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I have tried to solve this exercise:

Let $H$ be a Hilbert space and let $T = T^* \in B_{\infty}(H)$ a compact operator. Given $\psi_0 \in H$ consider the equations: \begin{align*} (1)\quad &T\psi = \psi,\\ (2)\quad &T\psi' = \psi' + \psi_0. \end{align*}

Show that if the only solution of (1) is $\psi = 0$, then equation (2) has a unique solution.

When I was done solving it, I found out that I didn't know whether or not the hypothesis of T being compact was useful. Can you help me find out if it is needed somewhere in my proof?

Here is the proof:

$\exists !\psi \in H$ | $T\psi = \psi = 0 \implies$ $\psi \in$ ker($T$) $= \{0\} \implies (T-I)$ is injective.

ker$(T-I)^{\bot} = $ Ran$(T^{*}- I) = $ Ran$(T - I) = $ $\{0\}^{\bot} = H \implies T$ is surjective $\implies T$ is bijective.

Then $\forall \psi_0 \in H $, $\exists! \psi' \in H $ so that $(T - I)\psi' = \psi' + \psi_0 \implies$ (2) has a unique solution.

Thanks in advance for your help!!!

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1 Answer 1

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For a bounded operator you have $(ker(T- I))^\perp = \overline{Ran(T^*- I)}$. But , since $T=T^*$ is compact we have that $Ran(T^*- I)$ is closed, so $(ker(T- I))^\perp = Ran(T^*- I)$.

A faster way to solve this is: $(1)$ says that $1$ is not an eigenvalue of $T$. Since $T$ is compact and self-adjoint we conclude that $1$ is not in the spectrum of $T$, so $T-I$ is invertible.

Added. Claim: If $T$ is compact then $Ran(T-I)$ is closed.

Suppose $T(x_n)-x_n\rightarrow_n y.$ WLOG we can assume that $x_n \in Ker(T -I)^\perp$. The sequence $x_n$ is bounded. Indeed otherwise taking a subsequence we can assume $\lim_n \|x_n\|=\infty$. Then

$T(\frac{x_n}{\|x_n\|})-\frac{x_n}{\|x_n\|}\rightarrow_n 0.$

Since $T$ is compact WLOG we can assume that $\lim_n T(x_n/\|x_n\|) = z\in (ker(T- I))^\perp$. But this implies

$$\lim_n \frac{x_n}{\|x_n\|} =z\in (ker(T- I))^\perp$$

so $\|z\|=1$ and $T(z)-z=0$, that contradicts $z\in (ker(T- I))^\perp$ and $z\neq 0$. So $x_n$ is bounded.

Taking a subsequence we can assume that $\lim_n T(x_n) = w\in H$. This implies $\lim_n x_n = u := w-y.$ and $Tu -u = y$. So $Ran (T -I)$ is closed.

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  • $\begingroup$ Can you specify why I can say that the $Ran(T - I)$ is closed? $\endgroup$ Commented Nov 22 at 12:47
  • $\begingroup$ I added a proof. $\endgroup$ Commented Nov 22 at 13:17

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