Is norm $E[|X|^p]^{1/p}$ a continous function of $p$

Question

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}

Answer 1

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 $p_0>1$.

Answer 2

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

$$\mathbb{E}(|X|^{p_n}) \to \mathbb{E}(|X|^p).$$

Answer 3

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.