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Statement: Let $T$ be a bounded operator in a Hilbert space $\mathscr{H}$ Show that if $T-\lambda I$ is not dense in $\mathscr(H)$, then $\overline{\lambda}$ is an eigenvalue of $T^*$.

Attempted Proof 1: We know that $N(T-\lambda I)^{\perp} = \overline{R(T-\lambda I)}$ where N, and R represent the null space, and range respectively. Hence

$\overline{\lambda} \in \sigma_p(T^*) \Leftrightarrow N(T^*-\overline{\lambda}I) \neq \{0\} \Leftrightarrow \overline{R(T-\lambda I)} \neq \mathscr{H} \Leftrightarrow R(T-\lambda I)$ is not dense.

Q.E.D.

Now what I'm wondering is if this is sufficient? Alternatively I have:

Attempted proof 2: Since the range is not dense we know that $\exists \, y \in \mathscr{H}$ such that $(T-\lambda I)x \neq y$ let $y \perp (T-\lambda I)x$ then

$0 = \langle (T-\lambda I)x, y \rangle = \langle x, (T^* -\overline{\lambda}I)y \rangle$ so what we arrive at is that $(T^*-\overline{\lambda}I)y=0\Rightarrow \overline{\lambda}\in \sigma_p(T^*)$

Q.E.D.?

Now for my second attempt it seems somewhat incomplete. What I'm wondering is that this works for a specific $x$, but what if we have a sequence $\{x_n\}$ would we need to show that $(T-\lambda I)x_n$ does not converge to $y$, and if so how?

nonameswereavailable
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1 Answers1

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Attempt 1 looks perfect.

Attempt 2: The vector $x$ is arbitrary, hence you obtain $(T^*-\bar\lambda I)y=0$. For this conclusion you do not need that range of $T-\lambda I$ is dense.

For this attempt, also the backward conclusion is valid: $\bar\lambda\in \sigma_p(T^*)$, then exists $y\ne 0$, $y\in N(T^*-\bar \lambda I)$. This implies $y \in R(T-\lambda I)^\perp$. That is, the orthogonal complement of $R(T-\lambda I)$ contains more than the null vector. Hence, $R(T-\lambda I)$ cannot be dense.

daw
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