# A completely continuous operator carries weakly Cauchy sequences into norm-convergent ones

Let $X,Y$ be Banach spaces. Show that if $T:X\to Y$ is a completely continuous operator, then $T$ carries weakly Cauchy sequences into norm-convergent sequences.

Let $(x_n)_{n=1}^\infty$ be a weakly Cauchy sequence. Let us suppose that $(Tx_n)_{n=1}^\infty$ is not norm-convergent, i.e., it is not a Cauchy sequence. Then I don't know what can I do.

• A sequence $(x_n)_{n=1}^\infty$ is said to converge weakly to $x$ if for every $f\in X'$, $f(x_n)\to f(x)$.
• A sequence $(x_n)_{n=1}^\infty$ is said to be a weak Cauchy sequence if for every $f\in X'$ the scalar sequence $(f(x_n))_{n=1}^\infty$ is Cauchy, or, equivalently, it is convergent (because the scalar field is complete). These notions are in general different because $(f(x_n))_{n=1}^\infty$ may converge to different points when different functionals $f$ are applied.
• @Tomás Read the EDIT. I wrote the difference. – Castaroth Nov 2 '13 at 18:14
• Sorry to ask @DavidMitra, but what is a weakly null sequence? – Tomás Nov 2 '13 at 19:09
• @Tomás Converges to zero weakly. – David Mitra Nov 2 '13 at 19:10
• Wow @DavidMitra, with this hint the problem became so easy. – Tomás Nov 2 '13 at 19:11

Hint: If $(x_n)_{n=1}^\infty$ is weakly Cauchy, then for any two strictly increasing sequences of integers, $(n_k)_{k=1}^\infty$ and $(m_k)_{k=1}^\infty$, the sequence $(x_{n_k}−x_{m_k})_{k=1}^\infty$ is weakly null. This implies $(T(x_n))_{n=1}^\infty$ is norm-Cauchy.