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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

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

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  • $\begingroup$ Okay, I know this result. I'm editing my original post $\endgroup$ – VoB Jan 28 at 16:58
  • $\begingroup$ Fixed. Let me know if it's alright $\endgroup$ – VoB Jan 28 at 17:16
  • $\begingroup$ Looks good to me $\endgroup$ – bitesizebo Jan 28 at 17:38

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