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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.

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  • $\begingroup$ 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. $\endgroup$ – gerw May 11 at 11:09
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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 $CB^{1/2}=A^{1/2}$.

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  • $\begingroup$ 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 $A = CB$ ? $\endgroup$ – Rebellos May 12 at 7:15
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    $\begingroup$ 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. $\endgroup$ – Martin Argerami May 12 at 8:02
  • $\begingroup$ 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 ! $\endgroup$ – Rebellos May 12 at 8:07
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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.

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  • $\begingroup$ 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. $\endgroup$ – Rebellos May 12 at 7:10
  • $\begingroup$ @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. $\endgroup$ – DisintegratingByParts May 12 at 15:43
  • $\begingroup$ 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. $\endgroup$ – Rebellos May 12 at 15:48

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