1. MSE link We have $\displaystyle\left\langle \frac{\varepsilon}{\pi(x^{2}+\varepsilon^{2})},\varphi \right\rangle=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{\varepsilon}{x^{2}+\varepsilon^{2}}\varphi(x)\mathrm{d}x\xlongequal{x=\varepsilon y}\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{\varphi(\varepsilon y)}{1+y^{2}}\mathrm{d}y$. Now let $\displaystyle \varepsilon\to0$ and use Dominated Convergence Theorem, then we have $\displaystyle \lim_{\delta\to 0}\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{\varphi(\varepsilon y)}{1+y^{2}}\mathrm{d}y=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{\varphi(0)}{1+y^{2}}\mathrm{d}y=\varphi(0)$, which equals to $\left\langle \delta_{0},\varphi \right\rangle$.

2. Considering $\displaystyle \left\langle\frac{1}{x+i\varepsilon},\varphi\right\rangle=\int_{\mathbb{R}}\frac{\varphi(x)\mathrm{d}x}{x+i\varepsilon}=\int_{\mathbb{R}}\frac{x\varphi(x)-i\varepsilon\varphi(x)}{x^{2}+\varepsilon^{2}}\mathrm{d}x=\left\langle \frac{x}{x^{2}+\varepsilon^{2}},\varphi \right\rangle-\langle i\pi\delta_{0},\varphi\rangle$ by Problem 1, we can use Dominated Convergence Theorem so that we have $\displaystyle \left\langle \lim_{\varepsilon\to0^+} \frac{1}{x+i\varepsilon},\varphi \right\rangle=\lim_{\varepsilon\to0^{+}}\left\langle \frac{1}{x+i\varepsilon},\varphi \right\rangle=\left\langle \text{vp}\frac{1}{x}-i\pi\delta_{0},\varphi \right\rangle$. Now we know $\displaystyle \lim_{\varepsilon\to0^{+}}\frac{1}{x+i\varepsilon}=\text{vp}\frac{1}{x}-i\pi\delta_{0}$.

3. MSE link Let $\displaystyle \varphi(x)=\varphi(0)\psi(x)+x\theta(x)$ where $\displaystyle \psi$ satisfying $\displaystyle \psi(x)=1$ when $\displaystyle x\in[-1,1]$ and $\displaystyle \text{supp}\psi\subset[-2,2]$. So we have $\displaystyle \left\langle \frac{\sin nx}{x},\varphi \right\rangle=\varphi(0)\int_{-1}^{1}\frac{\sin nx}{x}\mathrm{d}x+\left(\int_{-2}^{-1}+\int_{1}^2\right)\frac{\psi(x)}{x}\sin nx\mathrm{d}x+\int_{\text{supp}\theta}\theta(x)\sin nx\mathrm{d}x$. Then $\displaystyle \int_{-1}^{1}\frac{\sin nx}{x}\mathrm{d}x=\int_{-n}^{n}\frac{\sin x}{x}\mathrm{d}x\xlongequal{n\to+\infty}\pi$ and we know the other terms vanish when $\displaystyle n\to +\infty$ by Riemann's Lemma. Now we know $\displaystyle \left\langle \lim_{n\to +\infty}\frac{\sin nx}{x},\varphi \right\rangle=\pi\varphi(0)=\left\langle \pi\delta_0,\varphi \right\rangle$. That is, $\displaystyle \lim_{n\to +\infty}\frac{\sin nx}{x}=\pi\delta_0$.

4. We know $\displaystyle \left\langle \triangle E,\varphi \right\rangle=\left\langle E,\triangle \varphi \right\rangle=\frac{1}{(2-n)|\mathbf{S}^{n-1}|}\int_{\mathbb{R}^{n}}\frac{1}{|x|^{n-2}}\triangle \varphi(x)\mathrm{d}x$. Then notice that $\displaystyle \triangle\frac{1}{(2-n)|x|^{n-2}}=\sum_{i=1}^{n}\partial_{i}\left(\left(\sum_{i=1}^{n}x_{i}^{2}\right)^{-\frac{n}{2}}x_{i}\right)=\sum_{i=1}^{n}\left(-n\left(\sum_{i=1}^{n}x_{i}^{2}\right)^{-\frac{n+2}{2}}x_i^{2}+\left(\sum_{i=1}^{n}x_{i}^{2}\right)^{-\frac{n}{2}}\right)=0$. So we have $\displaystyle \left\langle \triangle E,\varphi \right\rangle=\frac{1}{(2-n)|\mathbf{S}^{n-1}|}\int_{\mathbb{R}^{n}}\frac{1}{|x|^{n-2}}\triangle \varphi(x)-\varphi(x)\triangle \frac{1}{|x|^{n-2}}\mathrm{d}x=\lim_{\varepsilon\to0}\frac{1}{(2-n)|\mathbf{S}^{n-1}|}\int_{|x|\ge\varepsilon}\frac{1}{|x|^{n-2}}\triangle \varphi(x)-\varphi(x)\triangle \frac{1}{|x|^{n-2}}\mathrm{d}x\xlongequal{\text{Green's formula}}-\lim_{\varepsilon\to0}\frac{1}{(2-n)|\mathbf{S}^{n-1}|}\int_{|x|=\varepsilon}\frac{1}{|x|^{n-2}}\partial_{r}\varphi(x)-\varphi(x)\partial_{r} \frac{1}{|x|^{n-2}}\mathrm{d}\sigma$. Consider that $\displaystyle 0\le\left|\int_{|x|=\varepsilon}\frac{1}{|x|^{n-2}}\partial_r\varphi(x)\mathrm{d}\sigma\right|\le \frac{\sup|\partial_r\varphi(x)|}{\varepsilon^{n-2}}\int_{|x|=\varepsilon}\mathrm{d}\sigma=\varepsilon|\mathbf{S}^{n-1}|\sup|\partial_{r}\varphi(x)|\xrightarrow{\varepsilon\to 0}0$ and $\displaystyle \lim_{\varepsilon\to 0}\frac{1}{(2-n)|\mathbf{S^{n-1}}|}\int_{|x|=\varepsilon}\varphi(x)\partial_{r}\frac{1}{|x|^{n-2}}\mathrm{d}\sigma=\lim_{\varepsilon\to0}\frac{1}{|\mathbf{S^{n-1}}|}\int_{|x|=\varepsilon}\frac{\varphi(x)}{|x|^{n-1}}\mathrm{d}\sigma=\varphi(0)$, thus we have $\displaystyle \left\langle \triangle E,\varphi \right\rangle=\varphi(0)$, which means $\displaystyle \triangle E=\delta_0$.

5. We know one solution to $\displaystyle \triangle u=f$ is $\displaystyle u_{0}=E*f$, which implies $\displaystyle u_{0}\in C^{\infty}(\mathbb{R}^{n})$ because $\displaystyle E\in \mathcal{D}'(\mathbb{R}^{n})$ and $\displaystyle f\in \mathcal{D}(\mathbb{R}^{n})$. And notice that $\displaystyle \triangle v=0$ means $\displaystyle v\in \mathcal{D}(\mathbb{R}^{n})$, we know the solution is $\displaystyle u=u_{0}+cv\in C^{\infty}(\mathbb{R}^{n})$.

6. When $\displaystyle n=3$, $\displaystyle E(x)=-\frac{1}{4\pi|x|}$. First, we can consider that $\displaystyle E*f=\left\langle f(y),E(x-y) \right\rangle$. If $\displaystyle f$ is a singular distribution, we can find a function cluster $\displaystyle \{f_{\varepsilon}\}\in C_{0}^{\infty}(\mathbb{R}^{3})$ satisfying $\displaystyle f_{\varepsilon}\xrightarrow[\varepsilon\to0]{\mathcal{D}'}f$. Therefore we only need to prove the case of $\displaystyle f$ is regular, that is, $\displaystyle\left|\int_{\mathbb R^3}\frac{f(y)dy}{|x-y|}\right|\le \frac{C_f}{|x|}$.
   Considering the support set of $f$ and assuming it's $\{a:|a|<R\}$. Now when $|x|\ge2R$, we have $\displaystyle\left|\int_{\mathbb R^3}\frac{f(y)dy}{|x-y|}\right|=\left|\int_{|y|\le R}\frac{f(y)dy}{|x-y|}\right|\le \left|\int_{|y|\le R}\frac{f(y)dy}{\left|\frac x 2\right|}\right| \le\frac{\frac83\pi R^3f_{\max}}{|x|}$. So the inequality holds.

7. Consider that $\displaystyle \left\langle \triangle (E*f),\varphi \right\rangle=\left\langle (\triangle E)*f,\varphi \right\rangle=\left\langle \delta_{0}*f,\varphi \right\rangle=\left\langle f,\varphi \right\rangle$, so we know $\displaystyle u=E*f$ is one solution. Then we need to prove that $\displaystyle \left|\left\langle f(y),E(x-y) \right\rangle+\frac{\left\langle f,1 \right\rangle}{4\pi|x|}\right|<\frac{C_{2}}{|x|^{2}}$ when $\displaystyle |x|\to \infty$. It's equivalent to $\displaystyle \left|\int_{\mathbb{R}^{3}}\frac{f(y)}{4\pi|x-y|}-\frac{f(y)}{4\pi|x|}\mathrm{d}y\right|<\frac{C_{2}}{|x|^{2}}$.
   Considering the support set of $f$ and assuming it's $\{a:|a|<R\}$. Now when $|x|\ge2R$, we have $\displaystyle \left|\int_{\mathbb{R}^{3}}\frac{f(y)}{4\pi|x-y|}-\frac{f(y)}{4\pi|x|}\mathrm{d}y\right|=\left|\int_{|y|<R}\frac{(|x|-|x-y|)f(y)}{4\pi|x-y||x|}\mathrm{d}y\right|\le \left|\frac{1}{4\pi}\int_{|y|<R}\frac{|y|f(y)}{\frac{|x|^{2}}{2}}\mathrm{d}y\right|<\frac{R}{2\pi}\left|\int_{|y|<R}\frac{f(y)}{|x|^{2}}\mathrm{d}y\right|<\frac{2R^{4}f_{\max}}{3|x|^{2}}$. So the inequality holds.

8. MSE link We have $\displaystyle \triangle \sqrt{u^{2}+\varepsilon^{2}}=\nabla\cdot\left(\frac{u}{\sqrt{u^{2}+\varepsilon^{2}}}\nabla u\right)=\nabla\left(\frac{u}{\sqrt{u^{2}+\varepsilon^{2}}}\right)\cdot\nabla u+\frac{u}{\sqrt{u^{2}+\varepsilon^{2}}}\triangle u=\frac{u\triangle u}{\sqrt{u^{2}+\varepsilon^{2}}}+\frac{\varepsilon^{2}|\nabla u|^{2}}{(\sqrt{u^{2}+\varepsilon^{2}})^{3}}\overset{\mathcal{D}'}{\ge}\frac{u\triangle u}{\sqrt{u^{2}+\varepsilon^{2}}}$. Now take the limit for $\displaystyle \varepsilon\to0$ and we have $\displaystyle \triangle|u|\overset{\mathcal{D}'}{\ge}\text{sign}(u)\cdot\triangle u$.