it's something I've been thinking about when I was trying to prove that the identity is a compact operator (I found out later that it actually wasn't compact.) but here's a little bit of what I was doing while trying to prove the false claim :

for example consider the sequence space $\ell^2$, and define the following sequence of operators for $k \geq 1$ :

$T_k((x_n)_{n \geq 1}) = (x_1,x_2,\dots,x_k,0,0,0,\dots)$

$T_k$ is of finite rank and therefore compact ($\dim T_k(\ell^2) = k < \infty$).

intuitively speaking, I'd say that $T_k$ converges to the identity operator in $\ell^2$, but if that were the case then the identity would be the limit of a sequence of compact operators, therefore compact itself, which is not true.

so to my surprise, the limit of $T_k$ in $\ell^2$ is actually not the $Id$.

it can be seen by computing the norm of $\|T_k - T\|$, on the one hand it's bounded above by $2$ and for $x = (0,0,0,0,\dots,1,0,0,\dots)$ where the $1$ is situated in the $k+1$-th position, we find that the operator norm is also bounded below by $1$. (then use squeeze theorem).

now what would be a non-trivial sequence of operators that actually converges to the identity operator ?


1 Answer 1


Hints: $Ax(t)=(t^n(1-t)+1)x(t)$ in $C[0,1]$,

$Ax=(\frac{x_1}{n}+x_1,\frac{x_2}{n}+x_2,...)$ in $l_2$.

  • 1
    $\begingroup$ I see, so the trick is to add any sequence that goes to zero, thanks. $\endgroup$ Commented Feb 16, 2020 at 14:02

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