I would like to prove the following,

Let $X$,$Y$ be infinite dimensional Banach-Spaces and $T$ a compact, linear and bounded operator. Then there exists a sequence $(x_n)_{n\in\mathbb N}$ with $||x_n||=1$ and $||Tx_n||\rightarrow\;0\quad n\rightarrow\infty$

Clearly if we consider a sequence with $||x_n||=1$ then $(||Tx_n||)_{n\in\mathbb N}$ contains a Cauchy subsequence and hence a converging subsequence.

Then by taking just the index set of the subsequence we have $||x_{n_k}||=1$ and $||Tx_{n_k}||\rightarrow\;y\quad k\rightarrow\infty$ for some $y\in TY$. But how to construct a sequence where the limit is 0? Could someone help me?

Thanks in advance!

  • 3
    $\begingroup$ choose $v_n$'s in $X$ such that $\Vert v_n\Vert=1$, $\Vert v_n-v_m\Vert\geq 1/2$, replace $v_n$'s by a subsequence $u_n$ s.t. $Tu_n$ converges, look at $u_n-u_m$'s $\endgroup$ – user8268 May 26 '14 at 7:50
  • $\begingroup$ @user8268 I somehow do not see how to obtain a sequence which converges to zero after all...? $\endgroup$ – Thorben May 26 '14 at 20:28
  • 1
    $\begingroup$ how about $x_n=(u_n-u_{n+1})/\Vert u_n-u_{n+1}\Vert$ ? $\endgroup$ – user8268 May 26 '14 at 20:31
  • $\begingroup$ great! I totally forgot I can norm those guys! Thanks you very much $\endgroup$ – Thorben May 26 '14 at 20:33

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