Existence of minimum for $\inf_{k \in \mathbb{R}} E[|X-k|^p]$

Question

Suppose $E[|X|^p ]< \infty$ for the given $p \in \mathbb{R}^{+}$.

How to show that the following expression has a minimum \begin{align} \inf_{k \in \mathbb{R}} E[|X-k|^p] \end{align}

That is $\inf_{k \in \mathbb{R}} E[|X-k|^p]=\min_{k \in \mathbb{R}} E[|X-k|^p]$ and the minimizing $k$ exists. Observe, that I am not looking for what $k$ is just that it exists. However, knowing what $k$ is a plus.

Clearly the infimum exists since \begin{align} 0 \le \inf_{k \in \mathbb{R}} E[|X-k|^p] \le E[|X-0|^p]=E[|X|^p] < \infty \end{align}

Here is my solution for when $p$ is an even integer

\begin{align} \inf_{k \in \mathbb{R}} E[|X-k|^p]&=\inf_{k \in \mathbb{R}} E[(X-k)^p] \text{ used the fact that $p$ is even}\\ &= \inf_{k \in \mathbb{R}} E \left[ \sum_{n=0}^p { p \choose n} X^{p-n}k^{n} \right] \text{ Binomial formula}\\ &=\inf_{k \in \mathbb{R}} \sum_{n=0}^p { p \choose n} E\left[ X^{p-n} \right] k^{n} \end{align} where the expectation $E\left[ X^{p-k} \right]$ exists since $E\left[ |X|^{p} \right]$ exist.

Now the $\inf_{k \in \mathbb{R}} \sum_{n=0}^p { p \choose n} E\left[ X^{p-n} \right] k^{n}$ is a polynomial of an even degree and therefore has a minimum. So, for $p$ even \begin{align} \inf_{k \in \mathbb{R}} \sum_{n=0}^p { p \choose n} E\left[ X^{p-n} \right] k^{n}=\min_{k \in \mathbb{R}} \sum_{n=0}^p { p \choose n} E\left[ X^{p-n} \right] k^{n} \end{align}

and there exist a minimizing $k$.

My question is how to extend this proof to any real $k>0$.

Thank you looking forward to you input.

Edit

Based on the discussion in the comments: We have been able to show that for $p \ge 1$ the function $f(y)=E[|X-y|^p]$ is continuous. This is done by using dominated convergence theorem.

Another, property we have been able to show is that if $y_n$ is such that \begin{align} \lim_{n\to \infty} E[(X-y_n)^p]=\inf_{y}E[|X-y|^p] \end{align} then $y_n$ is bounded.

However, I am still not sure how this implies that there exists $y_0$ that minimizes $f(y)=E[|X-y|^p]$? Please help.

Answer 1

Collected from the comments.

Fact 1. $f(y) = E[|X-y|^p]$ is continuous. Indeed, for any sequence $\{y_n,n\ge 1\}$ such that $y_n\to y_0$, $n\to\infty$, we have $f(y_n)\to f(y_0)$ in view of the dominated convergence theorem (thanks to convergence, $\{|y_n|\}$ is bounded by some $k$).

Fact 2. $f\to+\infty$, $f\to\pm\infty$. Hint: use the Fatou lemma. Therefore, there exists some $R$ such that $f(y) > f(0)$ whenever $|y|>R$. Thus, $\inf_{y\in\mathbb{R}} f(y) = \inf_{|y|\le R} f(y) = \min_{|y|\le R} f(y)$ thanks to continuity.

From these two facts the existence of minimum follows.