$ ? – Rebellos May 12 at 7:15
• For starters, the proof wouldn't work! Yes, in finite dimension, one can always get $A=CB$; but, as opposed to the construction above, you cannot guarantee that $C$ is a contraction. In infinite dimension, I'm not sure. – Martin Argerami May 12 at 8:02
• Okay, got it ! It really wasn't as simple an exercise as it thought it would be, but your guidance (and also DisintegrateByParts' guidance) has taught me some very nice elaborations and handlings. I couldn't appreciate it more ! – Rebellos May 12 at 8:07

Because $$B-A \ge 0$$, then $$[x,y]=\langle (B-A)x,y\rangle$$ is a pseudo inner product, lacking only positive definiteness. As such, the Cauchy-Schwarz inequality holds $$|[x,y]|^2 \le [x,x][y,y] \\ |\langle (B-A)x,y\rangle|^2 \le \langle (B-A)x,x\rangle\langle (B-A)y,y\rangle.$$ Now set $$y=(B-A)x$$ in the above to obtain $$\|(B-A)x\|^4 \le \langle (B-A)x,x\rangle\langle(B-A)(B-A)x,(B-A)x\rangle \\ \|(B-A)x\|^4 \le \langle (B-A)x,x\rangle\|B-A\|\|(B-A)x\|^2 \\ \|(B-A)x\|^2 \le \|B-A\|\langle(B-A)x,x\rangle \\ \|(B-A)x\|^2 \le \|B-A\|\langle Bx,x\rangle.$$ Suppose $$\{ x_n \}$$ is a bounded sequence. Because $$B$$ is compact, there is a subsequence $$\{ x_{n_k} \}$$ such that $$\{ Bx_{n_k} \}$$ converges. By the above, $$\{ (B-A)x_{n_k}\}$$ is a Cauchy sequence and, hence, converges to some $$y$$. But $$\{ Bx_{n_k} \}$$ also converges. Therefore $$\{ Ax_{n_k} \}$$ converges, leading to the conclusion that $$A$$ is compact.

• Great answer. The pseudo-inner product statement, we never mentioned something like this. Could I still make such claim, lets say ? I do understand that if it was $B-A >0$ it would be a standard inner product process with no issues. – Rebellos May 12 at 7:10
• @Rebellos : If it is not definite, then consider $B-A+\epsilon I$, obtain the inequality, and then let $\epsilon \downarrow 0$. You'll have the inequality you want. – DisintegratingByParts May 12 at 15:43
• Thanks for clearing up. The answer is great, I only accepted the other (great one as well), since its closer to the handlings I am familiar with. – Rebellos May 12 at 15:48