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Part I: $CN^{1/3}$ is enough.

Start with $P(x)=\prod_{k\le N^{1/3}}(1-k^2x^2)$. Notice that it vanishes at $1/k$ with $k\le N^{1/3}$, is bounded by $1$ on $[0,N^{-1/3}]$, the degree of $P$ is $2N^{1/3}$ and on the interval $[0,2N^{-1}]$ we have $\log P\ge -2\sum_{k\le N^{1/3}}\frac{4k^2}{N^2}=o(1)$. Now just take the Chebyshev polynomial $Q$ of degree $N^{1/3}$ adjusted to the interval $[0,N^{-1/3}]$. It will jump as you requested between $N^{-1}$ and $2N^{-1}$ and stay below $1$ on $[0,N^{-1/3}]$. The product $P(x)Q(x)$ will then satisfy all your conditions.

Part 2: $n<cN^{1/3}$ is insufficient

One thing that is written in every decent textbook addressing such inequalities is that, in principle, you need no fancy tools for them. Everything can be derived directly from the Lagrange interpolation formula if you can guess the right nodes to use. This is just the general linear programming mumbo-jumbo.

So, the question is how to get the right nodes in your setting. If you do not care about constant factors, then the canonical set of nodes for human consumption is just $\{k^2:k=0,1,\dots,n\}$. Let's recall how it works. Suppose you know the values $P(0),P(1),P(4),\dots,P(n^2)$ of a polynomial of degree $N$ and want to estimate $P'(-y), 0<y<1$ or something like that. You just write $$ P(x)=\sum{k=0^n}P(k)L_k(x) $$ where $$ L_k(x)=\frac {\prod_{m:\ne k}(x-m^2)}{\prod_{m:m\ne k}(k^2-m^2)} $$ and estimate $$ |P'(-y)|\le\max_k|P(k^2)|\sum_k |L'_k(-y)| $$ Now let us estimate $|L'_k(-y)|$. It is just $$ |L_k(-y)|\sum_{m:m\ne k}\frac 1{y+m^2} $$ Now, $$ |L_k(y)|=\frac{\prod_{m:m\ne k}(y+m^2)}{\prod_{m:m\ne k}(|m-k|(m+k))} $$ If $k=0$, then the numerator is $\le (n!)^2\prod_{m=1}^n(1+\frac y{m^2})\le C(n!)^2$ while the denominator is exactly $(n!)^2$. Thus $|L_0(-y)|\le C$ whence $$ |L_0'(-y)|\le C\sum_{m=1}^n\frac 1{y+m^2}\le C\,. $$ If $k>0$, then the numerator in $L_k(-y)$ is at most $Cy\frac 1{k^2}(n!)^2$ (the same count except now $y=|-y-0|$ is present and $y+k^2=|-y-k^2|$ is missing) while the denominator is $$ \frac{k!(n-k)!(n+k)!}{2k(k-1)!}=\frac{(n-k)!(n+k)!}{2} $$ whence $$ |L_k'(-y)|\le \left(\frac 1y+\sum_{m\ge 1}\frac 1{y+m^2}\right)|L_k(-y)|\le \frac C{k^2}\frac{(n!)^2}{(n-k)!(n+k)!}\le \frac C{k^2}e^{-k^2/n}\,. $$ Thus we can easily derive from here that $$ |P'(-y)|\le C\sum_{k=0}^n |P(k^2)|\frac 1{k^2+1}e^{-k^2/n}\,. $$

Now choose an integer $M$ such that $n^3<c M\le c^2N$ where $c>0$ is some small constant. Consider the nodes $x_k=\frac 1M+\frac{k^2}{M}, k=0,1,\dots,n$. If they were available, that would be the end of the story (we would get an estimate $CM$ for the derivative on $[0, M^{-1}]$). Unfortunately only $\frac 1M$ is surely there and the rest have to be approximated. Note that for any $x\in[0,1]$, the nearest to $x$ inverse integer is within $x^2$ from $x$. So, if we choose $z_k$ to be the closest inverse integers to $x_k$, we'll have $|x_k-z_k|\le x_k^2\le 4\frac{k^4}{M^2}$ for $k>0$. Note that the distance from $x_k$ to any neighboring node is about $\frac kM$, so the shift $4\frac{k^4}{M^2}\le 4\frac{n^3}M\frac kM\le 4c\frac kM$ is small compared to it and we, indeed, get $n+1$ different nodes.

Let us now look at how different the corresponding elementary Lagrange polynomials $$ \widetilde L_k(x)=\frac{\prod_{m:m\ne k}(x-z_m)}{\prod_{m:m\ne k}(z_k-z_m)} $$ are from the polynomials $L_k(x)$ built on the canonical nodes $x_k$ on the interval $[0,\frac 1M]$.

The numerator:

We have $\frac {|x-z_m|}{|x-x_m|}\le 1+\frac {4m^2}{M}$, so the numerator can grow at most $\prod_{m=1}^n (1+\frac {4m^2}{M})\le e^{4n^3/M}\le 2$ times.

The denominator:

If $k=0$, then we have $\frac{z_m}{x_m}\ge 1-\frac {4m^2}{M}$, so, as above, we see that it can drop at most twice.

If $k>0$, then $$ \frac {|z_k-z_m|}{|x_k-x_m|}\ge 1-\frac {4(m^4+k^4)}{M|k^2-m^2|}\ge 1-\frac {4(m^3+k^3)}{M|k-m|} $$ We also have $$ \sum_{m:0\le m\le 2k,m\ne k}\frac {4(m^3+k^3)}{M|k-m|}\le C\frac{k^3}M(1+\log k) $$ and $$ \sum_{m:2k\le m\le n}\frac {4(m^3+k^3)}{M|k-m|}\le C\sum_{m:2k\le m\le n}\frac {m^2}{M}\le C\frac {n^3}{M}<1 $$ Thus, the ratio of the denominators we can have against us is at most $Ck^{Ck^3/M}$. That is played against the decay $k^{-2}e^{-k^2/n}$ and it is clear that the decay wins (say, because the power on $k$ in the growth factor is always below $1/2$; we have plenty of leeway here).

For the transition from the value to the derivative, we can change the distances $x-x_m$ even twice without changing the corresponding factor too much, so we are done.

Part I: $CN^{1/3}$ is enough.

Start with $P(x)=\prod_{k\le N^{1/3}}(1-k^2x^2)$. Notice that it vanishes at $1/k$ with $k\le N^{1/3}$, is bounded by $1$ on $[0,N^{-1/3}]$, the degree of $P$ is $2N^{1/3}$ and on the interval $[0,2N^{-1}]$ we have $\log P\ge -2\sum_{k\le N^{1/3}}\frac{4k^2}{N^2}=o(1)$. Now just take the Chebyshev polynomial $Q$ of degree $N^{1/3}$ adjusted to the interval $[0,N^{-1/3}]$. It will jump as you requested between $N^{-1}$ and $2N^{-1}$ and stay below $1$ on $[0,N^{-1/3}]$. The product $P(x)Q(x)$ will then satisfy all your conditions.

Part 2: $n<cN^{1/3}$ is insufficient

One thing that is written in every decent textbook addressing such inequalities is that, in principle, you need no fancy tools for them. Everything can be derived directly from the Lagrange interpolation formula if you can guess the right nodes to use. This is just the general linear programming mumbo-jumbo.

So, the question is how to get the right nodes in your setting. If you do not care about constant factors, then the canonical set of nodes for human consumption is just $\{k^2:k=0,1,\dots,n\}$. Let's recall how it works. Suppose you know the values $P(0),P(1),P(4),\dots,P(n^2)$ of a polynomial of degree $N$ and want to estimate $P'(-y), 0<y<1$ or something like that. You just write $$ P(x)=\sum_{k=0}^n P(k)L_k(x) $$ where $$ L_k(x)=\frac {\prod_{m:\ne k}(x-m^2)}{\prod_{m:m\ne k}(k^2-m^2)} $$ and estimate $$ |P'(-y)|\le\max_k|P(k^2)|\sum_k |L'_k(-y)| $$ Now let us estimate $|L'_k(-y)|$. It is just $$ |L_k(-y)|\sum_{m:m\ne k}\frac 1{y+m^2} $$ Now, $$ |L_k(y)|=\frac{\prod_{m:m\ne k}(y+m^2)}{\prod_{m:m\ne k}(|m-k|(m+k))} $$ If $k=0$, then the numerator is $\le (n!)^2\prod_{m=1}^n(1+\frac y{m^2})\le C(n!)^2$ while the denominator is exactly $(n!)^2$. Thus $|L_0(-y)|\le C$ whence $$ |L_0'(-y)|\le C\sum_{m=1}^n\frac 1{y+m^2}\le C\,. $$ If $k>0$, then the numerator in $L_k(-y)$ is at most $Cy\frac 1{k^2}(n!)^2$ (the same count except now $y=|-y-0|$ is present and $y+k^2=|-y-k^2|$ is missing) while the denominator is $$ \frac{k!(n-k)!(n+k)!}{2k(k-1)!}=\frac{(n-k)!(n+k)!}{2} $$ whence $$ |L_k'(-y)|\le \left(\frac 1y+\sum_{m\ge 1}\frac 1{y+m^2}\right)|L_k(-y)|\le \frac C{k^2}\frac{(n!)^2}{(n-k)!(n+k)!}\le \frac C{k^2}e^{-k^2/n}\,. $$ Thus we can easily derive from here that $$ |P'(-y)|\le C\sum_{k=0}^n |P(k^2)|\frac 1{k^2+1}e^{-k^2/n}\,. $$

Now choose an integer $M$ such that $n^3<c M\le c^2N$ where $c>0$ is some small constant. Consider the nodes $x_k=\frac 1M+\frac{k^2}{M}, k=0,1,\dots,n$. If they were available, that would be the end of the story (we would get an estimate $CM$ for the derivative on $[0, M^{-1}]$). Unfortunately only $\frac 1M$ is surely there and the rest have to be approximated. Note that for any $x\in[0,1]$, the nearest to $x$ inverse integer is within $x^2$ from $x$. So, if we choose $z_k$ to be the closest inverse integers to $x_k$, we'll have $|x_k-z_k|\le x_k^2\le 4\frac{k^4}{M^2}$ for $k>0$. Note that the distance from $x_k$ to any neighboring node is about $\frac kM$, so the shift $4\frac{k^4}{M^2}\le 4\frac{n^3}M\frac kM\le 4c\frac kM$ is small compared to it and we, indeed, get $n+1$ different nodes.

Let us now look at how different the corresponding elementary Lagrange polynomials $$ \widetilde L_k(x)=\frac{\prod_{m:m\ne k}(x-z_m)}{\prod_{m:m\ne k}(z_k-z_m)} $$ are from the polynomials $L_k(x)$ built on the canonical nodes $x_k$ on the interval $[0,\frac 1M]$.

The numerator:

We have $\frac {|x-z_m|}{|x-x_m|}\le 1+\frac {4m^2}{M}$, so the numerator can grow at most $\prod_{m=1}^n (1+\frac {4m^2}{M})\le e^{4n^3/M}\le 2$ times.

The denominator:

If $k=0$, then we have $\frac{z_m}{x_m}\ge 1-\frac {4m^2}{M}$, so, as above, we see that it can drop at most twice.

If $k>0$, then $$ \frac {|z_k-z_m|}{|x_k-x_m|}\ge 1-\frac {4(m^4+k^4)}{M|k^2-m^2|}\ge 1-\frac {4(m^3+k^3)}{M|k-m|} $$ We also have $$ \sum_{m:0\le m\le 2k,m\ne k}\frac {4(m^3+k^3)}{M|k-m|}\le C\frac{k^3}M(1+\log k) $$ and $$ \sum_{m:2k\le m\le n}\frac {4(m^3+k^3)}{M|k-m|}\le C\sum_{m:2k\le m\le n}\frac {m^2}{M}\le C\frac {n^3}{M}<1 $$ Thus, the ratio of the denominators we can have against us is at most $Ck^{Ck^3/M}$. That is played against the decay $k^{-2}e^{-k^2/n}$ and it is clear that the decay wins (say, because the power on $k$ in the growth factor is always below $1/2$; we have plenty of leeway here).

For the transition from the value to the derivative, we can change the distances $x-x_m$ even twice without changing the corresponding factor too much, so we are done.

Part I: $CN^{1/3}$ is enough.

Start with $P(x)=\prod_{k\le N^{1/3}}(1-k^2x^2)$. Notice that it vanishes at $1/k$ with $k\le N^{1/3}$, is bounded by $1$ on $[0,N^{-1/3}]$, the degree of $P$ is $2N^{1/3}$ and on the interval $[0,2N^{-1}]$ we have $\log P\ge -2\sum_{k\le N^{1/3}}\frac{4k^2}{N^2}=o(1)$. Now just take the Chebyshev polynomial $Q$ of degree $N^{1/3}$ adjusted to the interval $[0,N^{-1/3}]$. It will jump as you requested between $N^{-1}$ and $2N^{-1}$ and stay below $1$ on $[0,N^{-1/3}]$. The product $P(x)Q(x)$ will then satisfy all your conditions.

Part 2: $n<cN^{1/3}$ is insufficient

One thing that is written in every decent textbook addressing such inequalities is that, in principle, you need no fancy tools for them. Everything can be derived directly from the Lagrange interpolation formula if you can guess the right nodes to use. This is just the general linear programming mumbo-jumbo.

So, the question is how to get the right nodes in your setting. If you do not care about constant factors, then the canonical set of nodes for human consumption is just $\{k^2:k=0,1,\dots,n\}$. Let's recall how it works. Suppose you know the values $P(0),P(1),P(4),\dots,P(n^2)$ of a polynomial of degree $N$ and want to estimate $P'(-y), 0<y<1$ or something like that. You just write $$ P(x)=\sum{k=0^n}P(k)L_k(x) $$ where $$ L_k(x)=\frac {\prod_{m:\ne k}(x-m^2)}{\prod_{m:m\ne k}(k^2-m^2)} $$ and estimate $$ |P'(-y)|\le\max_k|P(k^2)|\sum_k |L'_k(-y)| $$ Now let us estimate $|L'_k(-y)|$. It is just $$ |L_k(-y)|\sum_{m:m\ne k}\frac 1{y+m^2} $$ Now, $$ |L_k(y)|=\frac{\prod_{m:m\ne k}(y+m^2)}{\prod_{m:m\ne k}(|m-k|(m+k))} $$ If $k=0$, then the numerator is $\le (n!)^2\prod_{m=1}^n(1+\frac y{m^2})\le C(n!)^2$ while the denominator is exactly $(n!)^2$. Thus $|L_0(-y)|\le C$ whence $$ |L_0'(-y)|\le C\sum_{m=1}^n\frac 1{y+m^2}\le C\,. $$ If $k>0$, then the numerator in $L_k(-y)$ is at most $Cy\frac 1{k^2}(n!)^2$ (the same count except now $y=|-y-0|$ is present and $y+k^2=|-y-k^2|$ is missing) while the denominator is $$ \frac{k!(n-k)!(n+k)!}{2k(k-1)!}=\frac{(n-k)!(n+k)!}{2} $$ whence $$ |L_k'(-y)|\le \left(\frac 1y+\sum_{m\ge 1}\frac 1{y+m^2}\right)|L_k(-y)|\le \frac C{k^2}\frac{(n!)^2}{(n-k)!(n+k)!}\le \frac C{k^2}e^{-k^2/n}\,. $$ Thus we can easily derive from here that $$ |P'(-y)|\le C\sum_{k=0}^n |P(k^2)|\frac 1{k^2+1}e^{-k^2/n}\,. $$

To be continued...

Part I: $CN^{1/3}$ is enough.

Start with $P(x)=\prod_{k\le N^{1/3}}(1-k^2x^2)$. Notice that it vanishes at $1/k$ with $k\le N^{1/3}$, is bounded by $1$ on $[0,N^{-1/3}]$, the degree of $P$ is $2N^{1/3}$ and on the interval $[0,2N^{-1}]$ we have $\log P\ge -2\sum_{k\le N^{1/3}}\frac{4k^2}{N^2}=o(1)$. Now just take the Chebyshev polynomial $Q$ of degree $N^{1/3}$ adjusted to the interval $[0,N^{-1/3}]$. It will jump as you requested between $N^{-1}$ and $2N^{-1}$ and stay below $1$ on $[0,N^{-1/3}]$. The product $P(x)Q(x)$ will then satisfy all your conditions.

Part 2: $n<cN^{1/3}$ is insufficient

One thing that is written in every decent textbook addressing such inequalities is that, in principle, you need no fancy tools for them. Everything can be derived directly from the Lagrange interpolation formula if you can guess the right nodes to use. This is just the general linear programming mumbo-jumbo.

So, the question is how to get the right nodes in your setting. If you do not care about constant factors, then the canonical set of nodes for human consumption is just $\{k^2:k=0,1,\dots,n\}$. Let's recall how it works. Suppose you know the values $P(0),P(1),P(4),\dots,P(n^2)$ of a polynomial of degree $N$ and want to estimate $P'(-y), 0<y<1$ or something like that. You just write $$ P(x)=\sum{k=0^n}P(k)L_k(x) $$ where $$ L_k(x)=\frac {\prod_{m:\ne k}(x-m^2)}{\prod_{m:m\ne k}(k^2-m^2)} $$ and estimate $$ |P'(-y)|\le\max_k|P(k^2)|\sum_k |L'_k(-y)| $$ Now let us estimate $|L'_k(-y)|$. It is just $$ |L_k(-y)|\sum_{m:m\ne k}\frac 1{y+m^2} $$ Now, $$ |L_k(y)|=\frac{\prod_{m:m\ne k}(y+m^2)}{\prod_{m:m\ne k}(|m-k|(m+k))} $$ If $k=0$, then the numerator is $\le (n!)^2\prod_{m=1}^n(1+\frac y{m^2})\le C(n!)^2$ while the denominator is exactly $(n!)^2$. Thus $|L_0(-y)|\le C$ whence $$ |L_0'(-y)|\le C\sum_{m=1}^n\frac 1{y+m^2}\le C\,. $$ If $k>0$, then the numerator in $L_k(-y)$ is at most $Cy\frac 1{k^2}(n!)^2$ (the same count except now $y=|-y-0|$ is present and $y+k^2=|-y-k^2|$ is missing) while the denominator is $$ \frac{k!(n-k)!(n+k)!}{2k(k-1)!}=\frac{(n-k)!(n+k)!}{2} $$ whence $$ |L_k'(-y)|\le \left(\frac 1y+\sum_{m\ge 1}\frac 1{y+m^2}\right)|L_k(-y)|\le \frac C{k^2}\frac{(n!)^2}{(n-k)!(n+k)!}\le \frac C{k^2}e^{-k^2/n}\,. $$ Thus we can easily derive from here that $$ |P'(-y)|\le C\sum_{k=0}^n |P(k^2)|\frac 1{k^2+1}e^{-k^2/n}\,. $$

Now choose an integer $M$ such that $n^3<c M\le c^2N$ where $c>0$ is some small constant. Consider the nodes $x_k=\frac 1M+\frac{k^2}{M}, k=0,1,\dots,n$. If they were available, that would be the end of the story (we would get an estimate $CM$ for the derivative on $[0, M^{-1}]$). Unfortunately only $\frac 1M$ is surely there and the rest have to be approximated. Note that for any $x\in[0,1]$, the nearest to $x$ inverse integer is within $x^2$ from $x$. So, if we choose $z_k$ to be the closest inverse integers to $x_k$, we'll have $|x_k-z_k|\le x_k^2\le 4\frac{k^4}{M^2}$ for $k>0$. Note that the distance from $x_k$ to any neighboring node is about $\frac kM$, so the shift $4\frac{k^4}{M^2}\le 4\frac{n^3}M\frac kM\le 4c\frac kM$ is small compared to it and we, indeed, get $n+1$ different nodes.

Let us now look at how different the corresponding elementary Lagrange polynomials $$ \widetilde L_k(x)=\frac{\prod_{m:m\ne k}(x-z_m)}{\prod_{m:m\ne k}(z_k-z_m)} $$ are from the polynomials $L_k(x)$ built on the canonical nodes $x_k$ on the interval $[0,\frac 1M]$.

The numerator:

We have $\frac {|x-z_m|}{|x-x_m|}\le 1+\frac {4m^2}{M}$, so the numerator can grow at most $\prod_{m=1}^n (1+\frac {4m^2}{M})\le e^{4n^3/M}\le 2$ times.

The denominator:

If $k=0$, then we have $\frac{z_m}{x_m}\ge 1-\frac {4m^2}{M}$, so, as above, we see that it can drop at most twice.

If $k>0$, then $$ \frac {|z_k-z_m|}{|x_k-x_m|}\ge 1-\frac {4(m^4+k^4)}{M|k^2-m^2|}\ge 1-\frac {4(m^3+k^3)}{M|k-m|} $$ We also have $$ \sum_{m:0\le m\le 2k,m\ne k}\frac {4(m^3+k^3)}{M|k-m|}\le C\frac{k^3}M(1+\log k) $$ and $$ \sum_{m:2k\le m\le n}\frac {4(m^3+k^3)}{M|k-m|}\le C\sum_{m:2k\le m\le n}\frac {m^2}{M}\le C\frac {n^3}{M}<1 $$ Thus, the ratio of the denominators we can have against us is at most $Ck^{Ck^3/M}$. That is played against the decay $k^{-2}e^{-k^2/n}$ and it is clear that the decay wins (say, because the power on $k$ in the growth factor is always below $1/2$; we have plenty of leeway here).

For the transition from the value to the derivative, we can change the distances $x-x_m$ even twice without changing the corresponding factor too much, so we are done.

Part I: $CN^{1/3}$ is enough.

Start with $P(x)=\prod_{k\le N^{1/3}}(1-k^2x^2)$. Notice that it vanishes at $1/k$ with $k\le N^{1/3}$, is bounded by $1$ on $[0,N^{-1/3}]$, the degree of $P$ is $2N^{1/3}$ and on the interval $[0,2N^{-1}]$ we have $\log P\ge -2\sum_{k\le N^{1/3}}\frac{4k^2}{N^2}=o(1)$. Now just take the Chebyshev polynomial $Q$ of degree $N^{1/3}$ adjusted to the interval $[0,N^{-1/3}]$. It will jump as you requested between $N^{-1}$ and $2N^{-1}$ and stay below $1$ on $[0,N^{-1/3}]$. The product $P(x)Q(x)$ will then satisfy all your conditions.

To be continued...

Part I: $CN^{1/3}$ is enough.

Start with $P(x)=\prod_{k\le N^{1/3}}(1-k^2x^2)$. Notice that it vanishes at $1/k$ with $k\le N^{1/3}$, is bounded by $1$ on $[0,N^{-1/3}]$, the degree of $P$ is $2N^{1/3}$ and on the interval $[0,2N^{-1}]$ we have $\log P\ge -2\sum_{k\le N^{1/3}}\frac{4k^2}{N^2}=o(1)$. Now just take the Chebyshev polynomial $Q$ of degree $N^{1/3}$ adjusted to the interval $[0,N^{-1/3}]$. It will jump as you requested between $N^{-1}$ and $2N^{-1}$ and stay below $1$ on $[0,N^{-1/3}]$. The product $P(x)Q(x)$ will then satisfy all your conditions.

Part 2: $n<cN^{1/3}$ is insufficient

One thing that is written in every decent textbook addressing such inequalities is that, in principle, you need no fancy tools for them. Everything can be derived directly from the Lagrange interpolation formula if you can guess the right nodes to use. This is just the general linear programming mumbo-jumbo.

So, the question is how to get the right nodes in your setting. If you do not care about constant factors, then the canonical set of nodes for human consumption is just $\{k^2:k=0,1,\dots,n\}$. Let's recall how it works. Suppose you know the values $P(0),P(1),P(4),\dots,P(n^2)$ of a polynomial of degree $N$ and want to estimate $P'(-y), 0<y<1$ or something like that. You just write $$ P(x)=\sum{k=0^n}P(k)L_k(x) $$ where $$ L_k(x)=\frac {\prod_{m:\ne k}(x-m^2)}{\prod_{m:m\ne k}(k^2-m^2)} $$ and estimate $$ |P'(-y)|\le\max_k|P(k^2)|\sum_k |L'_k(-y)| $$ Now let us estimate $|L'_k(-y)|$. It is just $$ |L_k(-y)|\sum_{m:m\ne k}\frac 1{y+m^2} $$ Now, $$ |L_k(y)|=\frac{\prod_{m:m\ne k}(y+m^2)}{\prod_{m:m\ne k}(|m-k|(m+k))} $$ If $k=0$, then the numerator is $\le (n!)^2\prod_{m=1}^n(1+\frac y{m^2})\le C(n!)^2$ while the denominator is exactly $(n!)^2$. Thus $|L_0(-y)|\le C$ whence $$ |L_0'(-y)|\le C\sum_{m=1}^n\frac 1{y+m^2}\le C\,. $$ If $k>0$, then the numerator in $L_k(-y)$ is at most $Cy\frac 1{k^2}(n!)^2$ (the same count except now $y=|-y-0|$ is present and $y+k^2=|-y-k^2|$ is missing) while the denominator is $$ \frac{k!(n-k)!(n+k)!}{2k(k-1)!}=\frac{(n-k)!(n+k)!}{2} $$ whence $$ |L_k'(-y)|\le \left(\frac 1y+\sum_{m\ge 1}\frac 1{y+m^2}\right)|L_k(-y)|\le \frac C{k^2}\frac{(n!)^2}{(n-k)!(n+k)!}\le \frac C{k^2}e^{-k^2/n}\,. $$ Thus we can easily derive from here that $$ |P'(-y)|\le C\sum_{k=0}^n |P(k^2)|\frac 1{k^2+1}e^{-k^2/n}\,. $$

To be continued...