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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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