Re: compact operators, convergence pointwise / w.r.t. operator norm

David C. Ullrich wrote:

Because the identity is not compact. 

Yes, and there exist compact injective T, e.g. T(x_n) := (x_n/n)
on ell_2. For injective T we have T^r \to identity pointwise.
But not necessarily in norm, because else the identity would be
compact.

> (T^r _is_ a norm-continuous 
function of r for r > 0, right? Just a guess based on the
heuristic...)


Because the continuous functional calculus is, well, continuous.
(Is it called "continuous functional calculus" because it is
continuous, or because it is defined on the space of continuous
functions defined on the spectrum of T?)

Markus
.

Loading