Show that $I-K$ has a closed range when $K$ is compact on a Hilbert space.

Question

My proof: Consider $Ker(I-K)^\perp$. I claim that $\|x-K(x)\|\geq \beta\|x\|$ for some $\beta$. Assume otherwise, then there is a sequence $\|x_n\|=1$ s.t $\|x_n-k(x_n)\|<\frac{1}{n}$ since $\|x_n\|$ is bounded, it has a weakly convergent sub sequence (relabel as $x_n$). Thus $x_n-k(x_n)\to 0$ and $K(x_n) \to x$ thus $x_n \to x$ and so $x=K(x)$ thus hence $x \in Ker(I-K)$. contradiction. Thus $I-K$ is positive definite on $Ker(I-K)^\perp$ and so it has closed range when restricted to it. Let $x_n-K(x_n) \to y$ then the convergence also holds for projection onto $Ker(I-K)^\perp$ and since the range is closed on that set, $y \in Range(I-K)$ as desired. IS this correct?

Answer 1

Well you are essentially done simply by proving that there is a $\beta $ such that $\| x-K(x)\| \geq \beta \|x\|$, because this proves that $I-K$ is bounded below and any operator that is bounded below has closed range:

Indeed, suppose that $\|Tx\| \geq \beta \|x\|$ for all $x \in X$ (where $T:X\to Y$ is a bounded linear map between any two Banach spaces). If $Tx_n \to y \in Y$ we must have that $(x_n)_n$ is Cauchy and therefore $x_n \to x \in X$ and of course $y=Tx \in T(X)$, whence $T(X)$ is closed.