Proving the range of operator is closed

Question

I have a hard time understand (2) of the Fredholm alternative in Evan's Appendix.

To prove the image of $I-K$ is closed, what result from functional analysis is used? I am lost in understand the first sentence of the proof, up to equation (4). Could anyone explain to me what result in functional analysis is used? Which section in the relevant part of any functional analysis one shall study for?

I believe that the result, i.e. "The range of $K$ is closed iff there exists a constant $c>0$ such that $\|Kx\|\ge c\|x\|$ for all $x\in X$ ($K$ is bounded below) is used. But I do not know if so, and where does this result come from, in functional analysis text.

Please help.

Answer 1

The general result you might be looking for is the following: Assume $T \colon X \rightarrow Y$ is a bounded injective operator between Banach spaces. Then $R(T)$ is closed if and only if there exists $c > 0$ such that $||Tx|| \geq c||x||$ for all $x \in X$.

In one direction, if there exists such a constant and $Tx_n \rightarrow y$ then $Tx_n$ is Cauchy but then the inequality $||Tx_n - Tx_m|| = ||T(x_n - x_m)|| \geq c||x_n - x_m||$ implies that $x_n$ is also Cauchy. Since $X$ is complete, $x_n \rightarrow x$ and since $T$ is bounded, $Tx_n \rightarrow Tx = y$ and so $y \in R(T)$.

In the other direction, if $R(T)$ is closed then $T \colon X \rightarrow R(T)$ is a bounded bijective operator between Banach spaces and so by the inverse mapping theorem, it has a bounded inverse $T^{-1} \colon R(T) \rightarrow X$ and so there exists $d > 0$ such that $||x|| = ||T^{-1}(T(x))|| \leq d ||Tx||$ for all $x \in X$ and we can take $c = \frac{1}{d}$.

In your case, $T = I - K$ and $X = N(T)^{\perp}, Y = H$.

Answer 2

We prove the first estimate (1) $||u - Ku|| \geq \gamma ||u||$ $\forall u \in N(I-K)^\perp$ and for some $\gamma >0$. Reasoning by contradiction we have that $\exists \lbrace u_n \rbrace \subset N(I-K)^{\perp}$ such that $||u_n||=1$ and $(\star)$ $||u_n - K u_n || < 1/n$, by $(\star)$ we have $||u_n - Ku_n|| \rightarrow 0$ and also $||u_{n_j} - Ku_{n_j}|| \rightarrow 0$, therefore

$||u_{n_j} - u || \leq ||u_{n_j} - K u_{n_j}|| + ||K u_{n_j} - u || \rightarrow 0$

also $||K u_{n_j} - Ku || \rightarrow 0$, since $K:H \rightarrow H$ is continuos, and then $Ku=u$, i.e. $u \in N(I-K)$, but $u \in N(I-K)^\perp$ implies $u=0$ necessarily, this is a contradiction because $||u_n|| = 1$.

Now, let $\lbrace v_n \rbrace \subset R(I-K)$ such that $||v_n - v|| \rightarrow 0$, by definition $\exists \lbrace u_n \rbrace \subset H$ such that $u_n - Ku_n = v_n$ with $\lbrace u_n \rbrace \subset N(I-K)^\perp$, since if $u_n \in N(I-K)$ then $v_n=0$. By (1):

$|| v_n -v_m || = ||(u_n -u_m) -K(u_n - u_m) || \geq \gamma ||u_n - u_m||$

since $\lbrace v_n \rbrace$ is of Cauchy, also $\lbrace u_n \rbrace$ is of Cauchy, and $|| u_n - u || \rightarrow 0$, but $I-K$ is continuous and $u-Ku=v \in R(I-K)$. So $R(I-K)$ is a closed.