# gaussian lipschitz concentration about the $p$-th moment

Given $x \sim N(0, I)$ and $f : \mathbb{R}^{n} \longrightarrow \mathbb{R}$ which is $L$-Lipschitz, we have that $f - \mathbb{E}f$ is a sub-Gaussian random variable, specifically that $$\| f(x) - \mathbb{E} f(x) \|_{\psi_2} \leq C L \:.$$

I want to show that if in addition $f \geq 0$, then for all $p > 0$, the more general statement $$\| f(x) - (\mathbb{E} f(x)^p)^{1/p} \|_{\psi_2} \leq C_p L$$ holds, where $C_p$ is a constant which depends on $p$. This is Exercise 5.2.5 from Vershynin's book: http://www-personal.umich.edu/~romanv/teaching/2015-16/626/HDP-book.pdf (and the notation above is using his notation).

Using the fact that if $X$ is sub-gaussian then $$\| X\|_p \leq C \|X\|_{\psi_2} \sqrt{p} \:, \:\: p \geq 1 \:,$$ (Eq 2.16 in Vershynin's HDP book), I was able to show that (just by triangle inequality) $$\| X - \|X\|_p \|_{\psi_2} \leq (C_1 + C_2 \sqrt{p}) \| X \|_{\psi_2} \:.$$

I wanted then to apply this fact to the sub-Gaussian r.v. $Z := f(x) - \mathbb{E} f(x)$, but this doesn't work for $p \leq 1$ since the constant does not come out of the $p$-th moment of $Z$, so I don't think this approach is correct. Furthermore, it only applies when $p \geq 1$.

Any other hints/suggestions?

Ok here is a partial solution when $p \geq 1$ is an integer.
It uses the fact that if $X$ is sub-Gaussian, then every $p$-th moment is controlled as follows: $$(\mathbb{E} |X|^p)^{1/p} \leq C_1 \| X \|_{\psi_2} \sqrt{p} \:.$$
Now let $\mu := \mathbb{E} f(x)$. Suppose that $f$ is 1-Lipschitz (this is wlog by scaling) and $f \geq 0$. Then, $$\mathbb{E} f^p = \mathbb{E} (f - \mu + \mu)^p = \sum_{k=0}^{p} {p \choose k} \mu^{p-k} \mathbb{E}(f-\mu)^k \leq \sum_{k=0}^{p} {p \choose k} \mu^{p-k} \mathbb{E}|f-\mu|^k \:.$$ We know that $\|f - \mu\|_{\psi_2} \leq C_2$, and hence $\mathbb{E}| f - \mu|^k \leq C_3^k k^{k/2}$ for all $k \geq 0$. Hence, $$\mathbb{E} f^p \leq \sum_{k=0}^{p} {p \choose k} \mu^{p-k} (C_3 \sqrt{k})^k \leq \sum_{k=0}^{p} {p \choose k} \mu^{p-k} (C_3 \sqrt{p})^k = (\mu + C_3 \sqrt{p})^p \:.$$ Taking the $p$-th root of both sides, we conclude that $$(\mathbb{E} f^p)^{1/p} \leq \mu + C_3 \sqrt{p} \:.$$ Now we can make some progress. Using this inequality, for $t \geq 0$, $$f(x) - (\mathbb{E} f^p)^{1/p} \leq - C_3 \sqrt{p} - t \Longrightarrow f(x) - \mu \leq - t \:.$$ Hence, we have established the lower tail $$\mathbb{P}\left\{ f(x) - (\mathbb{E} f^p)^{1/p} \leq - C_3 \sqrt{p} - t \right\} \leq e^{-t^2/2} \:.$$ On the other hand, since $\mu \leq (\mathbb{E} f^p)^{1/p}$ by Jensen's inequality, we have $$f(x) - (\mathbb{E} f^p)^{1/p} \geq C_3 \sqrt{p} + t \Longrightarrow f(x) - \mu \geq t \:.$$ Therefore, $$\mathbb{P}\left\{ |f(x) - (\mathbb{E} f^p)^{1/p}| \geq C_3 \sqrt{p} + t \right\} \leq 2 e^{-t^2/2} \:.$$ This is not quite a sub-Gaussian tail, since when $t \leq C_3 \sqrt{p}$ it provides no information.