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[parent] $\operatorname{p.\!v.}(\frac{1}{x})$ is a distribution of first order (Proof)

(Following [1,2].)

Let $ u\in \mathcal{D}(U)$. Then $ \operatorname{supp}u \subset [-k,k]$ for some $ k>0$. For any $ \varepsilon>0$, $ u(x)/x$ is Lebesgue integrable in $ \vert x\vert\in[\varepsilon,k]$. Thus, by a change of variable, we have

$\displaystyle \operatorname{p.\!v.}(\frac{1}{x})(u)$ $\displaystyle =$ $\displaystyle \lim_{\varepsilon\to0+} \int_{[\varepsilon,k]} \frac{u(x)-u(-x)}{x} dx.$  

Now it is clear that the integrand is continuous for all $ x\in \mathbb{R}\setminus\{0\}$. What is more, the integrand approaches $ 2u'(0)$ for $ x\to 0$, so the integrand has a removable discontinuity at $ x=0$. That is, by assigning the value $ 2u'(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]} \big(u(x)-u(-x)\big)/x$ (where $ \chi$ is the characteristic function), and applying the Lebesgue dominated convergence theorem, we have
$\displaystyle \operatorname{p.\!v.}(\frac{1}{x})(u)$ $\displaystyle =$ $\displaystyle \int_{[0,k]} \frac{u(x)-u(-x)}{x} dx.$  

It follows that $ \operatorname{p.\!v.}(\frac{1}{x})(u)$ is finite, i.e., $ \operatorname{p.\!v.}(\frac{1}{x})$ takes values in $ \mathbb{C}$. Since $ \mathcal{D}(U)$ is a vector space, if follows easily from the above expression that $ \operatorname{p.\!v.}(\frac{1}{x})$ is linear.

To prove that $ \operatorname{p.\!v.}(\frac{1}{x})$ is continuous, we shall use condition (3) on this page. 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

$\displaystyle \vert\frac{u(x)-u(-x)}{x}\vert$ $\displaystyle =$ $\displaystyle \vert\frac{1}{x}\int_{(-x,x)} u'(t) dt\vert$  
  $\displaystyle \le$ $\displaystyle 2 \vert\vert u'\vert\vert _\infty,$  

where $ \vert\vert\cdot\vert\vert _\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
$\displaystyle \vert\operatorname{p.\!v.}(\frac{1}{x})(u)\vert \le 2k \vert\vert u'\vert\vert _\infty$
and $ \operatorname{p.\!v.}(\frac{1}{x})$ is a distribution of first order as claimed. $ \Box$

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.



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Cross-references: absolutely continuous, fundamental theorem of calculus for the Lebesgue integral, equality, supremum norm, compact subset, expression, vector space, Lebesgue dominated convergence theorem, characteristic function, Lebesgue measurable, removable discontinuity, continuous, clear, variable, Lebesgue integrable
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This is version 4 of $\operatorname{p.\!v.}(\frac{1}{x})$ is a distribution of first order, born on 2003-07-18, modified 2003-12-21.
Object id is 4473, canonical name is Operatornamepvfrac1xIsADistributionOfFirstOrder.
Accessed 2150 times total.

Classification:
AMS MSC: 46F05 (Functional analysis :: Distributions, generalized functions, distribution spaces :: Topological linear spaces of test functions, distributions and ultradistributions)
 46-00 (Functional analysis :: General reference works )
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