Linear compact operators Let $X$ be an infinite-dimensional Banach space, $Y$ be a Banach space, $A: X \to Y$ be a linear compact operator.
Is it true that there is always a sequence $\{x_n\}\subset X$ such that $\|x_n\| \to +\infty$ and $\|Ax_n\| \to 0 $ ?
If it is true how to prove it?
 A: Suppose $A$ is compact.
We can suppose that $AX$ is dense in $Y$, otherwise  let
$Y' = \overline{AX}$ and consider the operator $A: X \to Y'$.
If $\ker A$ is non trivial, then the result is immediate, so suppose
$\ker A$ is trivial, hence $A$ is injective.
Then we have $\inf_{\|x\|=1} \|Ax\| =0$. To see this, suppose $\epsilon = \inf_{\|x\|=1} \|Ax\|  >0$, that is,
$\|Ax\| \ge \epsilon \|x\|$ for all $x$. Note that this implies that if $A x_n$ is Cauchy, then so is $x_n$.
Then define $B:Y \to X$ as follows: 
If $y \in Y$, choose $x_n$ such that $A x_n \to y$. Since $A x_n$ is Cauchy, we have $x_n \to x$ for some $x$. It is straightforward to verify that this $x$ is unique and we can define $By = x$. It is straightforward to check that $B$ must be linear. Since $\|x_n\| \to \|x\|=\|By\|$
and $\|Ax_n\| \to \|y\|$, we have $\|y\| \ge \epsilon \|By\|$ and so $B$ is bounded. Finally, since the sequence $n \mapsto x$ converges to $x$, we have
$BAx = x$, that is, $BA = I$, which contradicts $A$ being compact.
Hence we have $\inf_{\|x\|=1} \|Ax\| =0$. So choose $z_n$ with
$\|z_n\| = 1$ such that $\|A z_n \| \to 0$. Since $\sqrt{\|A z_n \|} \to 0$, we can choose
$x_n = {1 \over \sqrt{\|A z_n\|}} z_n$ which has the desired properties.
A: Since $0$ is in the approximate point spectrum of $A$, there is a sequence $y_n$ of unit vectors such that $Ay_n \to 0$. Choose a subsequence $y_{n_k}$ such that
$$
\|Ay_{n_k}\| < 4^{-k}
$$
then $x_k := 2^ky_{n_k}$ does what you need it to do.
