1. Notice that $\displaystyle|\langle \delta^{(k)},\varphi \rangle|=|\varphi^{(k)}(0)|\le \left|\sup\left\|\varphi^{(k)}\right\|_{L^{\infty}(\mathbb{R})}\right|\le\left|\sup_{\alpha\le k}\left\|\varphi^{(\alpha)}\right\|_{L^{\infty}(\mathbb{R})}\right|$. If its order is less than $k$, we can assume that $\displaystyle \left|\varphi^{(n)}(0)\right|\le \sum_{i=0}^{n-1}\left\|\varphi^{(i)}\right\|_{L^{\infty}(\mathbb{R})}$.
   So we can take $\displaystyle \psi\in C_{[-1,1]}^{\infty}$ satisfying that $\displaystyle \psi(0)=1$. Considering $\displaystyle \psi_{t}(x)=\psi(tx)$, we have $\displaystyle \psi_{t}^{(t)}(x)=t^{n}\psi(tx)$ and $\displaystyle \psi_{t}(0)=t^{n}$. Let $\varphi=\psi_{t}$, we have $\displaystyle \left|t^n\right|\le |P(t)|$, where $\displaystyle \deg P\le n-1$. So there is the contradiction.

2. MSE link We can use the "estimate-equivalence", that is, $\displaystyle \sup_{\alpha\le k}\left\|\varphi^{(\alpha)}\right\|_{L^{\infty}(\mathbb{R})}\simeq \sum_{i=0}^{k}\left\|\varphi^{(i)}\right\|_{L^{\infty}(\mathbb{R})}$.
   Considering that there exist $\displaystyle m,n\in \mathbb{Z}$ in such that $\displaystyle K$, which is the support of $\displaystyle \varphi$, satisfies $\displaystyle K \subset[m,n]$. Assuming that $n>0$, then we can know it's a distribution from $\displaystyle \left|\sum_{i=0}^{\infty}\varphi^{(i)}(i)\right|\le \left|\sum_{i=0}^{n}\varphi^{(i)}(i)\right|\le \sum_{i=0}^{n}\left\|\varphi^{(i)}\right\|_{L^{\infty}(\mathbb{R})}$.
   If $\displaystyle u$ has a finite order, we can take $\displaystyle K=[n-\delta,n+\delta]$ then have $\displaystyle \left|\varphi^{(n)}(n)\right|\le \sum_{i=0}^{n-1}\left\|\varphi^{(i)}\right\|_{L^{\infty}(\mathbb{R})}$. We can translate it by $\displaystyle -n$. Thus we can just consider the case of $\displaystyle \left|\varphi^{(n)}(0)\right|$. Then proof has been finished by Problem 1.

3. MSE link First, we can know the order won't be greater than 1 by using the mean value theorem: $\displaystyle \exist \xi\in(-x,x)$ in such that $2\varphi'(\xi)=\frac{\varphi(x)-\varphi(-x)}{x}$ and $\displaystyle \left|\int_{0}^{+\infty}\frac{\varphi(x)-\varphi(-x)}{x}\mathrm{d}x\right|=\left|\int_{0}^{a}\frac{\varphi(x)-\varphi(-x)}{x}\mathrm{d}x\right|\le 2a\left\|\varphi'\right\|_{L^{\infty}(\mathbb{R})}$ if assuming that the support set of $\displaystyle \varphi$ is subset of $\displaystyle [-a,a]$.
   We only need to show the order cannot be 0. Considering $\displaystyle \varphi_{t}$ has support set of $[t,4t]$ and $\displaystyle \varphi_{t}([2t,3t])\equiv1$. Now $\displaystyle \left|\left\langle \text{vp}\frac{1}{x},\varphi_{t} \right\rangle\right|\ge \frac{1}{t}\sup\left\|\varphi_{t}\right\|_{L^\infty(\mathbb R)}$. Let $\displaystyle t\to \infty$, we know the order of $\displaystyle \text{vp}\frac{1}{x}$ cannot be 0.

4. We only need to prove that $\displaystyle \lim_{\varepsilon\to 0+}\int_{-\varepsilon}^{\varepsilon}\frac{\varphi(x)}{x}\mathrm{d}x=0$, which is equivalent to $\displaystyle \lim_{\varepsilon\to0^{+}}\left|\int_{-\varepsilon}^{\varepsilon}\frac{\varphi(x)}{x}\mathrm{d}x-2\varepsilon\varphi'(0)\right|=0$.
   Considering that $\displaystyle \lim_{x\to0^{+}}\frac{\varphi(x)}{x}=\varphi'(0)$, that is, $\displaystyle \left|\frac{\varphi(x)}{x}-\varphi'(0)\right|<\varepsilon_{0}$ when $\displaystyle 0<\left|x\right|<\delta$. Assuming that $\displaystyle \varepsilon<\delta$ we have $\displaystyle \left|\int_{-\varepsilon}^{\varepsilon}\frac{\varphi(x)}{x}\mathrm{d}x -2\varepsilon\varphi'(0)\right|\le \int_{-\varepsilon}^{\varepsilon}\left|\frac{\varphi(x)}{x}-\varphi'(0)\right|\mathrm{d}x<2\varepsilon\varepsilon_{0}$. Let $\displaystyle \varepsilon\to0$, we know the limit is 0.

5. MSE link We assume that it's a distribution. First, we can take a positive test function $\displaystyle \varphi$ which is supported in $[1,2]$. Define that $\displaystyle \varphi_{j}(x)=e^{-\frac{j}{3}}\varphi(jx)$, which is supported in $\displaystyle \left[ \frac{1}{j},\frac{2}{j}\right]$, then it's obvious that $\displaystyle \varphi^{(i)}_{j}(x)=j^{i}e^{-\frac{j}{3}}\varphi(jx)$. We can consider that $\displaystyle \sup\left|\varphi^{(i)}_{j}(x)\right|=\sup\left|j^{i}e^{-\frac{j}{3}}\varphi(jx)\right|\to 0(j\to \infty)$. Now we have $\displaystyle \int_{0}^{+\infty}e^{\frac{1}{x}}\varphi_{j}(x)\mathrm{d}x=\int_{\frac{1}{j}}^{\frac{2}{j}}e^{\frac{1}{x}}e^{-\frac{j}{3}}\varphi(jx)\mathrm{d}x=\int_{1}^{2}e^{\frac{j}{x}-\frac{j}{3}-\ln j}\varphi(x)\mathrm{d}x\ge \int_{1}^{2}\varphi(x)\mathrm{d}x=\left\|\varphi\right\|_{L^{1}}.(j\ge 100)$ Let $j\to \infty$, we get $\displaystyle \int_{0}^{+\infty}e^{\frac{1}{x}}\varphi_{j}(x)\mathrm{d}x\to0$ because of the definition of distribution. So $\displaystyle \left\|\varphi\right\|_{L^{1}}=0$, that is, $\varphi\equiv0$. It's a contradiction.

6. Consider $\displaystyle \left\langle f_{\varepsilon},\varphi \right\rangle=\int_{\mathbb{R}\backslash[-\varepsilon,2\varepsilon]}\frac{\varphi(x)-\varphi(0)}{x-0}\mathrm{d}x-\int_{\varepsilon}^{2\varepsilon}\frac{\varphi(0)}{x}\mathrm{d}x$. Notice that the first term is bounded by using mean value theorem. We only need to focus $\displaystyle \int_{\varepsilon}^{2\varepsilon}\frac{\varphi(0)}{x}\mathrm{d}x=\varphi(0)\ln2$. So the limit of distribution $\displaystyle f_{\varepsilon}$ exists.

7. Same to Problem 6, we just need to verify that $\displaystyle \int_{\varepsilon^{2}}^{\varepsilon}\frac{\varphi(0)}{x}\mathrm{d}x$ is divergent. The integral equals to $\displaystyle -\varphi(0)\ln\varepsilon$ and we know it's divergent by taking $\displaystyle \varepsilon\to 0^{+}$.

8. We only prove that $\displaystyle f'\mathrm{d}x=\mathrm{d}\mu$. If so, we have $\displaystyle \int f'\varphi\mathrm{d}x=\int\varphi\mathrm{d}\mu$.
   Because of Lebesgue measure is absolutely continuous to measure $\displaystyle \mu$. So we have unique function $\displaystyle h$ satisfying $\displaystyle h\mathrm{d}x=\mathrm{d}\mu$ by Radon-Nikodym Theorem. And we know $\displaystyle \int_{(-\infty,a)} \mathrm{d}\mu=\mu((-\infty,a))=\int_{(-\infty,a)} h\mathrm{d}x$. So $\displaystyle h=f'$.

9. By using the corollary of Riesz Representation Theorem for monotonic function, we know that: there exists an unique regular Borel measure $\mu$ satisfying that : $$\mu((a,b])=f(b)-f(a),\forall-\infty<a<b<\infty$$
   $$\mu((-\infty,b])=f(b)-f(-\infty),\forall-\infty<b<\infty$$
   $$\mu((a,\infty))=f(\infty)-f(a),\forall -\infty\le a<\infty$$
   So we proved it by Problem 8.

10. We have $\displaystyle \left\langle (x_{+}^{\alpha})',\varphi\right\rangle=-\left\langle x_{+}^{\alpha},\varphi' \right\rangle =-\int_{0}^{+\infty}x^{\alpha}\varphi'(x)\mathrm{d}x=-\int_{0}^{+\infty}x^{\alpha}\mathrm{d}\varphi(x)=\int_{0}^{+\infty}\alpha x^{\alpha-1}\varphi(x)\mathrm{d}x=\left\langle \alpha x_{+}^{\alpha-1},\varphi \right\rangle$. So as distributions, $\displaystyle (x^{\alpha}_{+})'$ equals to $\alpha x_{+}^{\alpha-1}$.

11. We have $\displaystyle \left\langle (x_{+}^{\alpha})^{(\alpha)},\varphi \right\rangle=(-1)^{\alpha}\left\langle x_{+}^{\alpha},\varphi^{(\alpha)} \right\rangle=(-1)^{\alpha}\int_{0}^{+\infty}x^{\alpha}\varphi^{(\alpha)}(x)\mathrm{d}x\xlongequal{\text{Use }\alpha\text{ times integration by parts}}\int_{0}^{+\infty}\alpha!\varphi(x)\mathrm{d}x=\alpha!\int_{\mathbb{R}}\chi_{x\ge0}\varphi(x)\mathrm{d}x=\left\langle \alpha!H(x),\varphi \right\rangle$. So as distributions, we know $\displaystyle (x_{+}^{\alpha})^{(\alpha)}=\alpha!H(x)$.

12. By using integration by parts, we have $\displaystyle \int_{\varepsilon}^{\infty}x^{\alpha}\varphi(x)\mathrm{d}x=\frac{1}{\alpha+1}\int_{\varepsilon}^{\infty}\varphi(x)\mathrm{d}x^{\alpha+1}=-\frac{\varepsilon^{\alpha+1}\varphi(\varepsilon)}{\alpha+1}-\frac{1}{\alpha+1}\int_{\varepsilon}^{\infty}x^{\alpha+1}\varphi'(x)\mathrm{d}x$. Thus we can define the distribution according to the problem.

13. It's obvious that the support set of $\displaystyle u$ is $\displaystyle \left\{0\right\}$. So we have $\displaystyle u=\sum_{k=0}^{n}C_{k}\delta_{0}^{(k)}$. Notice that, $\displaystyle \left\langle x\delta_{0},\varphi \right\rangle=0\cdot\varphi(0)=0$, $\displaystyle \left\langle x\delta_{0}',\varphi \right\rangle=-\left\langle \delta_{0},\varphi+x\varphi' \right\rangle=-\varphi(0)$. Further, for any $\displaystyle k\ge1$, $\displaystyle \left\langle x\delta_{0}^{(k)},\varphi \right\rangle\not\equiv0$. Thus we have $\displaystyle u=C\delta_{0}$.

14. Notice that the support of $u$ is $\{0\}$. Then $\displaystyle u=\sum_{k=0}^{n}C_{k}\delta_{0}^{(k)}$. By Problem 13, we know that $u=-\delta_{0}'$ is a particular solution. And the solutions for equation $x\cdot u=0$ are $\displaystyle u=C\delta_{0}$. So the solutions for $\displaystyle u\cdot x=\delta_{0}$ are $\displaystyle C\delta_{0}-\delta_{0}'$.

15. Take $\displaystyle \rho\in C_{c}^{\infty}(\mathbb{R})$ satisfying $\displaystyle \rho(0)=1$. Consider $\displaystyle \left\langle \tilde u,\varphi\right\rangle=\left\langle \tilde u,\varphi-\varphi(0)\rho \right\rangle+\varphi(0)\left\langle \tilde u,\rho \right\rangle\xlongequal{C=\left\langle \tilde u,\rho \right\rangle}\left\langle \tilde u,\varphi-\varphi(0)\rho \right\rangle+\left\langle C\delta_{0},\varphi \right\rangle=\left\langle u,\frac{\varphi-\varphi(0)\rho}{x} \right\rangle+\left\langle C\delta_{0},\varphi \right\rangle$. Notice that $x\cdot\delta_0=0$, so $\displaystyle \left\langle \tilde u,\varphi \right\rangle=\left\langle u,\frac{\varphi-\varphi(0)\rho}{x} \right\rangle$. Now we have the definition of $\displaystyle \tilde u$.

16. We just need to find a particular solution for the equation. Notice that $\displaystyle \text{pv}\frac{1}{x}$ is a solution. So $\displaystyle u=\text{pv}\frac{1}{x}+C\delta_{0}$.

17. First, we can notice that there exists a solution $\displaystyle u=\text{pv}\frac{1}{x^{2}}$. So we have $\displaystyle u=C\delta_{0}+\text{pv}\frac{1}{x}$.

18. We can find a solution $\displaystyle u=x_{+}^{\alpha-1}$ so all solutions are $\displaystyle x_{+}^{\alpha-1}+C\delta_{0}$.

19. MSE link 1, MSE link 2 We have $\displaystyle 0=u+xu'=(xu)'$, so $\displaystyle 0=\left\langle (xu)',\varphi \right\rangle=-\left\langle xu,\varphi' \right\rangle$. Notice that, $\displaystyle \left\langle v,\varphi' \right\rangle=0$ for all $\displaystyle \varphi\in \mathcal{D}$ means $\displaystyle \left\langle v,\phi \right\rangle=0$ for all $\displaystyle \phi\in \mathcal{D}$ satisfying $\displaystyle \int\phi=0$. So it's equivalent to $xu=k$. Then we fix $\displaystyle \psi_{0}\in \mathcal{D}$ with $\displaystyle \int\psi_{0}=1$. Let $\psi\in \mathcal{D},\displaystyle \alpha=\int \psi,c=\left\langle v,\psi_{0} \right\rangle$, now $\displaystyle \int\alpha\psi_{0}-\psi=0$. So $\displaystyle \left\langle v,\alpha\psi_{0}-\psi \right\rangle=0$, that is, $\displaystyle \left\langle v,\psi \right\rangle=\left\langle v,\alpha\psi_{0} \right\rangle=c\int\psi$.
    Then we know $\displaystyle v=c$. Back to this problem, we know $\displaystyle xu=k$. Therefore we know $\displaystyle u=C\delta_{0}+k\text{vp}\frac{1}{x}$ by Problem 13,17.

20. We can reduce the formula first: $\displaystyle \left\langle u_{\lambda},\varphi \right\rangle=\left\langle \lambda^{-d}u,\varphi \right\rangle$ equals to $\displaystyle \left\langle u,\lambda \varphi(\lambda x) \right\rangle$. Thus we have $\displaystyle \int_{0}^{+\infty}\lambda^{-d}\frac{\varphi(x)-\varphi(-x)}{x}\mathrm{d}x=\int_0 ^{+\infty}\lambda\frac{\varphi(\lambda x)-\varphi(-\lambda x)}{x}\mathrm{d}x\xlongequal{t=\lambda x}\int_{0}^{+\infty}\lambda\frac{\varphi(t)-\varphi(-t)}{t}\mathrm{d}t$ for $\displaystyle u=\text{vp}\frac{1}x$. Then we know $\displaystyle \text{vp}\frac{1}{x}$ is a homogeneous distribution and its degree is $\displaystyle -1$. And we have $\displaystyle \int_{0}^{+\infty}\lambda^{-d}x^{\alpha}\varphi(x)\mathrm{d}x=\int_{0}^{+\infty}\lambda x^{\alpha}\varphi(\lambda x)\mathrm{d}x\xlongequal{t=\lambda x}\int_{0}^{+\infty}\lambda^{-\alpha}t^{\alpha}\varphi(t)\mathrm{d}t$. So we know it's a homogeneous distribution and its degree is $\displaystyle \alpha$.

21. Similar to Problem 20, we have $\displaystyle \lambda^{-d}\varphi(0)=\lambda\varphi(0)$. Thus it's a homogeneous distribution and its degree is $\displaystyle -1$.

22. First, we have $\displaystyle \left\langle \lambda^{-d}u,\varphi \right\rangle = \left\langle u,\lambda^{n} \varphi(\lambda x) \right\rangle$. So $\displaystyle \left\langle \lambda^{-d+1}\partial_{x_{1}}u,\varphi \right\rangle=\left\langle \lambda^{-d+1}u,-\partial_{x_{1}}\varphi \right\rangle=\left\langle \lambda u, -\lambda^{n}(\partial_{x_{1}}\varphi)(\lambda x)\right\rangle=\left\langle u,-\lambda^{n}\partial_{x_{1}}(\varphi(\lambda x)) \right\rangle=\left\langle \partial_{x_{1}}u,\lambda^{n}\varphi(\lambda x) \right\rangle$. Then we know its degree is $\displaystyle d-1$.

23. If they're $\mathbb{C}$-linear independence, we have $\displaystyle 0=\left\langle a_{1}u_{1},\lambda^{n}\varphi(\lambda x) \right\rangle+\dots+\left\langle a_{N}u_{N},\lambda^{n}\varphi(\lambda n) \right\rangle=a_{1}\lambda^{-d_{1}}\left\langle u_{1},\varphi \right\rangle+\dots+a_{N}\lambda^{-d_{N}}\left\langle u_{N},\varphi \right\rangle$, $\displaystyle a_{i}\in \mathbb{C}$ and we assume $d_{1}<\dots<d_{N}$ are degrees of $\displaystyle u_{1},\dots,u_{N}$. Multiply by $\displaystyle \lambda^{d_{1}}$ and let $\displaystyle \lambda\to \infty$, we will have $\displaystyle a_{1}\left\langle u_{1},\varphi \right\rangle=0$. But $u_{1}\neq 0$, then we know $\displaystyle a_{1}=0$. Similarly, we know $\displaystyle a_{i}=0,\forall i$. That causes the contradiction.

24. Let $[-m,m]^{n}$ be the supset of the support set of $\displaystyle \varphi$. Now consider $\displaystyle \psi_{\lambda}(x)=\varphi'(\xi),\xi_{i}\in (\lambda_{0}x_{i},\lambda x_{i})$ or $\displaystyle (\lambda x_{i},\lambda_{0}x_{i}) ,\forall i$. We only need to show $\displaystyle \xi_{i}\to \lambda_{0}x_{i} , \forall x_{i}\in[-m,m]$ because of the continuity of $\displaystyle \varphi'$. We will show $\displaystyle \lambda x_i\to \lambda_{0} x_{i}$ when $\displaystyle \lambda\to\lambda_{0}$. $\displaystyle \forall \varepsilon>0,\ \exist \delta=\frac{\varepsilon}{m}$, if $\displaystyle \left|\lambda -\lambda_{0}\right|<\delta$, then $\displaystyle \left|\lambda x-\lambda_{0}x\right|<\varepsilon$. So $\displaystyle \lim_{\lambda\to\lambda_{0}}\psi_{\lambda}(x)=\varphi'(\lambda_{0} x)$.

25. Consider that $\displaystyle u$ being homogeneous distribution with degree of $\displaystyle d$ is equivalent to $\displaystyle \lambda^{-(n+d)}\left\langle u,\varphi \right\rangle=\left\langle u,\varphi(\lambda x) \right\rangle$. Then we take the derivative of $\displaystyle \lambda$, we will get $\displaystyle -(n+d)\lambda^{-(n+d+1)}\left\langle u,\varphi \right\rangle=\left\langle u,\sum_{k=1}^{n}x_{k}\cdot\frac{\partial \varphi}{\partial x_{k}}(\lambda x) \right\rangle=\sum_{k=1}^{n}\left\langle x_{k}u,\frac{\partial\varphi}{\partial x_{k}}(\lambda x) \right\rangle=-\sum_{k=1}^{n}\left\langle \frac{\partial (x_{k}u)}{\partial x_{k}},\varphi(\lambda x) \right\rangle =-\sum_{k=1}^{n}\left\langle u+x_{k}\cdot\frac{\partial u}{\partial x_{k}},\varphi(\lambda x) \right\rangle=-n\left\langle u,\varphi(\lambda x) \right\rangle-\sum_{k=1}^{n}\left\langle \frac{\partial u}{\partial x_{k}},\varphi(\lambda x) \right\rangle$. Now let $\displaystyle \lambda=1$, so we get $\displaystyle \left\langle d\cdot u-\sum_{k=1}^{n}x_{k}\cdot\frac{\partial u}{\partial x_{k}},\varphi \right\rangle=0$ for any $\displaystyle \varphi\in \mathcal{D}(\mathbb{R}^{n})$. Thus $\displaystyle \sum_{k=1}^{n}x_{k}\cdot\frac{\partial u}{\partial x_{k}}=d\cdot u$.

26. For the distribution with degree of $\displaystyle 0$, we have $\displaystyle x\cdot u'=0$ by Problem 25. And we know $\displaystyle u'=c\delta_{0}$ by Problem 13. Because of $\displaystyle H'(x)=\delta_{0}$, it has a particular solution $\displaystyle u=cH(x)$. Then consider the general solution for $\displaystyle u'=0$, which is $\displaystyle u=k$ by Problem 19. So the solutions are $\displaystyle u=cH(x)+k$.
    For the distribution with degree of $\displaystyle 1$, we have $\displaystyle x\cdot u'=u$ by Problem 25. Then we know $\displaystyle x\cdot u''=0$. So we know $\displaystyle u'=cH(x)+k$ by the degree of $0$. And we know $\displaystyle u=0$ when $x=0$. Thus we know $\displaystyle u=\begin{cases}ax,x>0\\bx,x<0\end{cases}$.