Properties of a function $C_\ell(\ell)$ which checks an inequality in ideal case (decreasing assumption) and after estimating impact in general case

Question

Suppose that $X=2 \ell+1, Y=C_{\ell}$, both $X$ and $Y$ are function of $\ell$, $X$ is increasing and $Y$ is assuming to be decreasing.

But in reality, my data follow a $C_\ell$ increasing for a small part of range ($\ell = [10,40]$) and decreasing after up to $\ell=3000$, so it is difficult to estimate the impact of this short increasing behavior compared to all the other and large decreasing behavior. Here the real behavior of $C_\ell$ :

1. I would like to know under which conditions ( characteristics of $C_\ell$ decreasing ?, properties of $C_\ell$ function ?, conditions on $\ell_{min}$ and $\ell_{max}$ ?) the below inequality is true :

$$\sigma_{0,2}^{2} - \sigma_{0,1}^{2} = \frac{2\sum_{}^{}(2l + 1)}{({\sum_{}^{}{(2l + 1)C_{l})}}^{2}} \frac{1}{{(f}_{\text{sky}}N_{p}^{2})} - \frac{\sum_{}^{}\frac{2}{(2l + 1)}}{({\sum_{}^{}{C_{l})}}^{2}} \frac{1}{{(f}_{\text{sky}}N_{p}^{2})} = \frac{2}{{(f}_{\text{sky}}N_{p}^{2})}\left\lbrack \frac{\sum_{}^{}(2l + 1)}{({\sum_{}^{}{(2l + 1)C_{l})}}^{2}} - \frac{\sum_{}^{}\frac{1}{(2l + 1)}}{({\sum_{}^{}{C_{l})}}^{2}} \right\rbrack > 0$$

Give that

$\ell \in \lbrack 10,2990\rbrack\quad$ $\text{and}$ $C_{\ell}$ $\in \left\lbrack e^{-9},e^{-7}\right\rbrack$, $\text{$\ell$ is increasing and $C_{l}$ is decreasing}$

Therefore, we only need to check that

$$\frac{\sum_{}^{}(2l + 1)}{({\sum_{}^{}{(2l + 1)C_{l})}}^{2}} - \frac{\sum_{}^{}\frac{1}{(2l + 1)}}{({\sum_{}^{}{C_{l})}}^{2}} > 0$$

Suppose that

$X = 2\ell + 1,Y = C_{\ell}$,both $X$ and $Y$ are function of $\ell$, $X$ is increasing and $Y$ is decreasing.

Then we must show that

$$\frac{\sum_{}^{}(2l + 1)}{({\sum_{}^{}{(2l + 1)C_{l})}}^{2}} - \frac{\sum_{}^{}\frac{1}{(2l + 1)}}{({\sum_{}^{}{C_{l})}}^{2}} = \frac{\sum_{}^{}X}{({\sum_{}^{}{XY)}}^{2}} - \frac{\sum_{}^{}X^{- 1}}{({\sum_{}^{}{Y)}}^{2}} > 0$$

I.E., we need to show that

$$\frac{\sum_{}^{}X}{({\sum_{}^{}{XY)}}^{2}} > \frac{\sum_{}^{}X^{- 1}}{({\sum_{}^{}{Y)}}^{2}} \rightarrow \sum_{}^{}X({\sum_{}^{}{Y)}}^{2} > \sum_{}^{}X^{- 1}({\sum_{}^{}{XY)}}^{2}$$

We then apply Chebyshevs inequality: since the Chebyshevs inequality had different version, we apply one of the classic result stated as the following theorem (note that this one can be found in the paper link enter link description here, and the reference can be tracked as well):

Theorem:$g:\lbrack a,b\rbrack \rightarrow R$ and p:$\lbrack a,b\rbrack \rightarrow R$ be integrable functions. If $f$ and $g$ are monotonic in the same direction, then

$$\int_{}^{}{p(x)\text{dx}}\int_{}^{}{p(x)f(x)g(x)dx} \geq \int_{}^{}{p(x)f(x)dx}\int_{}^{}{p(x)g(x)dx}$$

In mathematics, we can translate the integrable function to sum function. Therefore, given P, if F and G are monotonic in the same direction, i.e., F and G are both increasing or decreasing, then we have that

$$\sum_{}^{}P\sum_{}^{}{P(FG)} \geq \sum_{}^{}\text{PF}\sum_{}^{}\text{PG}$$

Now let us go back to our prove process:

$$\sum_{}^{}X({\sum_{}^{}{Y)}}^{2} > \sum_{}^{}X^{- 1}({\sum_{}^{}{XY)}}^{2}$$

For the term in parentheses of right side of the inequality, since $Y$ and $X$ is increasing, we then have that

$${(\sum_{}^{}\text{XY})}^{2} = \sum_{}^{}\text{XY}\sum_{}^{}\text{XY} < \sum_{}^{}Y\sum_{}^{}{X^{2}Y}\ (Here,\ suppose\ P = Y\ and\ F = G = X)$$

Then we have the right side of inequality

$$\sum_{}^{}X^{- 1}({\sum_{}^{}{XY)}}^{2} < \sum_{}^{}X^{- 1}\sum_{}^{}Y\sum_{}^{}{X^{2}Y} = \sum_{}^{}Y(\sum_{}^{}{X^{2}Y}\sum_{}^{}X^{- 1})$$

Then for the term in parentheses of right side of the inequality, we change it to

$$\sum_{}^{}{X^{2}Y}\sum_{}^{}X^{- 1} = \sum_{}^{}{Y(X^{2})}\sum_{}^{}Y\frac{1}{\text{XY}} = \sum_{}^{}{Y(X^{2})}\sum_{}^{}Y{(XY)}^{- 1}$$

Since $X^{2}$ and $(XY)^{-1}$ are both increasing, then we have that

$$\sum_{}^{}{X^{2}Y}\sum_{}^{}X^{- 1} = \sum_{}^{}{Y(X^{2})}\sum_{}^{}Y{(XY)}^{- 1} < \sum_{}^{}Y\sum_{}^{}Y{(X^{2})(XY)}^{- 1} = \sum_{}^{}Y\sum_{}^{}X$$

$$(Here,\ suppose\ P = Y\ and\ F = X^{2}\ and\ G = {(XY)}^{- 1})$$

Therefore, we showed that

$$\sum_{}^{}X^{- 1}({\sum_{}^{}{XY)}}^{2} < \sum_{}^{}Y\left( \sum_{}^{}{X^{2}Y}\sum_{}^{}X^{- 1} \right) < \sum_{}^{}Y\sum_{}^{}Y\sum_{}^{}X = \sum_{}^{}X({\sum_{}^{}{Y)}}^{2}$$

QUESTION :

2. If someone could tell me if the demonstration is right or not.

3. If I consider $XY$ decreasing, that means that $Y=C_\ell$ decreases muore than $X=2\ell+1$ increases. Unfortunately, I can't check it from a coding and numerical point of view (if I take for example $C_\ell=(\dfrac{1}{2\ell+1} + \text{factor})$ and see the differences with $\text{factor}$ value (positive or negative to be lower or greater than $X=(2\ell+1)$.

I realize that a stronger decreasing for $C_\ell$ than $\dfrac{1}{2\ell+1}$ function decreasing is not enough.

I have to find properties for $Y=C_\ell$ that checks, with $X=(2\ell+1)$ and $XY$ decreasing :

$$\left\lbrack \frac{\sum_{}^{}(2l + 1)}{({\sum_{}^{}{(2l + 1)C_{l})}}^{2}} - \frac{\sum_{}^{}\frac{1}{(2l + 1)}}{({\sum_{}^{}{C_{l})}}^{2}} \right\rbrack > 0$$

Answer 1

It is not enough to assume that $Y=C_l$ is decreasing in $l$.

Indeed, if your inequality were true for all such $Y$, then its non-strict version would be true for all constant $Y$, say for $Y=1$ for all $l$. So, we would have $$ \left(\sum_{\ell=\ell_\text{min}}^{\ell_\text{max}} 1\right)^{2} \ge \sum_{\ell=\ell_\text{min}}^{\ell_\text{max}} X^{-1} \sum_{\ell=\ell_\text{min}}^{\ell_\text{max}} X. $$ However, by the inequality between the arithmetic and harmonic means, exactly the opposite inequality holds (assuming all the values of $X$ are $>0$): $$ \left(\sum_{\ell=\ell_\text{min}}^{\ell_\text{max}} 1\right)^{2} < \sum_{\ell=\ell_\text{min}}^{\ell_\text{max}} X^{-1} \sum_{\ell=\ell_\text{min}}^{\ell_\text{max}} X. $$