Approximation property and bounded approximation property

Let $X$ be a Banach space. It has the approximation property (AP) if for every norm-compact set $C\subseteq X$ and every $\epsilon >0$ there is a bounded finite-rank operator $T:X\to X$ such that

${\sup }_{x\in C}\mathrm{\parallel }Tx-x\mathrm{\parallel }<\epsilon \mathrm{.}$

Here finite rank means that $T(X)$ is finite-dimensional.

For $\lambda \ge 0$, $X$ has the $\lambda$-bounded approximation property ($\lambda$-BAP) if the same assertion holds with the additional operator-norm bound $\mathrm{\parallel }T\mathrm{\parallel }\le \lambda$, where the norm is that of The operator norm as the least bound and as the unit-sphere or unit-ball supremum. It has the bounded approximation property (BAP) if it has $\lambda$-BAP for some finite $\lambda$.

No sequence of approximating operators is required; this matters in nonseparable spaces. Plainly $\lambda$-BAP implies AP.