$\left( \sum_{i=1}^N a_i^2\right)^{1/2} \leq \left(\mathbb{E} \left| \sum_{i=1}^N a_iX_i\right|^p \right)^{1/p}$ for all $p \geq2$? Let  $X_1,X_2, \dots, X_N $ be i.i.d sub-guaussian random variable ,with $E[X_i] = 0$ and $\operatorname{Var}[X_i] =1$  $a_1,a_2,\dots ,a_N \in R$
$$\left(\sum_{i=1}^N a_i^2\right)^{1/2} \leq \left( E \left| \sum_{i=1}^N a_iX_i\right|^p\right)^{1/p}$$
for all  $p \geq2$ ? 
is that possible to do that? 
 A: Yes, it is possible. Actually, an upper bound holds as well -- namely, this is the Khintchine inequality for subgaussian random variables. See e.g. these lecture notes.
The lower bound you seek is the simple part, and id actually a simple consequence of Jensen's inequality:
$$\begin{align*}
\mathbb{E}\left[\left\lvert\sum_i a_i X_i\right\rvert^p\right]^{1/p}
&= \mathbb{E}\left[\left(\left|\sum_i a_i X_i\right|^2\right)^{p/2}\right]^{1/p}
\geq \left(\mathbb{E}\left[\left(\sum_i a_i X_i\right)^2\right]\right)^{1/2}
= \left(\operatorname{Var}\left[\sum_i a_i X_i\right]\right)^{1/2}\\
&= \left(\sum_i a_i^2 \operatorname{Var}\left[X_i\right]\right)^{1/2}
= \left(\sum_i a_i^2 \right)^{1/2}
\end{align*}$$
using the fact that $p\geq 2$ (i.e., $p/2\geq 1$) for Jensen's inequality, then the fact that $\mathbb{E}\sum_i a_i X_i = 0$ to rewrite the expectation of the square as the variance, and finally independence of the $X_i$'s.
A: Consider a Random Variable $$Z=\sum_{i=1}^{N} a_iX_i$$
Then $Z$ is Gaussian with  mean $E[Z]=0$ and variance $Var[Z] = \sigma^2=\sum_{i=1}^{N} a_i^2 $
$E[|Z|^p] = \sigma^p[(p-1)!!]\sqrt{\dfrac{2}{\pi}}$ if p is odd, and
$E[|Z|^p] = \sigma^p[(p-1)!!]$ if n is even. https://en.wikipedia.org/wiki/Normal_distribution#Moments
$\implies E[|Z|^p] = \sigma^p k$, let $k$ be a constant depending on whether n is even or odd. But note that $|k|\geq1$ for $p \geq 2$.
$\therefore$ $(E[|Z|^p])^{1/p} = (\sigma^p k)^{1/p} \geq \sigma$. But $\sigma = (\sum_{i=1}^{N}a_i^2)^{1/2}$
$\implies$ $(\sum_{i=1}^{N}a_i^2)^{1/2} \leq (E[|Z|^p])^{1/p}$
$$(\sum_{i=1}^{N}a_i^2)^{1/2} \leq (E[|\sum_{i=1}^{N} a_iX_i|^p])^{1/p} $$
