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Real Analysis HW 04-26 (II)

Fri, 21 May 2010 06:31:41 +0800 | Comments(845) | Category:Math | Tags:

From Folland p.289
Problem 2.
Let [tex]f[/tex] be a continuous function on [tex]\mathbb{R}^n\backslash\{0\}[/tex] that is homogeneous of degree [tex]-n[/tex] (i.e., [tex]f(rx)=r^{-n}f(x)[/tex]) and has mean zero on the unit sphere (i.e., [tex]$\int f\ d\sigma=0$[/tex] where [tex]\sigma[/tex] is surface measure on the sphere). Then [tex]f[/tex] is not locally integrable near the origin (unless [tex]f=0[/tex]), but the formula
[tex] \begin{align*} \left<\mathrm{PV}(f),\phi\right>=\lim_{\epsilon\rightarrow 0}\int_{\left\lvert x\right\rvert>\epsilon}{f(x)\phi(x)\ dx}\qquad (\phi\in C_c^\infty) \end{align*} [/tex]
defines a distribution [tex]\mathrm{PV}(f)[/tex] — “PV” stands for “principal value” — that agrees with [tex]f[/tex] on [tex]\mathbb{R}^n\backslash\{0\}[/tex] and is homogeneous of degree [tex]-n[/tex] in the sense of Exercise 9.
(Hint: For any [tex]a>0[/tex], the indicated limit equals
[tex] \begin{align*} \int_{\left\lvert x\right\rvert\leq a}{f(x)[\phi(x)-\phi(0)]\ dx} +\int_{\left\lvert x\right\rvert> a}{f(x)\phi(x)\ dx}, \end{align*} [/tex]
and these integrals converge absolutely.)
Sol:
  1. I'll show first that [tex]f[/tex] is not locally integrable near the origin. Given any [tex]r>0[/tex], consider
    [tex] \begin{align*} \int_{\left\lvert x\right\rvert\leq r}{\left\lvert f(x)\right\rvert\ dx}&= \int_{s=0}^r\int_{\left\lvert x\right\rvert=s}{\left\lvert f(x)\right\rvert\ d\sigma ds}\\&=\int_{s=0}^r s^{-1}\ ds\ \int_{\left\lvert x\right\rvert=1}{f(x)\ d\sigma}. \end{align*} [/tex]
  2. Next I'll show that [tex]\mathrm{PV}(f)\in \mathcal{D}'[/tex]. Given [tex]\phi\in C_c^\infty(\mathbb{R}^n)[/tex] and [tex]\epsilon>0[/tex], then
    [tex] \begin{align*} &\int_{\left\lvert x\right\rvert>\epsilon}{\left\lvert f(x)\phi(x)\right\rvert\ dx}\\\leq\quad&\int_{\epsilon<\left\lvert x\right\rvert\leq 1}{\left\lvert f(x)\right\rvert\left\lvert\phi(x)-\phi(0)\right\rvert\ dx}+\int_{\left\lvert x\right\rvert>1}{\left\lvert f(x)\right\rvert\left\lvert\phi(x)\right\rvert\ dx}\\=\quad&\mathrm{I}+\mathrm{II} \end{align*} [/tex]
    For part I,
    [tex] \begin{align*} \mathrm{I}&=\int_{\epsilon<\left\lvert x\right\rvert\leq 1}{\left\lvert f(x)\right\rvert\left\lvert x\cdot \nabla\phi(\xi(x))\right\rvert\ dx}\\&\leq\sup\left\lvert\nabla\phi\right\rvert\ \int_{\epsilon<\left\lvert x\right\rvert\leq 1}{\left\lvert x\right\rvert\left\lvert f(x)\right\rvert\ dx}\\&=\sup\left\lvert\nabla\phi\right\rvert\ \int_{s=\epsilon}^1 ds\ \int_{\left\lvert x\right\rvert=1}{\left\lvert f(x)\right\rvert\ d\sigma}\\&\leq\sup\left\lvert\nabla\phi\right\rvert\ \int_{\left\lvert x\right\rvert=1}{\left\lvert f(x)\right\rvert\ d\sigma}. \end{align*} [/tex]
    On the other hand, for part II,
    [tex] \begin{align*} \mathrm{II} \leq\left\lvert\mathrm{supp}\ \phi\right\rvert\sup_{x\in\mathbb{R}^n}{\left\lvert\phi(x)\right\rvert}\ \sup_{\left\lvert x\right\rvert\geq 1}{\left\lvert f(x)\right\rvert}. \end{align*}[/tex]
  3. From the estimate above, by dominated convergence theorem, [tex]\mathrm{PV}(f)[/tex] is indeed a distribution.
    The remaining is obvious.

Real Analysis HW 04-26 (I)

Wed, 19 May 2010 21:52:46 +0800 | Comments(205) | Category:Math | Tags:

From Folland p. 289
Problem 1.
A distribution [tex]F[/tex] on [tex]\mathbb{R}^n[/tex] is called homogeneous of degree [tex]\lambda[/tex] if [tex]F\circ S_r=r^\lambda F[/tex] for all [tex]r>0[/tex], where [tex]S_r(x)=rx[/tex].
  1. [tex]\delta[/tex] is homogeneous of degree [tex]-n[/tex].
  2. If [tex]F[/tex] is homogeneous of degree [tex]\lambda[/tex], then [tex]\partial^\alpha F[/tex] is homogeneous of degree [tex]\lambda-\left\lvert\alpha\right\rvert[/tex]
  3. The distribution [tex](d/dx)[\chi_{(0,\infty)}(x)\log x][/tex] discussed in the text is not homogeneous, although it agree on [tex]\mathbb{R}\backslash\{0\}[/tex] with a function that is homogeneous of degree [tex]-1[/tex].
Sol:
  1. Recall the fact that if [tex]T:\mathbb{R}^n\rightarrow\mathbb{R}^n[/tex] is an invertible linear transform, then
    [tex]\begin{align*} \int f(x)\ dx &= \int f\circ T(x)\ dT(x) \\ &= \det(T)\int f\circ T(x)\ dx. \end{align*}[/tex]
    This simple example helps us understand how composition of a distribution with an invertible linear transform behaves.
  2. For a, consider that given any [tex]\phi\in C_c^\infty(\mathbb{R}^n)[/tex],
    [tex]\begin{align*} \left<\delta\circ S_r,\phi\right>&=\det(S_r)^{-1} \left<\delta,\phi\circ {S_r}^{-1}\right> \\ &=r^{-n}\left<\delta,\phi\right>. \end{align*}[/tex]
  3. For b, consider that given an [tex]F\in\mathcal{D}'[/tex] which is homogeneous of degree [tex]\lambda[/tex], then for any [tex]\phi\in C_c^\infty(\mathbb{R}^n)[/tex] we have
    [tex]\begin{align*} \left<\partial^\alpha F\circ S_r,\phi\right>&= \det(S_r)^{-1}\left<\partial^\alpha F,\phi\circ {S_r}^{-1}\right>\\ &=r^{-(n+\left\lvert\alpha\right\rvert)}\left<F, (-1)^{\left\lvert\alpha\right\rvert}\partial^\alpha\phi\circ {S_r}^{-1}\right>\\ &=r^{-\left\lvert\alpha\right\rvert}\left<F\circ S_r,(-1)^{\left\lvert\alpha\right\rvert}\partial^\alpha\phi\right>\\ &=r^{\lambda-\left\lvert\alpha\right\rvert}\left<F,(-1)^{\left\lvert\alpha\right\rvert}\partial^\alpha\phi\right>\\ &=r^{\lambda-\left\lvert\alpha\right\rvert}\left<\partial^\alpha F,\phi\right> .\end{align*}[/tex]
  4. For c, as same as text, we define
    [tex]\begin{align*} L_\epsilon(x)=\chi_{(\epsilon,\infty)}(x)\log x \end{align*}[/tex]
    for some [tex]\epsilon>0[/tex].
    Given any [tex]\phi\in C_c^\infty(\mathbb{R})[/tex], we have
    [tex]\begin{align*} \left<{L_\epsilon}'\circ S_r,\phi\right>&=r^{-2}\left<L_\epsilon,-\phi'\circ {S_r}^{-1}\right>\\ &=r^{-1}\left<L_\epsilon\circ S_r,\phi'\right>\\ &=r^{-1}\left<(\log r+\log x)\chi_{(\epsilon/r,\infty)},-\phi'\right> \end{align*}[/tex]
    If we let [tex]\epsilon\rightarrow 0[/tex], then the equation above becomes
    [tex]\begin{align*} \left<L'\circ S_r,\phi\right>&=r^{-1}\left<\log r\ \delta+L',\phi\right> \end{align*}[/tex]
    Therefore, [tex]L'[/tex] is not homogeneous.