Let $X, Y$ be Banach spaces and $T : Y \to X$ contractive, i.e., $\lVert T \rVert \leq 1$. Suppose furthe that $Y$ is densely embedded into $X$. For an application, there are two compactness properties I am interested in:
- $T$ is called weakly compact if for each bounded sequence $(y_n)_{n \in \mathbb N}$ in $Y$ the sequence $(Ty_n)_{n \in \mathbb N}$ has a weakly convergent subsequence, i.e. there exists $(f_{n_k})_{k \in \mathbb N}$ such that $x'(Ty_{n_k}) \to x'(x)$ for some $x \in X$ and $x' \in X'$.
- For each $\epsilon > 0$ there exists $m \in \mathbb N$ and $K : Y \to X$ compact such that $\lVert T^m - K \rVert < \epsilon$.
I want to understand if, and if, how these two notions are related. I started with the second one.
Suppose the second assertion holds and let $(y_n)_{n \in \mathbb N}$ be a bounded sequence in $Y$, $\epsilon >0$ and $m \in \mathbb N$ such that $\lVert T^m - K \rVert < \epsilon$ for a compact operator $K$. Since $K$ is compact, the sequence $(K y_n)_{n \in \mathbb N}$ has a strongly convergent subsequence $(x_{n_k})_{k \in \mathbb N}$ such that $x_{n_k} := Ky_{n_k}$ and $x_{n_k} \to x \in X$. Thus, $$\lVert T^m y_{n_k} - x \rVert \leq \lVert T^m y_{n_k} - x_{n_k} \rVert + \lVert x_{n_k} - x \rVert = \lVert T^m y_{n_k} - Ky_{n_k} \rVert + \lVert x_{n_k} - x \rVert < 2 \epsilon $$ for $k$ big enough. Thus $T^m y_{n_k} \to x$. This would imply that $T^m$ is compact which i find confusing. Is there any theorem that yields the weak compactness of $T$ from the fact that $T^m$ is compact for some $m \in \mathbb N$ (note that the case $m = 1$ is trivial)?
Regarding the other implication I got no idea how to construct $K$ or how to find such a $m$.
I really would like to get a reference if this should be common knowledge from the existing literature.
None of your two properties imply the other.
On the one hand, the identity $I$ on an infinite dimensional separable Hilbert space is weakly compact (because all bounded sequences in that space admit a weakly convergent subsequence) and yet $I^n$ is never close to a compact operator.
On the other hand, if $X$ is an infinite dimensional Banach space for which the identity operator is not weakly compact (e.g. a separable, non-reflexive space such as $\ell ^1$) then the operator $T$ on $X\oplus X$ given by $T(x, y)=(0, x)$ satisfies $T^2=0$, hence a compact operator, but clearly $T$ is not weakly compact.
PS: Your proof that $T^m$ is compact is not correct because $m$ depends on $\varepsilon $. The hypothesis is stating that the $T^m$ become closer and closer to $\mathscr K(X, Y)$, the space of compact operators, but this can happen without any $T^m$ being compact.