Folland real analysis 9.11 This comes from question 9.11 of Folland's Real analysis textbook. Unfortunately, I have no idea to how to start with this question. So can some one help me with part $a$?
For part $a$, I can not come up with any theorems which relate the distribution to the derivative of the test function, except the definition of derivative on distribution, which does not seems to be useful in solving this question.
BTW, the support is defined to be the complement of the maximal open subset of $R^n$ on which $F$ is $0$.

Update: I can solve part $a, b, c$ now. Can someone point out how to deal with part $d$?   
I think I forgot an important fact that $<\delta,\phi>= \phi(0)$ in proving part d, and that is how the question get the $\delta$ function...
 A: Since nobody answer this question formally, I post my solutions here. Any comments are welcomed.
(a). Choose function $\psi$ such that $\psi(x)=1$ for $|x|\le \frac{1}{2}$ and $\psi(x)=0$ for $|x|\ge 1$, then $$|\langle F,\phi\rangle|=|\langle F,\psi\phi\rangle+\langle F,(1-\psi)\phi\rangle|=|\langle F,\psi\phi\rangle|,$$ since $supp(F)=\{0\}$. And $$ |\langle F,\psi\phi\rangle|\le C\sum_{|\alpha|\le N}\sup_{|x|\le 1}|\partial^{\alpha}(\psi\phi)(x)|\le C^{'}\sum_{|\alpha|\le N}\sup_{|x|\le 1}|\partial^{\alpha}\phi(x)|$$ for some constants $C,C^{'},N$, by Proposition 5.15 and the product rule for derivatives.
(b). Obviously, $\phi_k(x)=\phi(x)$ when $x\ge\frac{2}{k}$. When $x\le\frac{2}{k}$, we can calcualte by the hint and the product rule for derivatives as follows, \begin{eqnarray*}
\partial^{\alpha}(\phi_k-\phi)&=&\sum_{\beta+\gamma=\alpha}A_{\beta,\gamma}k^{|\gamma|}\partial^{\beta}\phi(x)\partial^{\gamma}\psi(kx)\\
&\le&\sum_{\beta+\gamma=\alpha}B_{\beta,\gamma}k^{|\gamma|}|x|^{N+1-|\beta|}\partial^{\gamma}\psi(kx)\\
&\le&\sum_{\beta+\gamma=\alpha}C_{\beta,\gamma}k^{|\alpha|-(N+1)}\\
&\le&k^{-1}\sum_{\beta+\gamma=\alpha}D_{\beta,\gamma}\\
\end{eqnarray*}
where the $A_{\beta,\gamma},B_{\beta,\gamma},C_{\beta,\gamma},D_{\beta,\gamma}$ are some constants. The results follows.
(c). By (a),  \begin{eqnarray*}
|\langle F,\phi\rangle|&=&|\langle F,\phi-\phi_k\rangle|\\
&\le&C\sum_{|\alpha|\le N}\sup_{|x|\le 1}|\partial^{\alpha}(\phi-\phi_k)(x)|\rightarrow0\quad(k\rightarrow+\infty).
\end{eqnarray*}
(d).(I AM NOT SURE ABOUT THIS)
It would be easy to construct functions $\{f_{\alpha}\}_{|\alpha|\le N}$ such that $$\partial^{\alpha}f_{\beta}(0)=\delta_{\alpha\beta},\quad|\alpha|,|\beta|\le N $$
For example, the function $h=\int_0^x e^{-1/|t|^2}dt$ satisfy $h(0)=0, h^{'}(0)=1, h^{(k)}(0)=0$ for $k\ge 2$.
Since $\partial^{\beta}(\phi-\sum_{|\alpha|\le N}\partial^{\alpha}\phi(0)f_{\alpha})=0$ when $|\beta|\le N $ ,by (c) we have $$0=\langle F,\phi-\sum_{|\alpha|\le N}\partial^{\alpha}\phi(0)f_{\alpha}\rangle=\langle F,\phi\rangle-\sum_{|\alpha|\le N}\langle F, f_{\alpha}\rangle\partial^{\alpha}\delta(\phi)$$
Thus, $\langle F,\phi\rangle=\sum_{|\alpha|\le N}\langle F, f_{\alpha}\rangle\partial^{\alpha}\delta(\phi).$ Define $c_{\alpha}=\langle F, f_{\alpha}\rangle$ and we complete the proof.
