- What does it mean to say that $u$ is a distribution on $(a, b)$ of order $m β β_{0}$ ?

Let $V: π(a, b) β β$ be a linear functional on the space of real-valued test functions. Assume that $V$ is positive: $V(Ο) β©Ύ 0$ holds whenever $Ο β π(a, b)$ satisfies $Ο(x) β©Ύ 0$ for all $x β(a, b)$. Prove that $V$ is a distribution on $(a, b)$ of order 0. - Define the notion of distributional derivative for distributions on $(a, b)$ and explain how it is consistent with the usual notion for $\mathrm{C}^{1}$ functions on $(a, b)$.

Let $u β π'(a, b)$ be a distribution of order $m β β_{0}$. Prove that its distributional derivative $u'$ has order at most $m+1$. - Let $d: β β β$ be the function $$ d(x) β \dist(x, β€)\left(β \inf _{y β β€}|x-y|\right) $$ Calculate the distributional derivative $d'$. How is it related to the almost everywhere defined usual derivative of $d$ ?
- Put for $n β β_{0}$, $d_{n}(x) β 2^{-n} d(2^{n} x)$, and define for $x β β$,
$$
t(x) β \sum_{n=0}^{β} d_{n}(x)
$$
Explain why $t: β β β$ is a continuous and periodic function with period 1. Let $x_{0} β[0,1)$. Fix $m β β$ and take $k β β_{0}$ such that $k 2^{-m} β©½ x_{0}<(k+1) 2^{-m}$. Put $u_{m}=k 2^{-m}, v_{m}=(k+1) 2^{-m}$. Show that
$$
\frac{t(v_{m})-t(u_{m})}{v_{m}-u_{m}}=\sum_{n=0}^{m-1} \frac{d_{n}(v_m)-d_{n}(u_m)}{v_{m}-u_{m}}
$$
and
$$
\frac{d_{n}(v_m)-d_{n}(u_m)}{v_{m}-u_{m}} β\{ Β± 1\} \text{ for } n<m
$$
Conclude that $t$ is not differentiable at $x_{0}$. Calculate the distributional derivative of $t$.

[Results about differentiation in the sense of distributions can be used without proofs but must be clearly stated.]

- The space of test functions is $π(a, b)$ consists of all $\mathrm{C}^{β}$ functions $Ο:(a, b) β β$ with compact support in $(a, b)$. A distribution on $(a, b)$ is a linear functional $u: π(a, b) β β$ with the boundedness property: For each compact subset $K$ of $(a, b)$ there exist constants $c=c_{K} β₯ 0$, $m=m_{K} β β_{0}$ such that $|β¨u, Οβ©| β€ c \sum_{k=0}^{m} \sup \left|Ο^{(k)}\right|$ for all $Ο β π(a, b)$ with $\operatorname{supp}(Ο) β K$. If we can choose the constant $m β β_{0}$ independently of $K$, then $u$ has order at most $m$. For $m=0, u$ has order 0 if $u$ has order at most 0, and for $m β β, u$ has order $m$ if it has order at most $m$, but not order at most $m-1$.

Let $K$ be a compact subset of $(a, b)$. Put $ΞΎ=Ο_{Ξ΅} * π_{[a+Ξ΅, b-Ξ΅]}$ for $Ξ΅<\dist(K, β β(a, b)) / 3$. Then $ΞΎ β π(a, b), ΞΎ=1$ near $K$. Let $Ο β π(a, b)$ be supported in $K$ and note that $(\max |Ο|-Ο) ΞΎ β π(a, b)^{+}$, so by assumption $V((\max |Ο|-Ο) ΞΎ) β₯ 0$. By linearity and since $Ο ΞΎ=Ο$ we arrive at $V(Ο) β€V(ΞΎ) \max |Ο|$. Applying this to $-Ο$ yields by linearity of $V$ the bound $V(Ο) β₯-V(ΞΎ) \max |Ο|$, and consequently, $|V(Ο)| β€ c_{K} \max |Ο|$ where $c_{K}=|V(ΞΎ)|$. This is the boundedness property with $m=0$, and so $V$ is a distribution of order 0. - Let $u β π'(a, b)$. Then $u'$ is defined for $Ο β π(a, b)$ by $β¨u', Οβ©=-β¨u, Ο'β©$. Since differentiation is a linear map of test functions, it follows by linearity of $-u$ that also $u'$ is linear. Using the boundedness property of $u$ we obtain the boundedness property of $u'$ : For a compact subset $K$ of $(a, b)$ let the constants $c=c_{K}, m=m_{K}$ be as above. Then if $Ο β π(a, b)$ is supported in $K$ also $Ο'$ is supported there and so $$ \left|\leftβ¨u', Ο\rightβ©\right|=\left|\leftβ¨u, Ο'\rightβ©\right| β€ c \sum_{k=0}^{m} \sup \left|Ο^{(k+1)}\right| $$ This bound shows that $u' β π'(a, b)$. It also shows that if $u$ has order $m ββ_{0}$, then $u'$ has order at most $m+1$. Next, if $u β \mathrm{C}^{1}(a, b)$, then we consider $u$ as a distribution by the rule $β¨u, Οβ©=β«_{a}^{b} u Ο \mathrm{d} x$. Now by integration by parts $β¨u,-Ο'β©=-β«_{a}^{b} u Ο' \mathrm{d} x=-[u Ο]_{x β a+}^{x β b-}+β«_{a}^{b} u' Ο \mathrm{d} x=β«_{a}^{b} u' Ο \mathrm{d} x$, consequently the distributional derivative of $u$ can be identified with the usual derivative $u'$ in this case.
- The function $d$ is a 1-periodic function with $$ d(x)= \begin{cases}x &\text{if } x β\left[0, \frac{1}{2}\right) \\ 1-x &\text{if } x β\left(\frac{1}{2}, 1\right)\end{cases} $$ Note that it is piecewise $\mathrm{C}^{1}$, so we may perform partial integration as for $\mathrm{C}^{1}$ functions. Hereby we find $$ β¨d', Οβ©=β«_{β}(π_{I}-π_{J}) Ο \mathrm{d} x $$ where the function $π_{I}-π_{J}$ is the a.e. defined usual derivative of $d$. Here $$ I=\bigcup_{j β β€}\left(j, j+\frac{1}{2}\right) \text{ and } J=\frac{1}{2}+I $$
- The function $d_{n}$ is piecewise $\mathrm{C}^{1}$ and $2^{-n}$-periodic. Because $0 β€ d_{n}(x) β€2^{-1-n}$ for all $x$, the Weierstrass $M$-test shows that the series defining $t$ is uniformly convergent. It is therefore the uniform limit of continuous functions and so is continuous. Because all $d_{n}$ in particular are 1-periodic, it follows that also $t$ is 1-periodic.

Since for $n β₯ m$ the function $d_{n}$ is $2^{-m}$-periodic and $v_{m}-u_{m}=2^{-m}$ we have $$ \frac{t(v_m)-t(u_m)}{v_{m}-u_{m}}=\sum_{n=0}^{m-1} \frac{d_{n}(v_m)-d_{n}(u_m)}{v_{m}-u_{m}} \tag{1} $$ For $n<m$ the function $d_{n}$ is affine with slope $Β± 1$ on the interval $[u_{m}, v_{m}]$, hence $\frac{d_{n}(v_m)-d_{n}(u_m)}{v_{m}-u_{m}}= Β± 1$ as required. The sum in (1) can therefore not converge to a finite limit as $m β+β$. Since $$ \frac{t(x_{0}+h)-t(x_{0})}{h}=\sum_{n=0}^{m-1} \frac{d_{n}(x_{0}+h)-d_{n}(x_{0})}{h}=\sum_{n=0}^{m-1} \frac{d_{n}(v_m)-d_{n}(u_m)}{v_{m}-u_{m}} $$ when $h=v_{m}-x_{0}$, it follows that $t$ isn't differentiable at $x_{0}$.

Since uniform convergence implies convergence in the sense of distributions the series defining $t$ in particular converges in $π'(β)$. Because differentiation in the sense of distributions is a continuous operation in $π'(β)$ we get $t'=\sum_{n=0}^{β} d_{n}'$ in $π'(β)$. Here we get from (iii) above $d_{n}'=2^{-n}2^nd'(2^nx)=d'(2^nx)$, where the right-hand side is the a.e. defined usual derivative of $d$ evaluated at $2^{n} x$.

- Let $f: β β β$ be a continuous and even function. Define for each $Ο βπ(β)$ the principal value integral, $$ \leftβ¨\pv\left(\frac{f(x)}{x}\right), Ο\rightβ© β \lim _{Ξ΅ β 0} β«_{β β(-Ξ΅, Ξ΅)} \frac{f(x)}{x} Ο(x) \mathrm{d} x $$ Show that hereby $\pv\left(\frac{f(x)}{x}\right)$ is a distribution on $β$ and that $$ \leftβ¨\pv\left(\frac{f(x)}{x}\right), Ο\rightβ©=β«_{β} \frac{f(x)}{x} Ο(x) \mathrm{d} x $$ holds for all $Ο β π(β)$ with $Ο(0)=0$.
- Let $f: β β β$ be a continuous and even function with $f(0)=0$. Does it follow that the principal value distribution $\pv\left(\frac{f(x)}{x}\right)$ is a regular distribution? [Justify your answer carefully.]

- Fix $r>0$. Let $Ο β π(β)$ with support in $[-r, r]$. Then we have for each $Ξ΅ β(0, r)$ because $f(x) / x$ is a continuous and odd function on $β β\{0\}$ that $$ \left(β«_{-r}^{-Ξ΅}+β«_{Ξ΅}^{r}\right) \frac{f(x)}{x} \mathrm{~d} x=0 $$ and consequently, \begin{aligned} \left|\left(β«_{-r}^{-Ξ΅}+β«_{Ξ΅}^{r}\right) \frac{f(x)}{x} Ο(x) \mathrm{d} x\right| &=\left|\left(β«_{-r}^{-Ξ΅}+β«_{Ξ΅}^{r}\right) \frac{f(x)}{x}(Ο(x)-Ο(0)) \mathrm{d} x\right| \\ &β€\left(β«_{-r}^{-Ξ΅}+β«_{Ξ΅}^{r}\right)|f(x)| \frac{|Ο(x)-Ο(0)|}{|x|} \mathrm{d} x \\ &\stackrel{\text{FTC}}{β€}\left(β«_{-r}^{-Ξ΅}+β«_{Ξ΅}^{r}\right)|f(x)| \max \left|Ο'\right| \mathrm{d} x \\ &β€ 2 β«_{0}^{r}|f(x)| \mathrm{d} x β \max \left|Ο'\right|. \end{aligned} Read the inequality resulting from the second and fourth lines in the above display: take $Ξ΅ β 0$ and use Fatou's lemma to see that $f(x) \frac{Ο(x)-Ο(0)}{x}$ is integrable over $[-r, r]$. It therefore follows from Lebesgue's DCT that $$ \leftβ¨\pv\left(\frac{f(x)}{x}\right), Ο\rightβ©=β«_{-r}^{r} f(x) \frac{Ο(x)-Ο(0)}{x} \mathrm{~d} x $$ is well-defined and hereby also that $\pv\left(\frac{f(x)}{x}\right): π(β) β β$ is a well-defined linear functional. Since for a given compact set $K$ in $β$ we can find $r>0$ such that $K β[-r, r]$ it follows that the principal value integral satisfies the boundedness property: $$ \left|\leftβ¨\pv\left(\frac{f(x)}{x}\right), Ο\rightβ©\right| β€ c \max \left|Ο'\right|\text{ for all }Ο β π(K) $$ where $c=2 β«_{0}^{r}|f(x)| \mathrm{d} x$. It is therefore a distribution on $β$. Finally, when $Ο(0)=0$ the function $f(x) Ο(x) / x$ is integrable, so in this case the above yields $\leftβ¨\pv\left(\frac{f(x)}{x}\right), Ο\rightβ©=β«_{β} f(x) Ο(x) / x \mathrm{~d} x$ as required in that case.
- The answer is no:
any continuous and even function $f: β β β$ with $f(0)=0$ for which $f(x) / x β \mathrm{L}_\text{loc}^{1}(β)$ is a counter-example

If $\pv\left(\frac{f(x)}{x}\right)$ were a regular distribution on $β$, then we would have for some $g β \mathrm{L}_\text{loc}^{1}(β)$ that $\leftβ¨\pv\left(\frac{f(x)}{x}\right), Ο\rightβ©=β«_{β} g Ο \mathrm{d} x$ holds for all $Ο β π(β)$. But then it follows in particular from (b)(i) that $β«_{β β\{0\}} \frac{f(x)}{x} Ο(x) \mathrm{d} x=β«_{β β\{0\}} g(x) Ο(x) \mathrm{d} x$ for all $Ο β π(β β\{0\})$ and so by the fundamental lemma of the calculus of variations that $f(x) / x=g(x)$ a.e., a contradiction proving the claim.

For instance we can take $$ f(x)= \begin{cases}\frac{1}{|\log | x||+1} &\text{if } x β β β\{0\} \\ 0 &\text{if } x=0\end{cases} $$ Then $f: β β β$ is continuous, even and $f(0)=0$. But $$ β«_{-1}^{1}\left|\frac{f(x)}{x}\right| \mathrm{d} x=2 β«_{0}^{1} \frac{\mathrm{d} x}{(1-\log x) x}=2 β«_{1}^{β} \frac{\mathrm{d} y}{y}=+β $$ so $f(x) / x β \mathrm{L}_\text{loc}^{1}(β)$.