# Show compactness of an operator

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 definitin 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 that 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 op. 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

• How did you leap from convergent to finite rank? That is the essence of the question and it is not true. – copper.hat Jan 28 at 16:29
• Okay, now I see the problem. So I have to change strategy. Could I try with Ascoli Arzelà? – VoB Jan 28 at 16:44
• Perhaps showing T to be the limit of finite rank operators would be less of a fishing trip? – copper.hat Jan 28 at 16:51
• edited my original post ! – VoB Jan 28 at 17:25