= Solution
The <space of test functions> is
$$
\mathcal D(\mathbb R)=C_c^\infty(\mathbb R).
$$
A sequence $\varphi_m$ converges to $\varphi$ in $\mathcal D(\mathbb R)$ when all supports lie in one <compact set> $K$ and
$$
\sup_{x\in K}|\varphi_m^{(j)}(x)-\varphi^{(j)}(x)|\longrightarrow0
$$
for every <nonnegative integer> $j$. The <distribution> space $\mathcal D'(\mathbb R)$ is the <continuous dual space> of $\mathcal D(\mathbb R)$, and $u_m\to u$ in $\mathcal D'$ means
$$
\langle u_m,\varphi\rangle\longrightarrow\langle u,\varphi\rangle
$$
for every test function $\varphi$.
If a linear form $u$ is continuous, it clearly maps every <null sequence> to a scalar sequence tending to zero. Conversely, suppose it has this sequential property. For each compact $K$, its restriction to the <Fréchet space> $\mathcal D_K$ must be continuous: otherwise, for every $m$ one could choose $\varphi_m\in\mathcal D_K$ such that
$$
\max_{0\leq j\leq m}\|\varphi_m^{(j)}\|_\infty\leq\frac1m,
\qquad
|\langle u,\varphi_m\rangle|\geq1.
$$
Then $\varphi_m\to0$ in $\mathcal D$ but its images do not tend to zero, a contradiction. Continuity on every $\mathcal D_K$ is precisely continuity for the <strict inductive limit topology> of $\mathcal D$, so $u\in\mathcal D'$.
Use the convention $\tau_h\varphi(x)=\varphi(x-h)$. Translation and the <distributional derivative> are defined by
$$
\langle\tau_hu,\varphi\rangle
=\langle u,\tau_{-h}\varphi\rangle,
\qquad
\langle u',\varphi\rangle=-\langle u,\varphi'\rangle.
$$
Translation, differentiation, and multiplication by $-1$ are continuous maps on $\mathcal D$, so these formulas define continuous linear functionals and hence distributions.
For fixed $\varphi$, the <difference quotient> satisfies
$$
\frac{\tau_h\varphi-\varphi}{h}\longrightarrow-\varphi'
\quad\text{in }\mathcal D.
$$
Therefore
$$
\left\langle\frac{\tau_{-h}u-u}{h},\varphi\right\rangle
=\left\langle u,\frac{\tau_h\varphi-\varphi}{h}\right\rangle
\longrightarrow-\langle u,\varphi'\rangle
=\langle u',\varphi\rangle,
$$
which proves $u'=\lim_{h\to0}(\tau_{-h}u-u)/h$ in $\mathcal D'$.
For $-1<\lambda<0$, integration by parts after subtracting the value at zero gives
$$
\boxed{
\langle(x_+^\lambda)',\varphi\rangle
=\int_0^\infty[\varphi(x)-\varphi(0)]\lambda x^{\lambda-1}\,dx}.
$$
The subtraction makes the integrand locally integrable at zero, and $x^\lambda\to0$ handles the other boundary.
The analogous <Hadamard finite-part integral> is
$$
\boxed{
\langle(\log x_+)',\varphi\rangle
=\int_0^1\frac{\varphi(x)-\varphi(0)}x\,dx
+\int_1^\infty\frac{\varphi(x)}x\,dx}.
$$
Indeed, integrating from $\varepsilon$ and combining the boundary term $\varphi(\varepsilon)\log\varepsilon$ with the divergent constant part of the integral gives this limit.
Because $x_+^\lambda$ is a <locally integrable function>, its first distributional derivative has order at most one. It is not of order zero. Choose $\psi\in\mathcal D((0,1))$ with $\int_0^1\lambda t^{\lambda-1}\psi(t)\,dt\ne0$ and put $\psi_\varepsilon(x)=\psi(x/\varepsilon)$. The sup norms stay bounded while
$$
\langle(x_+^\lambda)',\psi_\varepsilon\rangle
=\varepsilon^\lambda
\int_0^1\lambda t^{\lambda-1}\psi(t)\,dt
$$
is unbounded as $\varepsilon\downarrow0$. This contradicts the local sup-norm estimate required of an <order-zero distribution>. Hence $(x_+^\lambda)'$ has order exactly $\boxed{1}$.
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