Let $T: C^0[0,1] \rightarrow l^1$, $(Tf)_n=a_n \int_0^{1/n} f(x)dx$, for an $f$ in $C^0[0,1]$. Prove that $T$ is compact when $\{\frac{a_n}{n} \}_n \in l^1$.
I know the definition of compact operator, but in this case I'd like to state that if $\{ \frac{a_n}{n} \} \in l^1$, then $T$ is a finite rank operator. Indeed,
$||(Tf)_n||_{l^1} = \sum_n |\frac{a_n}{n} \int_0^1 f(\frac{t}{n})dt|$. But $f$ is continuous in $[0,1]$, then this integral is finite, and, particularly, it's bounded from $||f||_{\infty}$, thus the last sum is less or equal than $||f ||_{\infty} \sum_n |\frac{a_n}{n}|$, thus it's convergent, and then $T$ has finite rank.
Is it okay?
EDIT
Set $T_m(f)=(a_1 \int_0^1f(x)dx,\ldots,a_m \int_0^{1/m}f(x)dx, 0, 0, \ldots)$.
I want to show that $|| T_m-T|| \rightarrow _m 0$ in the operator norm. I have to compute $\sup \{ ||T_m(f) – T(f)||_{l^1}: ||f||_{\infty} \leq 1 \}=\sup \{ \sum_{k=m+1} |a_k \int_0^\frac{1}{k}f(x)dx|: ||f||_{\infty} \leq 1 \}$
Now, $\sum_{k=m+1} |a_k \int_0^\frac{1}{k}f(x)dx| \leq \sum_{k=1}^{\infty} \frac|{a_k}{k} \int_0^1 f(\frac{t}{n})dt| \leq ||f||_{\infty} \sum_k |\frac{a_k}{k}|$.
Thus, by taking the supremum I have that $||T-T_m|| = \sum_{k=m+1}^\infty |\frac{a_k}{k}|$. But for $\{ \frac{a_k}{k} \} \in l^1$, this is the remainder of a convergent series, and then the limit over $m$ goes to $0$. So $T$ is compact, since it's limit of compact (they're finite rank) operators
Best Answer
While this operator is not of finite rank, can you show it is a limit of finite rank operators? (In this case you're done since a limit of compact operators is compact).