1
$\begingroup$

It is a result that if $||T-T_n|| \rightarrow 0$ in the norm operator an that the $T_n \in \mathcal{L}(X,Y)$ (were $Y$ is a Banach space) are compact operators, then $T$ is compact. I found from here that pointwise convergence is not enough. But is there an easy counter-example for the case when $||T-T_n|| \rightarrow 0$ but $Y$ is not Banach and when then $T$ is not compact?

$\endgroup$
  • $\begingroup$ What is that norm if $Y$ is not Banach? $\endgroup$ – lcv Jan 6 at 14:47
  • $\begingroup$ $||T||=sup_{||x \le 1||}||Tx||_Y$ no? $\endgroup$ – roi_saumon Jan 6 at 15:02
  • $\begingroup$ If you redefine compact operator so that the image of the unit ball is precompact (instead of relatively compact, for Banach spaces this is equivalent) then the limit is indeed compact. $\endgroup$ – Jochen Jan 7 at 8:17
2
$\begingroup$

Consider $X=Y=d$ the space of finite sequences with the supremum norm. Then consider $$T_n(x_1,x_2,...,x_n,x_{n+1},...)=\left(\frac{x_1}{1},\frac{x_2}{2},...,\frac{x_n}{n},0,...\right).$$ This sequence of compact operators converges to $$T(x_1,x_2,...)=\left(\frac{x_1}{1},\frac{x_2}{2},...\right).$$ However this is not a compact operator since $T(B_1)$ is not complete.

$\endgroup$
  • $\begingroup$ I am not sure I understand the "However this is not a compact operator since T(B1) is not complete.". What would be a sequence for which the image has no converging subsequence? $\endgroup$ – roi_saumon Jan 6 at 23:38
  • $\begingroup$ Consider $x^{(n)}=(2^{-1},2^{-2},...,2^{-n},0,...)$. Then $T(x^{(n)})=(2^{-1}/1,2^{-2}/2,...,2^{-n}/n,0,...)$ is a non-convergent Cauchy sequence in $T(B_1)$. (Because $d$ only contains finite sequences.) $\endgroup$ – SmileyCraft Jan 7 at 3:06
  • $\begingroup$ @SimileyCraft I am not sure to understand why we have the $1/n$ and the $2^{-n}$. If I take $T_n \in \mathcal L(l^p)$ with $T_n : x \mapsto (x_1, x_2, ..., x_n, 0, 0, ...)$ then $T_n$ is compact and $T_n \rightarrow Id_{l^p}$ in the operator norm, no? But then $Id_{l^p}$ is not compact because the sequence $(1,0,0,0,...), (0,1,0,0,...),(0,0,1,0,0,...)$ has no converging subsequence even though it is bounded. Also, you conclude by saying that we have a Cauchy sequence in $T(B_1)$ that doesn't converge meaning that $T(B_1)$ is not complete. Does it imply that it is not relatively compact? $\endgroup$ – roi_saumon Jan 7 at 12:17
  • $\begingroup$ @roi_saumon Your $T_n$ does not converge to the identity operator. Remember we must use the operator norm. Compact implies sequentially compact. If a Cauchy sequence does not converge, then it also has no converging subsequence. $\endgroup$ – SmileyCraft Jan 8 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.