$\mathrm{p}.\text{tmspace}-.1667\mathrm{emv}.(\frac{1}{x})$ is a distribution of first order
(Following [1, 2].) Let $u\in \mathcal{D}(U)$. Then $\mathrm{supp}u\subset [-k,k]$ for some $k>0$. For any $\epsilon >0$, $u(x)/x$ is Lebesgue integrable^{} in $|x|\in [\epsilon ,k]$. Thus, by a change of variable, we have
$\mathrm{p}.\mathrm{v}.({\displaystyle \frac{1}{x}})(u)$ | $=$ | $\underset{\epsilon \to 0+}{lim}{\displaystyle {\int}_{[\epsilon ,k]}}{\displaystyle \frac{u(x)-u(-x)}{x}}\mathit{d}x.$ |
Now it is clear that the integrand is continuous^{} for all $x\in \mathbb{R}\setminus \{0\}$. What is more, the integrand approaches $2{u}^{\prime}(0)$ for $x\to 0$, so the integrand has a removable discontinuity at $x=0$. That is, by assigning the value $2{u}^{\prime}(0)$ to the integrand at $x=0$, the integrand becomes continuous in $[0,k]$. This means that the integrand is Lebesgue measurable on $[0,k]$. Then, by defining ${f}_{n}(x)={\chi}_{[1/n,k]}\left(u(x)-u(-x)\right)/x$ (where $\chi $ is the characteristic function^{}), and applying the Lebesgue dominated convergence theorem^{}, we have
$\mathrm{p}.\mathrm{v}.({\displaystyle \frac{1}{x}})(u)$ | $=$ | ${\int}_{[0,k]}}{\displaystyle \frac{u(x)-u(-x)}{x}}\mathit{d}x.$ |
It follows that $\mathrm{p}.\mathrm{v}.(\frac{1}{x})(u)$ is finite, i.e., $\mathrm{p}.\mathrm{v}.(\frac{1}{x})$ takes values in $\u2102$. Since $\mathcal{D}(U)$ is a vector space, if follows easily from the above expression that $\mathrm{p}.\mathrm{v}.(\frac{1}{x})$ is linear.
To prove that $\mathrm{p}.\mathrm{v}.(\frac{1}{x})$ is continuous, we shall use condition (3) on this page (http://planetmath.org/Distribution4). For this, suppose $K$ is a compact subset of $\mathbb{R}$ and $u\in {\mathcal{D}}_{K}$. Again, we can assume that $K\subset [-k,k]$ for some $k>0$. For $x>0$, we have
$|{\displaystyle \frac{u(x)-u(-x)}{x}}|$ | $=$ | $|{\displaystyle \frac{1}{x}}{\displaystyle {\int}_{(-x,x)}}{u}^{\prime}(t)\mathit{d}t|$ | ||
$\le $ | $2{||{u}^{\prime}||}_{\mathrm{\infty}},$ |
where $||\cdot |{|}_{\mathrm{\infty}}$ is the supremum norm. In the first equality we have used the fundamental theorem of calculus^{} for the Lebesgue integral (valid since $u$ is absolutely continuous^{} on $[-k,k]$). Thus
$$|\mathrm{p}.\mathrm{v}.(\frac{1}{x})(u)|\le 2k{||{u}^{\prime}||}_{\mathrm{\infty}}$$ |
and $\mathrm{p}.\mathrm{v}.(\frac{1}{x})$ is a distribution^{} of first order (http://planetmath.org/Distribution4) as claimed. $\mathrm{\square}$
References
- 1 M. Reed, B. Simon, Methods of Modern Mathematical Physics: Functional Analysis^{} I, Revised and enlarged edition, Academic Press, 1980.
- 2 S. Igari, Real analysis - With an introduction to Wavelet Theory, American Mathematical Society, 1998.
Title | $\mathrm{p}.\text{tmspace}-.1667\mathrm{emv}.(\frac{1}{x})$ is a distribution of first order |
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Canonical name | operatornamepvfrac1xIsADistributionOfFirstOrder |
Date of creation | 2013-03-22 13:46:07 |
Last modified on | 2013-03-22 13:46:07 |
Owner | Koro (127) |
Last modified by | Koro (127) |
Numerical id | 7 |
Author | Koro (127) |
Entry type | Proof |
Classification | msc 46F05 |
Classification | msc 46-00 |