Let $X$ and $Y$ be some infinite dimensional Banach spaces. Let $T:X\longrightarrow Y$ be some compact linear operator. It is easy to understand that $T$ cannot be surjective: the Open Mapping Theorem due to Banach, states that surjective operators should be open, and it follows that the image of the unit ball $\mathbb{B}_{1,X}$ should contain an open ball $\{||y||<r\}$ for some $r>0$, so its closure can't be compact. Is it true that the image of $T$ cannot contain any closed subspace of infinite dimension?
[Math] Subspaces in the image of compact operator
banach-spacescompact-operatorsfunctional-analysisoperator-theory
Related Solutions
Hint: If an operator can be approximated by finite rank operators, then it is compact. Try to show that $\|T-T_k\| \to 0$ for a suitable chosen sequence $\{ T_k\}$ of finite rank operators.
Try with $$ T_k: (x_1, x_2, \ldots)\mapsto (x_1, \frac{x_2}{2}, \ldots, \frac{x_k}{k}, 0, 0, \ldots ). $$ It is obvious that $rk(T_k)=k$, i.e., it is finite rank operator. Since $$ \| T-T_k\|^p=\sup\{ \sum_{n=k+1}^{\infty}\left|\frac{x_n}{n}\right|^p;\quad \sum_{n=1}^{\infty}|x_n|^p\leq 1\}\leq $$ $$\leq \frac{1}{(k+1)^p}\sup\{ \sum_{n=k+1}^{\infty}|x_n|^p;\quad \sum_{n=1}^{\infty}|x_n|^p\leq 1\} \leq \frac{1}{(k+1)^p}$$ we have $$ \| T-T_k\|\leq\frac{1}{k+1} \to 0\quad (k\to \infty). $$
Case 1: $T$ and $S$ both one-to-one. In this case define $W:Z\rightarrow X$ by $Wz=x$ if $Sz=Tx$. If $z\in Z$ then $Sz\in S(Z)\subseteq T(X)$ so there exists $x\in X$ such that $Sz=Tx$. Further $x$ is uniquely determined by $z.$ Thus $W$ is a well defined linear map. Note that $T\circ W=S$ so it is enough to show that $W$ is bounded. Suppose $z_{n}\rightarrow z$ and $% Wz_{n}\rightarrow x$. Let $x_{n}=Wz_{n}$. Then $z_{n}$ $\rightarrow z,x_{n}\rightarrow x$ and $Sz_{n}=Tx_{n}$. Since $T$ and $S$ are bounded, $% Tx_{n}=Sz_{n}\rightarrow Sz$ and $Tx_{n}\rightarrow Tx$. Hence $Sz=Tx$ which im plies $Wz=x$. Closed Graph Theorem shows that $W$ is bounded.
Case 2 (general case): let $M$ and $N$ be the null spaces of $T$ and $S$ respectively. Define $T_{1}:X|M\rightarrow Y$ and $S_{1}:Z|N\rightarrow Y$ by $T_{1}(x+M)=Tx$ and $S_{1}(z+N)=Sz$. Then $T_{1}$ and $S_{1}$ are bounded operators. Further they are one-to-one. If $\{x_{n}+M\}$ is a norm bounded sequence in $X|M$ then there exists $\{m_{n}\}\subseteq M$ such that $% \left\Vert x_{n}+m_{n}\right\Vert $ is bounded. Since $T$ is compact $% \{T_{1}(x_{n}+M)\}=\{Tx_{n}\}=\{T(x_{n}+m_{n})\}$ has a convergent subsequence. We have proved that $T_{1}$ is compact. Also $% S(Z|N)=S(Z)\subseteq T(X)=T(X|M)$. Case 1 now shows that $S_{1}$ is compact. However $S$ is the composition of the canonical map $z\rightarrow z+N$ from $% z\rightarrow Z|N$ and $S_{1}$. Hence $S$ is compact.
Best Answer
Let $F\subset T(X)$ be a closed (in $Y$) subspace. Then $E = T^{-1}(F)$ is a closed subspace of $X$, and $T\lvert_E \colon E \to F$ is a compact surjective operator. Since $F$ is closed in $Y$, the open mapping theorem implies that $F$ is finite-dimensional.
So: the image of a compact operator cannot contain an infinite-dimensional Banach space.