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I was reading these lecture notes http://www.stat.yale.edu/~yw562/teaching/598/lec17.pdf, where on the top of page 4 it is mentioned that the $L_1$ distance is bigger than (up to some constant) the $L_2$ distance for bounded pdfs. In other words if $\alpha \leq f,g \leq \beta$ are two pdfs, it holds that $\|f-g\|_1^2 \gtrsim \|f-g\|_2^2$ where the inequality $\gtrsim$ hides constants that may depend on $\alpha, \beta$. That's not obvious to me, and does anyone have a reference or a proof of this fact? (I can certainly see that the reverse implication holds when the densities are bounded i.e. $\|f-g\|_1^2 \lesssim \|f-g\|_2^2$)

Thanks!

PS. Let me clarify that $$ \|f-g\|_1 = \int |f-g| d\mu $$ and $$ \|f-g\|_2^2 = \int (f-g)^2 d\mu $$

EDIT: Following the counterexample of Q9y5 below, I am adding the additional assumption that $f, g$ have the same support.

spacetimewarp
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    Apply Hölder inequality, note that the boundedness gives $|f-g|_{\infty}\leq \beta-\alpha$. – Q9y5 Jul 18 '22 at 03:42
  • If I understand correctly you mean $\int (f-g)^2 d \mu \leq \int |f-g| d\mu |f-g|_\infty \lesssim \int |f-g| d\mu $. But note that this is not the required inequality which is $\int (f-g)^2 d \mu \lesssim (\int |f-g| d \mu)^2 $ – spacetimewarp Jul 18 '22 at 04:06
  • An answer to this question in the negative was given here. – spacetimewarp Mar 08 '24 at 19:27

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I am afraid the statement is not true without the condition $\mathrm{supp}(f)=\mathrm{supp}(g)=[0,1]$, which was implied by $f,g\in\mathcal{P}$ where "$\mathcal{P}$ be an arbitrary set of pdfs on $[0,1]$" in the link.

Counter example: $f=1_{[0,1]}$ is pdf of $\mathrm{Unif}(0,1)$, $g=1_{[\epsilon,1+\epsilon]}$ is pdf of $\mathrm{Unif}(\epsilon,1+\epsilon)$ for $\epsilon\leq 1$, then $\alpha=0$, $\beta=1$.

But in this case, $$\int(f-g)^2d\mu=2\epsilon\quad\overset{\epsilon\to0^{+}}{\gg}\quad\biggl(\int|f-g|d\mu\biggr)^2=4\epsilon^2.$$

Q9y5
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  • Oh yeah I buy that. Nice counterexample. I guess I should have mentioned that the supports have to be the same. – spacetimewarp Jul 18 '22 at 05:22
  • @spacetimewarp If $|f-g|$ is bounded away from zero, then the statement is a corollary of reverse Cauchy-Schwarz inequality, see for example https://math.stackexchange.com/questions/879832/reverse-cauchy-schwarz-for-integrals, not sure about the case where $|f-g|$ can be arbitrarily small, maybe have to use the condition $\int f d\mu=\int g d\mu=1$. – Q9y5 Jul 18 '22 at 09:34
  • That's a good point, and I agree. Maybe we can split it into cases when $|f-g|$ is small and large... – spacetimewarp Jul 18 '22 at 15:08