$\operatorname{Ran} \lambda I - T$ is closed for compact operator $T$ and $\lambda \neq 0$

Question

Let $T$ be a compact operator on a Hilbert space $\mathcal{H}$ and $\lambda \in \Bbb{C} - \{0\}$. I want to show that $\operatorname{ran} \lambda I - T$ is closed. So suppose we have $g_j = (\lambda I - T)f_j \in \operatorname{ran} \lambda I - T$ converging to some $f$. We need to show $f \in \operatorname{ran} \lambda I - T$.

First because $\lambda I - T$ is a bounded operator its kernel $V_\lambda$ is closed, hence $\mathcal{H} = V_\lambda^\perp \oplus V_\lambda$. With this we may assume wlog that $f_j \in V_\lambda^\perp$. Now the next hint in the problem is to show that $\{f_j\}$ is bounded under the assumption that $\{f_j\} \subseteq V_\lambda^\perp$. I have tried for about 1 hour and I can't show this.

How can I go around showing this? Here's a partial proof for this fact that I have. We may write $$\begin{eqnarray*}||\lambda f_j||& =& \|\lambda f_j - Tf_j + Tf_j\| \\ &\leq& \| (\lambda I - T)f_j\| + \|Tf_j\|\\ &\leq& \|g_j\| + \|T\|\|f_j\|\\ &\leq& N + \|T\|\|f_j\|\end{eqnarray*}$$ where we have used that $g_j \to g$ impies $\|g_j\| \leq N$. Now we may rearrange to get $\|f_j\| (|\lambda - \|T\|) \leq N$ and so I will be done if I know $\|\lambda \| - \|T\|) \neq 0$. However this may not be true in general. So how can I show $\{f_j\}$ is bounded using the hint in Stein and Shakarchi? Please do not give full answers.

Next, assuming that $\{f_j\}$ is bounded and that $T$ is compact we know $Tf_{n_j} \to f$ for some $f \in \mathcal{H}$. However I don't know how to show this implies $g$ that we started out with is in $\operatorname{ran} \lambda I - T$.

What's strange about all of this is they claim in Stein and Shakarchi that the result is not true if $\lambda = 0$. My guess is that we somehow have to use the result that if $\lambda \neq 0$ that $V_\lambda $ is finite dimensional. All this is very puzzling at the moment and I would appreciate some help.

Answer 1

The key is to show that $S = (\lambda I - T)\lvert_{V_\lambda^\perp}$ has the property

$$\bigl(\forall x \in V_\lambda^\perp\bigr)\bigl(\lVert Sx\rVert \geqslant \delta \lVert x\rVert\bigr)\tag{1}$$

for some $\delta > 0$. Everything follows quite easily from that.

To show the existence of such a $\delta > 0$, suppose the contrary. Then there is a sequence $(x_n)$ in $V_\lambda^\perp$ with $\lVert x_n\rVert = 1$ for all $n$ and $Sx_n \to 0$. Use the compactness of $T$ and then $\lambda \neq 0$ to obtain a contradiction.

I assume the OP has meanwhile filled in the details, so let me add them here too:

If we had a sequence $x_n \in V_\lambda^\perp$ with $\lVert x_n\rVert = 1$, and $z_n = Sx_n \to 0$, by the compactness of $T$ there is a subsequence such that $Tx_{n_k}$ converges. Suppose without loss of generality that the subsequence is the entire sequence, so $y_n = Tx_n \to y$. Then

$$\lambda x_n = Sx_n + Tx_n = z_n + y_n \to 0 + y = y.$$

Since $\lambda\neq 0$, we have $\lVert y\rVert = \lvert\lambda\rvert > 0$, it follows that $x_n \to \lambda^{-1}y \in V_\lambda^\perp\setminus\{0\}$ and

$$S(\lambda^{-1}y) = \lim_{n\to\infty} Sx_n = \lim_{n\to\infty} z_n = 0,$$

contradicting $V_\lambda^\perp \cap V_\lambda = \{0\}$.

So the existence of a $\delta > 0$ with $\lVert Sx\rVert \geqslant \delta \lVert x\rVert$ for all $x\in V_\lambda^\perp$ is established.

Since $\mathcal{H} = V_\lambda \oplus V_\lambda^\perp$, we have the (linear) bijection $S\colon V_\lambda^\perp \to \mathcal{R}(\lambda I - T)$. Now if $(y_n)$ is a Cauchy sequence in $\mathcal{R}(\lambda I-T)$, then $(S^{-1}y_n)$ is a Cauchy sequence in $V_\lambda^\perp$ by $(1)$, and we have

$$\lim_{n\to\infty} y_n = \lim_{n\to\infty} S(S^{-1}y_n) = S(\lim_{n\to\infty} S^{-1}y_n) \in \mathcal{R}(S) = \mathcal{R}(\lambda I - T).$$

If we had not assumed $\lambda\neq 0$, we couldn't have deduced the convergence of $x_n$ from the convergence of $Sx_n\to 0$ and $Tx_n \to y$, and in fact the range of a compact operator is in general not closed.

It should be noted that the result does not depend on $\mathcal{H}$ being a Hilbert space, that only makes the proof a little bit more easy. When working with an arbitrary Banach space, the proof is exactly the same if one can establish a closed complement of $V_\lambda$. Although generally closed subspaces need not be complemented in Banach spaces, the compactness of $T$ implies the finite-dimensionality of $V_\lambda$, and finite-dimensional subspaces (as well as subspaces of finite codimension) are always complemented in Banach spaces. A slight variation of the proof considers the quotient $\mathcal{H}/V_\lambda$ instead of a complementing subspace.