Setup
-----


To clarify, let constants $0 < a < b < \infty$, and $p \in \mathbb{N}$ be fixed. Further let $B \subset \mathbb{R}^{p}$ be a fixed compact support. We then define the space of bounded (probability) densities on $B$, as follows:


$$
\mathcal{F}\_{B}^{[a, b]} := \left\{f \colon B \to [a, b] \mid \int\_{B}{f} d\mu = 1, \text{$f$ measurable} \right\}
$$


Where $\mu$ is taken to be a normalizing (prob) measure. We can then define the $L\_{1}$- metric, and $L\_{2}$-metric on $\mathcal{F}\_{B}^{[a, b]}$, as follows:


\begin{aligned}
&\|f-g\|\_1=\int|f-g| d \mu \\
&\|f-g\|\_2^2=\int(f-g)^2 d \mu
\end{aligned}


Question
--------


The [lecture notes](http://www.stat.yale.edu/%7Eyw562/teaching/598/lec17.pdf) claim *without proof* (on top of page 4) that


\begin{equation}
\|f-g\|\_1^2 \gtrsim\|f-g\|\_2^2,
\end{equation}
where $\gtrsim$ means greater than or equal to, ignoring universal constants (which may depend on $a, b, B$).


This claim seems rather remarkable. Could anyone please prove this, if it is true? Or provide a suitable counterexample if it is false?


If it is true, what are the constants in the inequality (in terms of $a, b, B$)?


Additional Comments
-------------------


If this inequality is true, then I believe it would imply a [reverse Pinsker inequality](https://arxiv.org/pdf/1503.07118.pdf) on $\mathcal{F}\_{B}^{[a, b]}$ (see equation (10)), which is non-trivial since $B$ here is assumed to be compact, not just finite.


**Aside:** Although this has been [asked elsewhere](https://math.stackexchange.com/questions/4494978/l-1-distance-is-bigger-up-to-constants-than-l-2-distance-for-bounded-pdfs), the original specification there was slightly different and the counterexample given there does not apply (since they did not assume a *common* support set $B$). Moreover, the original math.SE question is now quite old and inactive. Since this is a research related question with no well citable proof, I believe it is fair to post here on math.overflow to settle the issue.



Lower bound $L_{1}$-metric with $L_{2}$-metric for bounded pdfs, on common support
This is false. If (let's say) $f-g=1$ on a set of measure $\epsilon$ and $|f-g|\simeq \epsilon$ otherwise (and note that we can still normalize them both), then $\|f-g\|\_1^2\simeq \epsilon^2$, $\|f-g\|\_2^2\simeq\epsilon$.


For a concrete example, you can take $\mu$ as Lebesgue measure on $B=[0,1]$ and then
$$
f(x) = \begin{cases} 2 & x<\epsilon \\ \frac{1-2\epsilon}{1-\epsilon} & x>\epsilon \end{cases} , \quad g(x)=1 .
$$


