Working on this question for exam preparation and am stumped. Clearly we will want to use the closed image theorem but I am stuck on how to proceed and would appreciate a hint. I am having issues with working with the linear operators rather than $\mathbb{R}^n$, and am not understanding how to show that a certain subspace is open.
Let $X$ and $Y$ be Banach spaces. Prove that the set of all injective bounded linear operators $A: X \to Y$ with closed image is an open subset of $\mathcal{L}(X,Y)$ with respect to the norm topology
Best Answer
Assume $X$ and $Y$ are Banach spaces. By the Banach inverse mapping theorem a bounded operator $T$ is injective and has a closed image iff there is $c>0$ such $$\|Tx\|\ge c\|x\|,\quad x\in X$$ If $\|S-T\|<{c\over 2}$ then $$\|Sx\|\ge \|Tx\|-{c\over 2}\|x\|\ge {c\over 2}\|x\| ,\quad x\in X$$ Hence $S$ is injective and has a closed image.
Remark The conclusion is not true if either $X$ or $Y$ is not complete.
Indeed, let $A:\ell^2\to \ell^2$ act by $$A\{x_n\}_{n=1}^\infty=\{2^{-n}x_n\}_{n=1}^\infty $$ Let $X=\ell^2$ and $Y={\rm Im}A,$ where both spaces are equipped with $\ell^2$-norm. Then $Y$ is not complete. The operator $T:\ell^2\to {\rm Im}\,A$ given by $Tx=Ax$ is injective and its image is closed. For fixed $k$ let $$T_kx=Tx-2^{-k}\langle x,\delta_k\rangle \delta_k$$ where $\delta_k$ denote the elements of the standard basis in $\ell^2.$ Then ${\rm Im}\,T_k\subset {\rm Im}\,T$ and $\|T-T_k\|=2^{-k}.$ The operator $T_k$ is not injective as $T_k\delta_k=0.$
Let $Y=\ell^2$ and $X$ be defined by $$X=\left \{ x\in \ell^2\,: \exists N\, \sum_{n=1}^Nx_n=0\right \}$$ Then $X\subsetneq \ell^2$ as $\delta_1\notin X.$ Let $T:X\to \ell^2$ act by $$Tx=x-\langle x,\delta_1\rangle \delta_1$$ Then ${\rm Im}\, T=\delta_1^\perp.$ The inclusion ${\rm Im}\, T\subset\delta_1^\perp$ is obvious. On the other hand if $x\perp\delta_1,$ i.e. $x_1=0,$ then $$T(-x_2,x_2,x_3,\ldots )=(0,x_2,x_3,\ldots )$$ hence ${\rm Im}\,T\supset \delta_1^\perp.$ Let $$\varphi_n=\delta_1-{\delta_2+\ldots +\delta_{n+1}\over n},\quad \psi_n= {\delta_2+\ldots +\delta_{n+1}\over n}$$ and $$S_nx={n\over n+1}\langle x,\varphi_n\rangle \psi_n$$ Then $$\|S_n\|\le {n\over n+1}\|\varphi_n\|\,\|\psi_n\|={1\over \sqrt{n+1}}\to 0$$ and $$(T+S_n)\varphi_n=-\psi_n+\psi_n=0 $$ Hence $T+S_n$ is not injective and $T+S_n\to T.$