- Let X and Y be Hilbert spaces and \(L : X \to Y\) be a bounded linear operator. Prove that the following two conditions are equivalent:
The image \(L ( \mathbf { B } )\) of the unit ball in X has compact closure in \(Y .\)
There is a sequence of bounded linear operators \(L_n:X\to Y\) with finite-dimensional range such that \(\|L_n-L\|\to0\), where \(\|\cdot\|\) is the operator norm.
Solution
1. Assume (a) and approximate \(L(B)\) by a finite-dimensional subspace.
Let \(B\) be the closed unit ball of \(X\), and assume \[ K:=\overline{L(B)} \] is compact in \(Y\).
Fix \(n\ge1\). By compactness, choose points \[ y_1,\dots,y_{N_n}\in K \] such that \[ K\subseteq\bigcup_{j=1}^{N_n}B_Y(y_j,1/n). \] Let \[ Y_n:=\operatorname{span}\{y_1,\dots,y_{N_n}\}. \] This is finite dimensional, hence closed. Let \(P_n:Y\to Y_n\) be the orthogonal projection and define \[ L_n:=P_nL. \] Then \(L_n\) is bounded and \(L_n(X)\subseteq Y_n\), so \(L_n\) has finite-dimensional range.
2. Prove \(L_n\to L\) in operator norm.
If \(\|x\|\le1\), then \(Lx\in K\), so there is some \(j\) with \[ \|Lx-y_j\|<1/n. \] Because \(P_nLx\) is the best approximation to \(Lx\) from the subspace \(Y_n\), \[ \|Lx-P_nLx\| \le \|Lx-y_j\|<1/n. \] Hence \[ \|L-L_n\| =\sup_{\|x\|\le1}\|(I-P_n)Lx\| \le\frac1n. \] Thus \(\|L-L_n\|\to0\), proving (b).
3. Assume (b) and prove that \(L(B)\) is relatively compact.
Suppose \(L_n\) has finite-dimensional range and \[ \|L_n-L\|\longrightarrow0. \] Fix \(\varepsilon>0\) and choose \(n\) so that \[ \|L-L_n\|<\varepsilon/3. \] The set \(L_n(B)\) is bounded in the finite-dimensional space \(L_n(X)\), so its closure is compact and therefore totally bounded. Hence there are points \[ z_1,\dots,z_N\in Y \] such that \[ L_n(B)\subseteq\bigcup_{j=1}^N B_Y(z_j,\varepsilon/3). \]
For every \(x\in B\), choose \(j\) with \[ \|L_nx-z_j\|<\varepsilon/3. \] Then \[ \|Lx-z_j\| \le\|(L-L_n)x\|+\|L_nx-z_j\| <\frac{2\varepsilon}{3}<\varepsilon. \] Thus \(L(B)\) is totally bounded. Its closure is also totally bounded; since \(Y\) is complete, the closure of a totally bounded set is compact. Therefore \[ \overline{L(B)} \] is compact, proving (a).