Timeline for For a bounded, measurable function $f$ defined on $[1,\infty]$, does the fact that $f$ is integrable imply that $\sum_n a_n$ converges?
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8 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2022 at 0:12 | comment | added | GReyes | Observe that $\sum_{n=1}^k\int_{n}^{n+1}|f|dx=\int_1^{k+1}|f|dx$. We apply MCT to the sequence $g_k(x)=|f|\chi_{[1,k+1)}(x)$. Clearly, $g_k$ approaches monotonically to $|f|$. Therefore, $\int g_k(x)\,dx\to\int|f|\,dx<\infty$ as $k\to\infty$. | |
| Mar 20, 2022 at 23:43 | comment | added | qqq123 | I still don't get which sequence of functions we are applying MCT to. Is it the sequence of functions $\{\sum_{n=1}^k \int_n^{n+1} |f| \}_k$? This gives $\lim_{k\to\infty}( \sum_{n=1}^k \int_n^{n+1} |f| )= \int_n^{n+1}( \lim_{k\to\infty} \sum_{n=1}^k |f| )$. I cannot see how the right hand side of the equality is $\int_1^\infty |f|$. | |
| Mar 20, 2022 at 23:39 | vote | accept | qqq123 | ||
| Mar 20, 2022 at 22:14 | comment | added | GReyes | Absolute convergence implies convergence. Yes, I mean Monotone convergence theorem. The theorem, for non-negative functions like $|f|$, says that the limit of the integrals is the integral of the limit. $|f|$ is the point-wise monotonic limit of its restrictions to $[1,n+1)$. Your limit is integrable, so it it finite. | |
| Mar 20, 2022 at 22:12 | comment | added | qqq123 | I get the fact that $\sum |a_n| $ is convergent, but this is what we are trying to prove. I didn't really get how you approach the statement in your answer by monotonic convergence. I assume that you meant Monotone Convergence Theorem. But this theorem requires that the sequence to be bounded. However, we are trying to prove boundedness. | |
| Mar 20, 2022 at 21:39 | comment | added | GReyes | No, it is not. What we are saying is that $\sum |a_n|$ is convergent, and its sum is bounded by $\int|f|\, dx$. Therefore $\sum a_n$ is also convergent. | |
| Mar 20, 2022 at 21:15 | comment | added | qqq123 | Isn't this statement equivalent to $\sum^\infty_{n=1} a_n = \int^\infty_1 f$? But $\sum^\infty_{n=1} a_n = \int^\infty_1 f$ is not correct. A counterexample is set $f:= \sin(2\pi x)$. $a_n$ is zero for every $n$ and therefore $\sum a_n$ is zero. But $\sin(2\pi x)$ is not integrable. Assume that this statement is correct, then the fact that $\sum a_n$ converges would imply $\int f$ is integrable, which contradicts to the example I made. | |
| Mar 20, 2022 at 21:05 | history | answered | GReyes | CC BY-SA 4.0 |