Timeline for Why can't I interchange integration and differentiation here?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| S Jan 8 at 9:53 | history | suggested | Felix Benning | CC BY-SA 4.0 |
fix broken link
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| Jan 8 at 9:21 | review | Suggested edits | |||
| S Jan 8 at 9:53 | |||||
| Jun 12, 2021 at 0:59 | comment | added | LSpice | WayBack to @ToddTrimble's link: $1 = 0$. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 15, 2012 at 14:50 | comment | added | Todd Trimble | See also Timothy Chow's "proof" that 0 = 1 at his website here math.mit.edu/~tchow/math.html, involving differentiation under the integral sign. | |
| Dec 15, 2012 at 13:56 | comment | added | Moritzplatz | thank you very much, you have been very helpfull and I think I understand this problem a lot better now! | |
| Dec 15, 2012 at 13:47 | vote | accept | Moritzplatz | ||
| Dec 15, 2012 at 13:45 | comment | added | Pietro Majer | Precisely: there is no $g\in L^1(\mathbb{R}_+)$ such that $y^2e^{-y^2}\le g(x)$ for all $x$ and $y$, because computing the supremum you get $1/x\le g(x)$. Note that even extracting a sequence $y_n\to0$, you would always get a supremum $\sup_ n y^2 e^{-y^2 _n x}$ not integrable on $\mathbb{R}_+$. You may verify that, by a direct computation, but the simplest reason is: otherwise the dominated convergence theorem would give convergence in $L^1$ to the pointwise limit, which is $0$, whereas these functions all have $L^1$ norm equal to $1$. | |
| Dec 15, 2012 at 12:49 | comment | added | Moritzplatz | Thanks pietro, I think I've got closer to understand the problem. Is the fact that for any $x$ big enough i can find an $y_0$ close enough to zero such that $y_0^2e^{-y_0^2x}=1/x$ and then use the fact that $1/x$ is not integrable on $(k,\infty)$ for any $k$ ? thank you very much! | |
| Dec 15, 2012 at 9:11 | answer | added | Peter Michor | timeline score: 4 | |
| Dec 15, 2012 at 8:43 | comment | added | Pietro Majer | Moritzplatz, note that $(F(y)-F(0))/y=\int_0^\infty y^2e^{-y^2x}dx$. It's the integral of $e^{-x}$ after rescaling. Rescaling is in fact the easiest way to make a counterxample to the dominated convergence theorem (then $F(y)$ has been defined consequently). If you make a picture of the graphs of $x\mapsto y^2e^{-y^2x}$ and compute $\sup_y y^2e^{-y^2x}$ it will be clear why the integrands are not dominated in $L^1$. | |
| Dec 15, 2012 at 1:44 | comment | added | Michael Renardy | How are you going to pick k and l in such a way that the limit y->0 is covered? This is not a research level question. | |
| Dec 14, 2012 at 23:37 | history | asked | Moritzplatz | CC BY-SA 3.0 |