Suppose $X_1,\ldots,X_n$ are i.i.d. random vectors in $\mathbb{R}^d$ ($d \geq 2$) from some density $f$. I want to show that $\sup_{n \geq 2} \mathbb{E} n^{1/d}\min_{2 \leq j \leq n} ||X_j-X_1||_2 < \infty$, under some assumption on $f$ of course.
I have proceeded so far to this point. For any $t >0$, we have \begin{align*} \mathbb{P} \left( n^{1/d}\min_{2 \leq j \leq n} ||X_1-X_j||_2 > t\right) &= \int_{\mathbb{R}^d} \mathbb{P} \left( ||y-X_j||_2 > tn^{-1/d}, \;\forall \; 2 \leq j \leq n \right)f(y)\,dy \\ & \int_{\mathbb{R}^d} \left( 1- \int_{B_d(0,tn^{-1/d})} f(x) \, dx \right)^{n-1} f(y)\,dy \\ & \leq \int_{\mathbb{R}^d} \exp \left(-(n-1)\int_{B_d(y,tn^{-1/d})} f(x) \, dx \right) f(y)\, dy, \end{align*} where $B_d(z,r)$ is the ball of radius $r$ around $z$ in $\mathbb{R}^d$. By Lebesgue Differentiation Theorem, we have $$ \dfrac{1}{C_dt^dn^{-1}}\int_{B_d(y,tn^{-1/d})} f(x) \, dx \stackrel{a.e.}{\longrightarrow} f(y),$$ where $C_d$ is the volume of unit ball in $\mathbb{R}^d$. We apply DCTto yield that \begin{equation}{\label{tightmin}} \limsup_{n \to \infty} \mathbb{P} \left( n^{1/d}\min_{2 \leq j \leq n} ||X_1-X_j||_2 > t\right) \leq \int_{\mathbb{R}^d} \exp \left(-C_df(y)t^d \right) f(y)\, dy. \end{equation} Since the right hand side goes to $0$ as $t \to \infty$ (by applying DCT), this shows that $\left\{n^{1/d}\min_{2 \leq j \leq n} ||X_j-X_1||_2 : n \geq 1 \right\}$ is a tight sequence. To prove $L^1$-boundedness we need to show the following. \begin{align*} \limsup_{n \to \infty} \mathbb{E} n^{1/d}\min_{2 \leq j \leq n} ||X_j-X_1||_2 &= \limsup_{n \to \infty} \int_{0}^{\infty} \mathbb{P} \left( n^{1/d}\min_{2 \leq j \leq n} ||X_1-X_j||_2 > t\right) \, dt \\ & \leq \limsup_{n \to \infty} \int_{0}^{\infty} \int_{\mathbb{R}^d} \exp \left(-(n-1)\int_{B_d(y,tn^{-1/d})} f(x) \, dx \right) f(y)\, dy \, dt \\ & \stackrel{?}{\leq} \int_{0}^{\infty} \limsup_{n \to \infty} \int_{\mathbb{R}^d} \exp \left(-(n-1)\int_{B_d(y,tn^{-1/d})} f(x) \, dx \right) f(y)\, dy \, dt \\ & = \int_{0}^{\infty} \int_{\mathbb{R}^d} \exp \left(-C_df(y)t^d \right) f(y)\, dy \, dt \\ & = \int_{\mathbb{R}^d} \int_{0}^{\infty} \exp \left(-C_df(y)t^d \right) f(y)\, dt \, dy \\ & = \text{constant} \int_{\mathbb{R}^d} f(y)^{(d-1)/d} \, dy. \end{align*} Therefore, if $f$ satisfies the condition that $ f^{(d-1)/d} \in L^1(\mathbb{R}^d)$, then we have proved the required assertion provided we can justify the interchange of $\limsup$ and integral in the step marked by $(?)$. I am stuck at this point and not sure how to proceed. Any suggestion at this point will be greatly helpful.
The assertion is known to be true under some more assumption. Assume that $f$ has compact support. Then it is known that $$ \lim_{n \to \infty} \mathbb{E} L(X_1,\ldots,X_n)/n^{(d-1)/d} = \text{constant}(d,f) \int_{\mathbb{R}^d} f(x)^{(d-1)/d} \, dx,$$ where $L(X_1,\ldots,X_n)$ is the length of the shortest Hamiltonian cycle through these $n$ points and $\text{constant}(d,f) $ is some constant depending upon $d,f$ only. See Equation (7.3) of Probability Theory of Classical Euclidean Optimization Problems (1998) by Joseph Yukich for this result. Since, $$ L(X_1,\ldots,X_n) \geq \sum_{i=1}^n \min_{j \neq i, 1 \leq j \leq n} ||X_i-X_j||_2,$$ we have $$ \mathbb{E} \min_{2 \leq j \leq n} ||X_1-X_j||_2 = \dfrac{1}{n} \mathbb{E} L(X_1, \ldots,X_n) = O(n^{-1/d}).$$