I am studying distributions on my own, and I was wondering if my solution to a question I've been working through looked okay.
Question. Prove that the Dirac measure in $\mathbb{R}^n$ is equal to a derivative of order $\le n$ of a bounded function (not having compact support). (Treves, 1969, q. 24.7)
Let $$ F = \begin{cases} \frac{(-1)^n}{2} & \text{on negative orthant} \\\\[1pt] \frac{1}{2} & \text{on positive orthant} \\\\[1.2pt] 0 & \text{otherwise} \end{cases} $$
Then $F$ is a bounded function not having compact support, and defines a distribution such that for $\varphi\in \mathcal{C}_c^{\infty}(\mathbb{R}^n)$:
$$ \begin{align*} &\langle (F^{\prime},\varphi\rangle \stackrel{def}{=} \langle F,(-1)^n\varphi ^{\prime}\rangle = (-1)^n\int_{\mathbb{R}^n}F\varphi^{\prime} \\\\ &= (-1)^n\int_{-\infty}^0\cdots\int_{-\infty}^0\frac{(-1)^n}{2} \varphi^{\prime} +(-1)^n\int_0^{\infty}\cdots\int_0^{\infty}\frac{1}{2}\varphi^{\prime} \\\\ &= \frac{(-1)^{2n}}{2}\int_{-\infty}^0\cdots\int_{-\infty}^0\varphi^{\prime}+\frac{(-1)^n}{2}\int_0^{\infty}\cdots\int_0^{\infty}\varphi^{\prime} \\\\ &= \frac{(-1)^{2n}}{2}\int_{-\infty}^0\cdots\int_{-\infty}^0\varphi^{\prime}+\frac{(-1)^n}{2}(-1)^n\int_{-\infty}^0\cdots\int_{-\infty}^0\varphi^{\prime}\\\\ &=\frac{1}{2}\varphi(0)+\frac{1}{2}\varphi(0)=\delta_0(\varphi) \end{align*} $$
where I’m using $\varphi^{\prime}$ to signify $(\partial/\partial x)^{(1,\cdots , 1)}\varphi$ and I use the Fundamental Theorem of Calculus in the last two lines.
Your solution looks okay. It can be written $F = \frac{1}{2^n}(\operatorname{sign}\otimes\cdots\otimes\operatorname{sign})$ with derivative $D = \operatorname{\partial}\otimes\cdots\otimes\operatorname{\partial}$ giving $$ DF = (\partial\otimes\cdots\otimes\partial)(\frac{1}{2^n}(\operatorname{sign}\otimes\cdots\otimes\operatorname{sign})) = \frac{1}{2^n}((\partial\operatorname{sign})\otimes\cdots\otimes(\partial\operatorname{sign})) \\ = \frac{1}{2^n}((2\delta)\otimes\cdots\otimes(2\delta)) = \frac{1}{2^n}2^n(\delta\otimes\cdots\otimes\delta) = \delta\otimes\cdots\otimes\delta, $$ which equals the $n$-dimensional delta distribution.