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Operators on $C([0,1])$ that is compact or not.
I feel a bit bad about raising this question since similiar question have been answered many times. But I couldn't find and answer so here I go.
Let $H:C[0,1]\to C[0,1]$ be the map defined by:
$H(f)={1\over x} \int_0^x f(t)dt$
Is H a bounded or compact operator?
I've proceeded by taking the space of polynomials and shown that if;
$p(x)= \sum_{n=0}^\infty a_nx^n$, where $a_n\neq 0 $ for only finetly many $n$
Then;
$H(p)(x) = \sum_{n=0}^\infty {a_n\over {n+1}}x^n$
It is clearly bounded. Here on out im on thin ice. I reason that the ring of polynomials is dense in C[0,1] and as such H can be extended as a bounded operator to all of C[0,1].
Next is the compactness and here I am truly lost. I do not really know how to go about proving or disproving it. It does not seem like a compact operator to me.
The sequence {(n+1)x^n} lies in the unit ball which of course get sent into the unit ball again, seeing as norm = 1, but it is infinite dimensional so it should not be compact.
Next trick i've got from the book is checking whether weak convergence gets sent to norm convergence, but frankly i do not not of any weakly convergent sequences in C[0,1]. In fact I'm not even sure about its dual, i have heard its all finit measures but i do not know how to use it.
I'd very much like to know how to answer the question but the general guidlines for solving problems like these would be much more appeachiated,
Hope you can help me, thanks in advance for taking the time, Tobias