Characterization of pre-compacts and application

Question

I have a problem consisting in this two parts (maybe connected each other):

1). Let X be a Banach space and let $A\subset X $ be bounded. Prove that $A$ is pre-compact iff for ever $\epsilon >0$ there exists a subspace $F_\epsilon $ of $X$ of finite dimension s.t. $dist(x,F_\epsilon)\leq \epsilon \quad \forall x \in A $.

2). Let $(\lambda_n)_{n\in \mathbb{N}}$ be a bounded sequence in $\mathbb{R}$ and consider the linear operator $T:l^2\rightarrow l^2$ s.t. $(Tu)_n \equiv \lambda_n u_n \quad \forall u=(u_n)_{n\in\mathbb{N}}\in l^2$. Prove that T is compact iff $\lim_{n\rightarrow\infty} \lambda_n=0$

Answer 1

1. $\implies: $

$A$ pre-compact $\implies$ $\bar{A}$ compact. Let $\epsilon>0$ and $\{B(x,\epsilon)\}_{x\in \bar{A}}$ an open cover of $\bar{A}$. Then there exists a finite open subcover $\{B(x_i,\epsilon)\}_{i=1,...,n}$ of $\bar{A}\supset A$. Then $F_\epsilon = \langle\{x_1,...,x_n\}\rangle$ verifies $dist(x,F_\epsilon)\leq \epsilon$ for all $x\in A$.

1. $\impliedby: $

In this part I will use the following characterization of pre-compactness, which appears, for instance, here:

Let $(Y,d)$ be a complete metric space. For a subset $M\subset Y$, the following are equivalent:

• $M$ is relatively compact in $Y$, i.e. its closure in $Y$ is compact.

• $M$ contains no infinite sequence $x_k$, satisfying $d(x_n,x_m)>p$ for some $p>0$.

Assume $\bar{A}$ not compact. Hence there is a sequence $x_k$ in $\bar{A}$, and some $\epsilon>0$ s.t. $n\neq m\implies \|x_n-x_m\|>\epsilon$.

We know $\forall\epsilon >0, k\in \mathbb{N},\: dist(x_k,F_\epsilon)<\epsilon$. Hence $\forall \epsilon >0, k\in \mathbb{N},\: \exists y_{k,\epsilon}\in F_\epsilon, dist(x_k,y_{k,\epsilon})<\epsilon$. Now, the sequences $\{y_{k,\epsilon}\}_k$ have convergent subsequences for all $\epsilon$.

For $\epsilon/3$, consider $\{y_{k_n,\epsilon/3}\}_n$, a convergent (hence, Cauchy) subsequence of $\{y_{k,\epsilon/3}\}_k$. For sufficiently large $n,m$ we have $$ \|x_{k_n}-x_{k_m}\|\leq \|x_{k_n}-y_{k_n,\epsilon/3}\| + \|y_{k_n, \epsilon/3}-y_{k_m,\epsilon/3}\| + \|x_{k_m}-y_{k_m,\epsilon/3}\|\leq \epsilon $$ which contradicts the definition of the sequence $\{x_k\}_k$.

2. $\implies: $

Consider the sequence $\{x_k\}_k\subset B(0,1)$ given by $$ (x_k)_n=\begin{cases} 1/2 & k=n \\ 0 & otherwise \end{cases} $$ Since $T$ compact, the sequence $y_k=Tx_k$ has convergent subsequences. Consider one such subsequence, converging to some $y\in \ell^2$. Its easy to see $y$ must be the zero sequence in $\ell^2$. Hence 0 is the only accumulation point of the sequence $y_k$, hence it converges to 0, which is equivalent to saying $\lambda_n\to 0$

2. $\impliedby: $

Let's prove $T(B(0,1))$ is compact. Consider $T_n$ the operator $T$ "truncated" at $n$, that is, $$ (T_nx)_j=\begin{cases} \lambda_jx_j & j\leq n \\ 0 & j>n \end{cases} $$ Since $\lambda_n\to 0$, for all $\forall\epsilon>0 \:\exists n_\epsilon,\: \forall k>n_\epsilon, |\lambda_j|<\epsilon/2$. Hence, for all $y=Tx\in T(B(0,1))$, $$ \|Tx-T_{n_\epsilon}x\|_2 = \left(\sum_{j>{n_\epsilon}}|\lambda_j|^2|x_j|^2\right)^{1/2}\leq \epsilon/2\left(\sum_{j>{n_\epsilon}}|x_j|^2\right)^{1/2}<\epsilon/2 $$ so $dist(x,R^{n_\epsilon})\leq \epsilon$. By the first part, this implies $T(B(0,1))$ pre-compact