I'm trying to show that, for a given random vector $X$ with values from $1$ to $n$ that is uniformly distributed on $K$ for $K= \{x \in \mathbb{R}^n : \|x\|_1 \leq r \}$, $$ \lim_{n \rightarrow \infty} \|X\|_{\psi2} = \infty $$
Assuming that $r \approx n $ is already proved.
The Question is from R. Vershynin High-dimensional probability Exercise 3.4.9 part b
ATTEMPT:
I want to show it by saying that $\|X_1\|_{L_p} \leq C \sqrt{p}$ does not hold.
$$\|X_1\|_{L_P}^{p} = \int_0^{\infty} p x^{p-1} P(|X_1| > x) = pr^p \frac{\Gamma(p) \Gamma(n+1))}{\Gamma(n+p+1)} = p r^p \frac{\Gamma(p) n!}{(n+p)!}$$
However, I'm unable to simplify it beyond this to show that the above statement doesn't hold.
I just realized that your attempt is different from the attempt in the other post (which is also kind of incomplete), so I'll address your question directly.
You want to show that $\lim_{n \to \infty} \|X^n_1\|_{\psi_2} = \infty$ where $X^n_1$ is the first coordinate of a uniform random vector $X^n$ over the unit $l_1$ ball in $\mathbb R^n$. As a consequence of this, we will also have $\lim_{n \to \infty} \|X^n\|_{\psi_2} = \infty$, which is exactly what Vershynin 3.4.9(ii) dictates.
The way you want to show this is to show that "$\|X_1\|_p \leq C \sqrt{p}\ $ does not hold". Now, this statement isn't the most clear, which is causing the problem. Indeed, $X_1$ is bounded hence subgaussian. The problem is that the same $C$ doesn't work for all $n$, and hence the limit becomes infinite.
Perhaps, what you want to argue is this:
Now, there are two problems. The first I already highlighted, namely the unclear statement. The second is that the proof that the non-existence of $C$ implies the infinite limit of the subgaussian norms is actually something that requires a bit of work. It is not merely obvious from the definitions of subgaussianity.
Indeed, to prove the implication, you need to go to Proposition 2.5.2 of Vershynin. In there, you will realize that when you don't scale $K_2$ and treat it as some constant in condition (ii), then $K_2$ and the constant $K_4$ which is obtained in condition (iv) (and which determines the subgaussian norm $\|\cdot\|_{\psi_2}$) are related in such a way that if $K_2 \to \infty$ as $n \to \infty$ (which is what will happen if the $L^p$ norm inequalities go through) then $K_4 \to \infty$ as $n \to \infty$(which is exactly the subgaussian norm going to infinity).
Thankfully, the proof of Proposition 2.5.2, when you look at it in detail, shows that $K_4$ is at least some constant times $K_2$, where this constant doesn't depend upon $n$. Therefore, the implication is actually true and I leave you to see that the relation between $K_2$ and $K_4$ is exactly as I described earlier.
With that in hand, let's look at $$ \|X^n_1\|^p_{L^p} = pr^p\frac{\Gamma(p)n!}{(n+p)!} $$ We know that $r \approx n$ from part(i) of the same problem (we've been asked to assume such a value of $r$ for this part), so we can say that $$ \|X^n_1\|^p_{L^p} \geq Cpn^p\frac{\Gamma(p)n!}{(n+p)!} $$ for some constant $C$ independent of $n$. The $L^p$ norm estimates that we want to show are equivalent to $$ \sup_{n \geq 1, p \geq 1} \frac{\|X^n_1\|_{L^p}}{\sqrt{p}} = \infty \iff \sup_{n \geq 1, p \geq 1} \frac{\|X^n_1\|^p_{L^p}}{\sqrt{p}^p} = \infty \iff \sup_{n \geq 1, p \geq 1} \frac{Cpn^p\Gamma(p)n!}{(n+p)!p^{p/2}} = +\infty $$
At this point, a quick glance at fixing a value of $p$ (say $p=2$) and letting $n \to \infty$ will result in a failure, since the top and bottom will then be polynomials of the same degree in $n$. So $p$ needs to vary with $n$. Let us ignore the constant $C$ in the supremum : after all, it plays no role in determining the finiteness/infiniteness of the supremum.
However, the very next substitution idea works, which is $p=n$. Indeed, suppose you put $p=n$ to get $$ \frac{Cpn^p\Gamma(p)n!}{(n+p)!p^{p/2}} \to \frac{Cn^{n+1}\Gamma(n)n!}{(2n)!n^{n/2}} = \frac{n^{n+1-(n/2)}(n+1)!n!}{(2n)!} = (n+1)n^{(n/2)+1}\frac{(n!)^2}{(2n)!} $$ You can now apply Stirling's approximation $$ \lim_{m \to \infty} \frac{m!}{\sqrt{2\pi}m^{m+\frac 12}e^{-m}} = 1 $$
We call $f$ asymptotic to $g$ if $\lim_{n \to \infty} \frac{f(n)}{g(n)}=1$ from now on.
As a consequence of Stirling's approximation, $(n!)^2$ is asymptotic to $2\pi n^{2n+1}e^{-2n}$ and $(2n)!$ is asymptotic to $\sqrt{2\pi}2^{2n+\frac 12}n^{2n+\frac 12}e^{-2n}$.
Taking their ratio, $\frac{(n!)^2}{(2n)!}$ is asymptotic to $\frac{\sqrt{\pi}n^{\frac 12}}{4^n}$.
Thus, $(n+1)n^{(n/2)+1}\frac{(n!)^2}{(2n)!}$ is asymptotic to $\frac{\sqrt{\pi}(n+1)n^{(n/2)+3/2}}{4^n}$. Since $n+1$ is asymptotic to $n$, the entire expression is asymptotic to $$ \frac{n^{n/2+5/2}}{4^n} = \exp\left((n/2+5/2)\log n -n\log 4\right)=\exp((n\log n)/2 - n \log 4 + 5\log n/2) $$ Now, the expression inside the exponential operation goes to infinity as $n \to \infty$ because the dominating term is the $n \log n$ term. As a consequence of all the approximations we've made, it follows that $$ \sup_{n \geq 1} \frac{(n+1)n^{n/2+1}(n!)^2}{(2n)!} = \infty \implies \sup_{n \geq 1, p \geq 1} \frac{Cpn^p\Gamma(p)n!}{(n+p)!p^{p/2}} = +\infty \implies \sup_{n \geq 1, p \geq 1} \frac{\|X^n_1\|_{L^p}}{\sqrt{p}} = \infty, $$ which is what you wanted to prove and which, by virtue of the argument connecting $K_2$ and $K_4$ that I spoke of earlier, shows that $\lim_{n \to \infty} \|X^n_1\|_{\psi_2} = +\infty$ and hence $\|X^n\|_{\psi_2} = +\infty$.