Weak convergence and distribution convergence I'm a bit confuse with weak convergence and distribution convergence :
1) $f_n\to f$ weakly in $L^p(\mathbb R^n)$ if for all $\varphi \in L^q$ (the conjugate exponent of $p$),
$$\lim_{n\to \infty }\int_{\mathbb R^n} f_n \varphi =\int_{\mathbb R^n}f\varphi .$$
2) $f_n\to f$ in distribution if for all $\varphi \in \mathcal C_0^\infty (\mathbb R^n)$, $$\lim_{n\to \infty }\int_{\mathbb R^n}f_n \varphi =\int_{\mathbb R^n}f\varphi .$$
Q1) I can see that $1)\implies 2)$, Is the converse true ? In what 2) is more interesting than 1) ? 
Q2) Now, if $\varphi _n$ is a family of measure s.t. $\varphi _n(A)\to \varphi(A) $ for all measurable $A$. What is the convergence : for all bounded and continuous $f$, $$\lim_{n\to \infty }\int_{\mathbb R^n}fd\varphi _n=\int_{\mathbb R^n}fd\varphi \ \ ?$$
I have the impression that in probability it's calls "convergence in distribution". But this one looks strange and very different than the previous one.
What is the idea behind this third "strange" convergence ?
 A: Q1) The implication $(2)\Rightarrow (1)$ does not hold. Fix $\alpha<0$ and take for instance
$$f_n(x)=n^{\alpha}\chi_{[-n,n]}(x) $$
Then for each $\varphi \in C_0^{\infty}(\mathbb{R})$ there is $m$ such that $\varphi(x)=0$ for all $|x|> m$. Therefore, for all $n\geq m$,
$$\int_{\mathbb{R}}f_n\varphi=\int_{-m}^{m}f_n\varphi=n^{\alpha}\int_{-m}^{m}\varphi\to 0 $$
 Hence $f_n\to 0$ in the sense of distributions. However, if we fix $\beta<0$ and take $$\varphi(x)=\begin{cases}1 & |x|\leq 1 \\ 
|x|^{\beta} & |x|>1\end{cases}$$
then $\varphi \in L^q(\mathbb{R})$ whenever $q\beta<-1$, and
$$\int_{\mathbb{R}}f_n\varphi =2n^{\alpha}\int_{0}^{n}\varphi(x)\,dx=2n^{\alpha}+n^{\alpha}\int_1^nx^{\beta}\,dx=2n^{\alpha}+n^{\alpha}\left(\frac{n^{\beta+1}-1}{\beta+1}\right)$$
Here $2n^{\alpha}\to 0$, but 
$n^{\alpha}(n^{\beta+1}-1)\to +\infty$ whenever $\alpha+\beta+1>0$. For instance, we might choose $p=2$, $\beta =-2/3$ (so that $q\cdot \beta=-4/3<-1$) and $\alpha>-\frac{1}{3}$. So, if the parameters $\alpha$ and $\beta$ satisfy this relation, weak convergence does not hold.
Q2) Every finite Borel measure $\varphi$ on $\mathbb{R}^n$ defines a continuous linear functional $\hat{\varphi}$ on $C_b(\mathbb{R}^n)$ by 
$$ \hat{\varphi}(f)=\int_{\mathbb{R}^n}f\,d\varphi$$
Indeed we have
$$|\hat{\varphi}(f)|\leq \varphi(\mathbb{R}^n)\|f\|_{\infty} $$
so $\hat{\varphi}$ is bounded. Now $\varphi_n\to \varphi$ in the convergence you described means that $\hat{\varphi_n}(f)\to \hat{\varphi}(f)$ for all $f\in C_b(\mathbb{R})$. Therefore if we call $\mathcal{M}(\mathbb{R}^n)$ the space of finite Borel measures on $\mathbb{R}^n$ we have that $\mathcal{M}(\mathbb{R}^n)$ may be identified as a subspace of the dual space of $C_b(\mathbb{R}^n)$. This makes it possible to define the weak-star topology on $\mathcal{M}(\mathbb{R}^n)$. In this sense, the convergence you described is precisely the weak-star convergence in $\mathcal{M}(\mathbb{R}^n)$. (recall that if $X$ is a normed space, we have that $\varphi_n$ converges weakly-star to $\varphi$ in $X^*$ when $\varphi_n(f)\to \varphi(f)$ for all $f\in X$. Here $X=C_b(\mathbb{R}^n)$).
