iff $E\Bigl(|X_1|\log ({1+|X_1|)}\Bigr)<\infty$ Question: $X_n$'s are i.i.d then
$$E\Bigl(\sup_{n\geq 1} \frac{|X_n|}{n}\Bigr)<\infty \iff E\Bigl(|X_1|\log ({1+|X_1|)}\Bigr)<\infty$$
My attempt: for $\Rightarrow$ part, because $\limsup \frac{|X_n|}{n} \leq \sup_{n\geq 1} \frac{|X_n|}{n}$ a.s. from which I get $\limsup \frac{|X_n|}{n} < \infty$ a.s. implying $X_1 \in L_1$. But the question says something stronger.
Motivation: 
(1)
$$0<p<1 \hspace{10pt} \text{then}\hspace{10pt} E\Bigl(\sup_{n\geq 1} \frac{|X_n|}{n^{1/p}}\Bigr)<\infty \iff X_1 \in L_1$$
(2)
$$1<p<\infty \hspace{10pt} \text{then}\hspace{10pt} E\Bigl(\sup_{n\geq 1} \frac{|X_n|}{n^{1/p}}\Bigr)<\infty \iff X_1 \in L_p$$
These two problems I was able to solve but then I saw a comment saying "what about the case $p=1$? Although it's much complicated to show, it turns out that...". I tried a lot but could not crack it. Any idea/hint/help? Thank you, 
 A: $(\Leftarrow)$ Let $S_n=\sum_{k\le n}X_k$. Then $n^{-1}S_n$ is a reversed MTG and so, using Doob's maximal inequality,
\begin{align}
\mathsf{E}\sup_{n}\frac{|X_n|}{n}&\le 2\mathsf{E}\sup_{n}\frac{|S_n|}{n} \\
&\le \frac{2e}{e-1}(1+\mathsf{E}|X_1|\ln^+|X_1|) \\
&\le \frac{2e}{e-1}(1+\mathsf{E}|X_1|\ln(1+|X_1|)) \\
&<\infty,
\end{align}
where $\ln^+(\cdot)=\ln(\cdot)\vee 0$.
($\Rightarrow$) Let $Y_n=1+|X_n|$. Then $Y_n\ge 1$ a.s., $\mathsf{E}\sup_n Y_n/n<\infty$, and
$$
\mathsf{E}|X_1|\ln(1+|X_1|)=\mathsf{E}Y_1\ln(Y_1)-\mathsf{E}\ln(Y_1).
$$
The second term is finite because $\mathsf{E}Y_1<\infty$ and
\begin{align}
\mathsf{E}Y_1\ln(Y_1)&=\int_1^\infty(\ln(y)+1)\mathsf{P}(Y_1>y)\,dy \\
&=\int_1^{\infty}\ln(y)\mathsf{P}(Y_1>y)\,dy+[\mathsf{E}Y_1-1].\tag{1}\label{1}
\end{align}
The second term of \eqref{1} is finite. Let $K\in\mathbb{N}$ be such that $Y_1\ge K-1$ a.s. and $\mathsf{P}(Y_1<K)>0$. Then $\sup_n Y_n/n\ge K-1$ a.s. and the first term is also finite because
\begin{align}
\infty&>\mathsf{E}\sup_n\frac{Y_n}{n}\ge \sum_{k=1}^\infty \mathsf{P}\!\left(\sup_n\frac{Y_n}{n}\ge k\right) \\
&=(K-1)+\sum_{k=K}^\infty \mathsf{P}\!\left(\sup_n\frac{Y_n}{n}\ge k\right) \\
&\ge^1 (K-1)+C_K\sum_{k=K}^\infty\sum_{n=1}^\infty \mathsf{P}(Y_1\ge nk),
\end{align}
where $C_K\equiv \prod_{k=K}^\infty \mathsf{P}(Y_1\le k)>0$), and
$$
\sum_{k=1}^\infty\sum_{n=1}^\infty \mathsf{P}(Y_1\ge nk)<^2\infty \Leftrightarrow \int_1^{\infty}\ln(y)\mathsf{P}(Y_1>y)\,dy <\infty.\tag{2}\label{2}
$$
The RHS of \eqref{2} converges iff
$$
\int_1^\infty\int_1^\infty \mathsf{P}(Y_1>xy)\,dxdy <\infty.
$$
By changing variables $u=x$ and $v=xy$, the latter becomes
$$
\int_1^\infty\int_1^v\frac{1}{u}\mathsf{P}(Y_1>v)\,dudv=\int_1^\infty \ln(v)\mathsf{P}(Y_1>v)\,dv 
$$
which is the LHS of \eqref{2}.

$^1$ For $k\ge K$,
\begin{align}
&P\!\left(\sup_n\frac{Y_n}{n}>k\right)=\sum_{n=1}^\infty P\!\left(\left\{\bigcap_{1\le i<n}Y_i\le ik\right\} \cap \{Y_{n}>nk\}\right) \\
&\qquad\quad=\sum_{n=1}^\infty \prod_{i=1}^{n-1}\mathsf{P}(Y_1\le ik)\mathsf{P}(Y_1> nk)\ge C_K\sum_{n=1}^\infty \mathsf{P}(Y_1>nk).
\end{align}

$^2$
$$
\sum_{k=1}^\infty\sum_{n=1}^\infty \mathsf{P}(Y_1\ge nk)\le (K-1)\sum_{n=1}^\infty \mathsf{P}(Y_1\ge n)+\sum_{k=K}^\infty\sum_{n=1}^\infty \mathsf{P}(Y_1\ge nk).
$$
