Why is this expectation finite? For a random vector $X$, if $E (|c \cdot X|) < \infty$ for any vector $c$, then how can we show $ E(\|X \|) < \infty$? Thanks. Note: $\cdot$ means inner product.
 A: EDIT: This is the answer to the edited question
Let $X=(X_1,X_2,\cdots,X_n)$. Take $c=(0,0,\cdots,1,0,\cdots,0)$ to conclude that $E(|X_i|) < \infty$ for $1 \leq i \leq n$. 
Now, $||X|| \leq n\cdot max\{|X_1|,\cdots,|X_n|\}$, and since $E(|X_i|) < \infty$, we have $E(max\{|X_1|,\cdots,|X_n|\}) < \infty$ which implies $E(||X||) < \infty$.
EDIT: This was the answer to the previous version of the question
This doesn't seem to be true. If your vector has just one corodinate (i.e. a random variable), then you are claiming that $E(cX) < \infty$ implies $E(X^2) < \infty$- which is false: see this. 
A: The induced norm (or operator norm) of a vector $X$ is defined as $$\|X\|=\sup_{c\neq0}\dfrac{\|Xc\|}{\|c\|}$$ In other words $\|X\|$ is the least upper bound of the set $$\left\{\dfrac{\|Xc\|}{\|c\|}:c\in\mathbb K, c\neq0\right\}$$ so that there exist $\epsilon>0$ and  $c_{\epsilon} \in \mathbb K$ such that $$\|X\|\le \dfrac{\|Xc_{\epsilon}\|}{\|c_{\epsilon}\|}+\epsilon=\dfrac{|c_{\epsilon}\cdot X|}{\|c_{\epsilon}\|}+\epsilon$$ Taking expectations in both sides yields the result $$E\|X\|\le E\dfrac{\|Xc_{\epsilon}\|}{\|c_{\epsilon}\|}+\epsilon=\dfrac{1}{\|c_{\epsilon}\|}E|c_{\epsilon}\cdot X|+\epsilon<+\infty$$ since $E|c_{\epsilon}\cdot X|<+\infty$, due to the given condition.
