One claim I find in many papers about identity testing, and closeness testing is that any distribution over $[n]$ can be approximated to within $\ell_1$ distance $\epsilon$ in $O\left(\frac{n}{\epsilon^2}\right)$ samples.
I do not seem to be able to find a proof of this anywhere. On trying to prove it, I seem to be able to prove it if I assume that there exists $\delta>0$, such that, $\min_{i \in [n]} P_i > \delta$, where $P$ is the true distribution. However, I do not seem to be able to prove it in the general case.
My attempt at a proof is the following:
Let $P$ be any distribution over $[n]$. Let $X_1,X_2,\cdots,X_{\frac{n}{\epsilon^2}}$ be iid samples of $P$. Let $m=c\frac{n}{\epsilon^2}$. Define $C_i$, $1 \leq i \leq n$, as
$$C_i= \sum_{j=1}^{m} 1(X_j=i).$$
We note that $\mathbb{E}[C_i]= mP_i$.
Further from Chernoff bounding, we get that,
$$P[C_i > (1+\epsilon)mP_i] \leq \exp(-\epsilon^2 \frac{mP_i}{3})= \exp(-\frac{cnP_i}{3}).$$
Similarly, $$P[C_i < (1-\epsilon)mP_i] \leq \exp(-\frac{cnP_i}{2}).$$
Thus defining $\hat{P}_i= \frac{C_i}{m}$, and $\hat{P}=(\hat{P}_1,\cdots,\hat{P}_n)$, by union bounding, we get that with probability at least $1-\sum_{i=1}^n 2\exp(- \frac{cnP_i}{3})$, we have that $||\hat{P}-P||_1 \leq \epsilon$. This gives a proof of the claim as long as there exists $\delta>0$, such that, $\min_{i \in [n]} P_i > \delta$ (as the probability can be made arbitrarily close to 1 by increasing the constant).
However, I do not know how I can generalise this result to the general case. I'd appreciate any help with respect to this.