Expression for Lipschitz constant for the $L^p$ norm function on $\mathbb R^n$ Let $p \ge 1$  and $f:\mathbb R^n \to \mathbb R$ is given by $f(x):= \|x\|_p.$ 
Then is $f$ a Lipschitz function, and if yes, what's its Lipschitz constant? 
For $p=1,$ I see that it's $\sqrt{n}$ which follows from Cauchy-Schwartz inequality, for $p=2,$ it's just $1,$ but what is it for a general $p?$
Let's try to do some relevant computation below.
$$ \sup_{x \ne y} \frac{|\|x\|_p - \|y\|_p|}{\|x-y\|_2} \le  \sup_{x \ne y} \frac{\|  x- y\|_p}{\|x-y\|_2} = \sup_{x \ne 0} \frac{\|x\|_p}{\|x\|_2}, $$
Note that the first inequality is obtained from triangle inequality, but it's actually an equality because we take the supremum and we plug in $y=0$.
So we indeed have:
$$ \sup_{x \ne y} \frac{|\|x\|_p - \|y\|_p|}{\|x-y\|_2} =  \sup_{x \ne y} \frac{\|  x- y\|_p}{\|x-y\|_2} = sup_{x \ne 0} \frac{\|x\|_p}{\|x\|_2}, $$
Now note that, the last equantity is the operator norm of the identity operator from $(\mathbb R^n, \|\cdot\|_2) \to (\mathbb R^n, \|\cdot\|_p)$, and this operator norm is of course finite, because the spaces are finite dimensional.
So I guess my question translates to: what's $\sup_{x \ne 0} \frac{\| x\|_p}{\|x\|_2} = \sup_{\|x\|_2=1} \|x\|_p?$
 A: For $p > 2$ you can just use that $|x_i|^p \leq |x_i|^2$ for each $i$ and add over all $i$, to get that $||x||_p \leq 1$ whenever $||x||_2 = 1$. This is achieved when $x = (1,0,....0)$ for example.
When $p < 2$ it can be reduced to Holder's inequality, amongst other possibilities. The basic inequality is
$$\bigg({|x_1|^p + ... + |x_n|^p \over n}\bigg)^{1 \over p} \leq 
\bigg({|x_1|^2 + ... + |x_n|^2 \over n}\bigg)^{1 \over 2}$$
So if $||x||_2 = 1$ you end out with $||x||_p \leq n^{{1 \over p} - { 1\over 2}}$. This will be achieved when each $x_i$ is equal, that is, when $x_i = {1 \over \sqrt{n}}$ for each $i$.
There are other ways to get the $p < 2$ case, including Jensen's inequality or even Lagrange multipliers.
A: Observe that for $p\ge 1$, $f:\mathbb R^n \to \mathbb R$ is a continous function (you can show this just by sequential criteria for continuity).
$S^{n-1}=\{x\in \mathbb R^n:\|x\|_2=1\}$ is a compact subset of $\mathbb R^n$ and continuous image of a compact set is compact . Compact subset of $\mathbb R$ is bounded.
Hence you have $$M=\sup_{x\in S^{n-1}} |f(x)|=\sup_{\|x\|_2=1}\|x\|_p<\infty$$
So you have $f$ is bounded.
$x=(x_1,x_2,\cdots,x_n)\in \mathbb  R^n$ , then $\|x\|_2=1\implies |x_i|\le 1$ , for $i=1(1)n$
Then $$\|x\|_p^p=\sum_{i=1}^n|x_i|^p\le n \implies \|x\|_p\le n^{\frac{1}{p}} , \forall x\in S^{n-1}$$
And then you can have the constant to be $n^{\frac{1}{p}}$.

Remark: Every norm on $\mathbb R^n$ are equivalent.
