# what would be a non-trivial (non-constant) sequence of operators that converges to the identity in infinite dimensional Banach spaces?

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 ?

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 $$l_2$$.