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Suppose $A$ is a normal ($||Ax||=||A^*x||$ for all $x \in H$ ) bounded linear operator on a Hilbert space $H$. We want to show that if $A+A^*$ is a bijection then so is $A$.

A is injective is very easy since $kerA=kerA^*$ we have that if $Ax=0$ then $x \in ker(A)\cap ker(A^*)$ so $x \in ker(A+A^*)$ so $x=0_H$.

I'm stuck on surjective. My hint is to use the open mapping theorem (so $A+A^*$ is open); but I can't see how this helps me at all.

I thought maybe I could prove $Im(A)$ was dense and closed but I can't relate either to $A+A^*$.

EDIT In

Muselive
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    @runway44 If $A^\ast A=AA^\ast$, then $|Ax|^2=\langle A^\ast Ax,x\rangle=\langle AA^\ast x,x\rangle=|A^\ast x|^2$. Moreover, your statement on kernels and ranges is not true. What is true is $(\mathrm{im}(A))^{\perp}=\ker A^\ast$ and the same with $A$ and $A^\ast$ exchanged. – MaoWao May 06 '21 at 08:21
  • @MaoWao I'm wrong on both counts! Thanks. – anon May 06 '21 at 08:25
  • You want to prove that $A+A^$ is open, but in order to use the open mapping theorem, you need to prove that $A+A^$ is continuous, how is that true? (I know why, it's just a hint) – Maxence1402 May 06 '21 at 08:31
  • I think I know $A+A^$ is open since it's bijective by hypothesis its certainly surjective and continuous since both $A$ and $A^$ are? – Muselive May 06 '21 at 08:38
  • Even knowing $A+A^*$ is open doesn't seem to help me get to $A$ is surjective? – Muselive May 06 '21 at 08:39
  • Then it suffices to prove that $A$ is open. – Maxence1402 May 06 '21 at 08:40
  • ...should this be obvious? I feel like I'm missing something. I think I can show $A$ is open [because $A(U)$ is merely a translate of $(A+A^*)(U)$] but $A$ open implies $A$ surjective is still mysterious to me. – Muselive May 06 '21 at 08:55
  • If $A$ is open then $A(B(0,1))$ is open ($B(0,1)$ is the open unit ball). Therefore, there exists $r>0$ s.t. $B(0,r)\subset A(B(0,1))$, as $0=A(0)\in A(B(0,1))$, and therefore $A$ is surjective. – Maxence1402 May 06 '21 at 09:04
  • @Muselive $(A+A^\ast)(U)$ is in general not a translate of $A(U)$. By the same argument $0=A-A$ would be open whenever $A$ is, which is clearly not the case. – MaoWao May 06 '21 at 09:09

2 Answers2

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Since $A$ and $A^\ast$ commute, we have $$ \frac 1 4(A+A^\ast)^2-\frac 14(A-A^\ast)^2=A^\ast A. $$ Since $A+A^\ast$ and $i(A-A^\ast)$ are self-adjoint operators, it follows that \begin{align*} \|A x\|^2&=\langle A^\ast Ax,x\rangle\\ &=\frac 1 4\langle (A+A^\ast)^2x,x\rangle-\frac 1 4\langle (A-A^\ast)^2x,x\rangle\\ &=\frac 1 4\|(A+A^\ast)x\|^2+\frac 1 4 \|i(A-A^\ast)x\|^2. \end{align*} By the open mapping theorem (or the bounded inverse theorem if youlike), there exists a constant $c>0$ such that $\|(A+A^\ast)x\|\geq c\|x\|$. Thus $$ \|Ax\|\geq \frac{c}{2}\|x\|. $$ Thus $A$ is a bijective operator from $H$ to $\operatorname{im}A$ with bounded inverse. In particular, $\operatorname{im}A$ is closed. Finally, if $y\perp\operatorname{im}A$, then $$ 0=\langle y,AA^\ast y\rangle=\|A^\ast y\|^2, $$ hence $y\in \ker A^\ast=\{0\}$. Therefore, $\operatorname{im}A$ is also dense in $H$.

Here is a shorter answer that requires a little more technology: By the spectral theorem, the unital $C^\ast$-algebra generated by $A$ is $\ast$-isomorphic to $C(K)$ for some compact $K$, so this question boils down to the invertibility of $f\in C(K)$ under the assumption that $\operatorname{Re}f$ is invertible. Of course, an element of $C(K)$ is invertible if and only if it has no zero. Furthermore, the zeros of $f$ coincide with the zeros of $|f|$. Now all we have to do is to notice that $$ |f|^2=(\operatorname{Re}f)^2+(\operatorname{Im}f)^2\geq (\operatorname{Re}f)^2>0. $$

MaoWao
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For finite dimensions: Isnt $A$ and $A^H$ simulataneously diagonalizable by unitary matrix cant u use that to prove this by simply saying $A+A^H = U (D+D^H) U^H$ and hence bijective iff $\lambda_i+conj(\lambda_i) \neq 0$ and hence $\lambda_i \neq 0$ and hence $A$ is bijective .

Balaji sb
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    But in finite dimensions, bijectivity is equivalent to injectivity, which was already proof in the question. – MaoWao May 06 '21 at 09:47