I'm trying to prove Lemma 4.8 of [1] online reading:
Notation: $\tau_C(X)$ means the topology of uniform convergence on the compact subsets of $X$.
Lemma 4.8. For a Banach space $X$ the following statements are equivalent.
(1) $X$ has the approximation property.
(2) The identity operator $I:X \longrightarrow X$ belongs to the $\tau_C(X)$-closure of the vector subspace $\mathtt{F}(X)$ of all finite-rank operators in $\mathtt{B}(X)$.
(3) If an arbitrary $\tau_C(X)$-continuous linear functional $\varphi$ on $\mathtt{B}(X)$ satisfies $\varphi(x^*\otimes x) = 0$ for all $x \in X$ and all $x^*\in X^*$, then $\varphi(I) = 0$.
-I can prove $1\Rightarrow 2\Rightarrow 3$ and also $2\Rightarrow 1$ but I don't have any idea about $3\Rightarrow 2$ or $3\Rightarrow1$!!
Thanks in advance for any help.
[1] Abramovich Y.A., Aliprantis C.D. An invitation to operator theory
Proof ($3\Rightarrow 2$): Use way of contradiction and then apply corollary [1] to $I\in \mathtt{B}(X)\setminus \overline{\mathtt{F}(X)}^{\tau_C(X)}$.
[1]: Corollary-IV.3.15 (J.B. Conway, A Course in Functional Analysis):
If $X$ is a LCS, $Y$ is a closed linear subspace of $X$, and $x_0\in X\setminus Y$, then there is a continuous linear functional $f:X\longrightarrow \mathbb{F}$ such that $f(y)=0$ for all $y\in Y$ and $f(x_0)=1$.