I have been reading about Fredholm operators and came across these notes.
My question is about Lemma 16.17 on the second page. This lemma states the following:
Lemma 16.17: Let $X$ and $Y$ be Banach spaces and $T:X \rightarrow Y$ a bounded linear map. The following are equivalent:
- $\text{Ker}(T)$ is finite dimensional and $\text{Im}(T)$ is closed
- Every bounded sequence $\left \{ x_{i} \right \} \subset X$ with $Tx_{i}$ convergent has a convergent subsequence.
I am interested in the forward (1. implies 2.) direction. The proof given in the attached notes states:
Suppose that 1. holds. Since $\text{ker}(T)$ is finite dimensional it admits a
closed compliment $C$. Since $\text{Ran}(T)$ is closed it is a Banach space so the Banach
isomorphism theorem implies $T |_C : C → \text{Ran}(T)$ is an isomorphism and the
result follows.
My question regards the last two sentences. I had thought the Banach isomorphism theorem implies a isomorphic mapping (one to one and onto) is also a topological isomorphism. Additionally, I had thought that $X$ mod the kernel of $T$ was isomorphic to $\text{Ran}(T)$, i.e. $X/ \text{ker}(T) \approx \text{Ran}(T)$. It is not clear to me how $T |_C : C → \text{Ran}(T)$ is an isomorphism, since $C$ can have multiple elements that map to the same point in $\text{Ran}(T)$, implying $T |_C$ is not surjective. Even if $T |_C$ is an isomorphism, I am not sure how the result directly follows. Since $T |_C$ is an isomorphism it seems to me convergence of $Tx_{i}$ in $\text{Ran}(T)$ would imply convergence of $x_{i}$ in $C$. Any help or reference would be appreciated.
Best Answer
$Ker (T)\oplus C=X$. If $Tc_1=Tc_2$, ($c_1,c_2 \in C)$ then $T(c_1-c_2)=0$ so $c_1-c_2 \in ker (T) \cap C=\{0\}$. So $c_1=c_2$. Thus $T|_C$ is injective. If $y \in Ran (T)$ the $y=Tx$ for some $x \in X$. We can write $x=x_1+c$ for some $x_1 \in \ker (T)$ and $c \in C$. So $y=T(x_1+c)=0+Tc$. This proves that $T|C$ is surjective. By the quote therom ther exists $a>0$ such that $\|Tx\| \geq a \|x\|$ for all $x$. [This is continuity of $(T|C)^{-1}$]. Now note that if $(Tx_i)$ is convergent then $\|x_i-x_j\|\leq \frac 1 a \|Tx_i-Tx_j\|\to 0$ . By completeness of $X$ we see that $x_i$ converges to some $x$.