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Let $ A\colon L_1[0,1] \to C[0,1] $

$$ Af(t) = \int_0^t f(s)ds,\quad f \in L_1[0,1] $$

Is $A$ a compact operator or not?

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This is not a duplicate of the proposed duplicate. Both the domain and the codomain are different. And the conclusion as well. – 1015 Jun 28 '13 at 16:27
@julien Whoops. – Pedro Tamaroff Jun 28 '13 at 17:19
A related problem. – Mhenni Benghorbal Jun 28 '13 at 18:32
@MhenniBenghorbal These are not Hilbert spaces. And the operator is not compact. This is quite unrelated. – 1015 Jun 28 '13 at 19:20
@julien: I know there is no Hilbert spaces! But there is the related definition of a compact operator "they map bounded sequences in X to sequences in Y with convergent subsequences" which has been used the answer as you see. – Mhenni Benghorbal Jun 28 '13 at 22:23
up vote 5 down vote accepted

No, it isn't.

The image of any bounded sequence under a compact operator must have a norm-convergent subsequence.

Consider the functions $f_n=n\chi_{[0,1/n]}$, $n=1,2,\ldots$. Each $f_n$ has $L_1$-norm one, but the sequence $(Af_n)$ has no subsequence which converges in $C[0,1]$.

To see this, note $f_n$ is the continuous function whose graph consists of the straight line segments connecting the points $(0,0)$, $(1/n,1)$, and $(1,1)$. Given any $n$, it is easy to see that there is an $M$ so that $\Vert Af_n-Af_m\Vert_{\infty}>1/2$ for all $m\ge M$.

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In your coment "To see this, note $f_n$ is the continuous function..." you mean "To see this, note $Af_n$ is the continuous function..." right? – O Empalador de Cabras Nov 19 '14 at 14:08
@OEmpaladordeCabras Yes, thanks for pointing out the typo. – David Mitra Nov 19 '14 at 14:10

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