Is norm $E[|X|^p]^{1/p}$ a continous function of $p$ Suppose $X$ is a random variable such that $E[|X|^p]<\infty$ is a function
\begin{align}
f(p)=E[|X|^p]^{1/p}
\end{align}
continuos function of $p$ for $1 \le p < \infty$.
Here is my attempt to show that the function is continuos as $c$.
I tried to use $\epsilon-\delta$ definitions. 
So, I want to show that for every $\epsilon>0$ there exists $\delta>0$ such that $|c-p| \le \delta$ implies
\begin{align}
|f(p)-f(c)| \le \epsilon.
\end{align}
But I could'n get the above in the form
\begin{align}
|f(p)-f(c)|=|E[|X|^p]^{1/p}-E[|X|^c]^{1/c}| \le K |c-p|
\end{align}
for some constant $K$.  If I could do the above the rest would follow by picking $\delta=\frac{\epsilon}{K}$.
Thanks for any help
Edit Based on the suggestion by  @Did 
Let $q$ be a conjugate exponent of $p$ and $r$ be conjugate exponent of $c$ then 
\begin{align}
\left|||X||_p-||X||_c \right|&=  \left| \sup_{Y: ||Y||_q \le 1}||XY||_1- \sup_{Y: ||Y||_r \le 1}||XZ||_1 \right|\\
& \le \sup_{Y: ||Y||_q \le 1}||XY||_1+ \sup_{Y: ||Y||_r \le 1}||XY||_1 \text{ by Triangle Inequality}\\
& \le 2\sup_{Y: ||Y||_{\max(q,r)} \le 1}||XY||_1 \text{ taking sup over the largest domain }
\end{align}
 A: It's enough to show that $E|X|^p$ is continuous in $p$. Why not use dominated convergence: $p\mapsto |X|^p$ is continuous, and if $p\in[1,p_0]$ then
$|X|^p\le 1+|X|^{p_0}$. (Think of $p_0>1$ as large but fixed.) In this way you show that $p\mapsto E|X|^p$ is continuous on $[1,p_0]$ for each $p_0>1$.
A: Since the mapping $$(y,p) \mapsto y^{1/p} = \exp( \frac{1}{p} \log y)$$ is clearly continuous, it suffices to show that
$$p \mapsto \mathbb{E}(|X|^p)$$
is continuous.
Fix $p \geq 1$ and some sequence $p_n \to p$. Choose $N$ sufficiently large such that $p_n \leq N$ for all $n$. Since
$$|X|^{p_n} \leq 1_{|X|\leq 1}+ |X|^{p_n} 1_{|X|>1} \leq 1+ |X|^N \in L^1$$
and $|X|^{p_n} \to |X|^p$ pointwise, an application of the dominated convergence theorem shows
$$\mathbb{E}(|X|^{p_n}) \to \mathbb{E}(|X|^p).$$
A: An alternative would to be note that
$$
|X|^p \leq (1 + |X|)^p \leq (1+|X|)^{p_0}
$$
for all $1 \leq p \leq p_0$. Note that the right-hand side is an integrable function.
Using the dominated convergence theorem, it thus follows that
$$
[1, p_0] \to \Bbb{R}, p \mapsto \Bbb{E}|X|^p
$$
is continuous. Since $p_0 >1$ is arbitrary, we see that $p \mapsto \Bbb{E}|X|^p$ is continuous and hence so is your desired map.
