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Let $\Omega=\{x\in \mathbb R^2: 0\lt\Vert x\Vert \lt1 \}$ and given the following functions

1.$u(x)=\Vert x\Vert^{-1}$
2.$u(x)=ln(\Vert x\Vert)$
3.$u(x)=ln(ln(\Vert x\Vert)+1)$

How do I prove that those functions are $L_2$ (square-integrable) integrable with respect to $\Omega$.

I would appreciate any help.

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    3. is not a well defined function. – 2017-01-16

2 Answers 2

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Hint: As stated, 3. is not a well defined function. But for 1. and 2. we can go to polar coordinates. Then 1., 2. turn into

  1. $\int_0^{2\pi } \int_0^1 r^{-2} r\, dr \,dt$

  2. $\int_0^{2\pi } \int_0^1 (\ln r)^2 r\, dr\,dt$

You're now in one variable.

2

Note as a preliminary that all three functions are even, that is, $u(x) = u(-x)$ so we can re-write the range as $0

The first function is not square-integrable, since $$ \int_0^1 \frac1{x^2} dx = -\frac11 + \frac10 \rightarrow \infty $$ The second function is square integrable since $$ \int_0^1 \left( \ln x \right)^2 dx = \lim_{x_0\to 0}\left[2x-2x\ln x + x(\ln x)^2\right]_{x_0}^1 = 2 $$ The third function is not even defined (as a real number) on the whole interval; consider any $x < \frac1e$. Are you sure the problem did not ask for $$ u(x) = \ln(\ln(x+1)) $$ That function is square integrable, although proving it is not so easy.

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    I dont understand why we integrate over the interval $(0,1)$ and not the open unit disk $\Omega$ could you explain ? – 2017-01-16