3楼: Originally posted by zaq123321 at 2015-09-12 12:12:13
Also can you tell who is the author if the book

2楼: Originally posted by zaq123321 at 2015-09-12 12:11:03
I think he wants to use the example g=1/r. If g=1/r, then g in L^2 but g not in L^1. Also |r|>1. If g=1/r^2, |r|>1, then g in L1 and g in L2
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5楼: Originally posted by hank612 at 2015-09-12 12:58:38
你针对的是一维情况， 而楼主说的是二维情况， 两个人说的都是正确的。

对R^2来说， 积分（极座标）面积元是dS=dxdy= rdrd\theta, 所以 对a>0, 有\int_{0}^{2\pi}\int_{a}^{\infty}\frac{1}{r^2}\cdot rdrd\ ...

5楼: Originally posted by hank612 at 2015-09-12 12:58:38
你针对的是一维情况， 而楼主说的是二维情况， 两个人说的都是正确的。

对R^2来说， 积分（极座标）面积元是dS=dxdy= rdrd\theta, 所以 对a>0, 有\int_{0}^{2\pi}\int_{a}^{\infty}\frac{1}{r^2}\cdot rdrd\ ...

7楼: Originally posted by shenyxtata at 2015-09-12 18:23:30
为什么要避开某个点？
我觉得既然已经定义了g(r)，也定义了domain (R^2)，就应该讨论实际的情况。

所以g(r)既不是L1，也不是L2...

8楼: Originally posted by zaq123321 at 2015-09-12 18:45:44
3x for discussion. It's not that rigorous
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