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$\newcommand\S{\mathcal S}\newcommand\R{\mathbb R}$For $x\in(0,1)$, let $g(x):=\frac1x$, and then let $$f(x):=e^{-g(x)-g(1-x)}$$ for $x\in(0,1)$, with $f(x):=0$ for real $x\notin(0,1)$.

Then $f\in\S_0^3(\R)$ -- see the details on this at the end of the answer.

As noted in Willie Wong's comment, you can rescale $f$ vertically and horizontally to get desired $C$ and $A$ (and you said you do not need $B$ to be small).

Details: Everywhere here, $x\in(0,1)$. We have $$f'(x)=P_1(x)f(x),$$ where $P_1(x):=\frac1{x^2}-\frac1{(1-x)^2}$. It is now easy to see by induction that for any $r=0,1,\dots$ $$f^{(r)}(x)=P_r(x)f(x),$$ where $P_r(x)$ is a polynomial in $\frac1x\,,\frac1{1-x}$ of total degree $\le2r$, containing $\le C^r$ monomials with coefficients $\le C^r r!$ in absolute value; here and elsewhere $C$ denotes various universal positive real constants. Also, for any nonnegative integers $m$ and $p$ such that $m+p\le2r\le2q$ we have $$\dfrac1{x^m(1-x)^p}\,e^{-g(x)-g(1-x)}\le e^{-m-p}m^m p^p \le q^{m+p}\le q^{2q}.$$ Thus, $$|f^{(q)}(x)|\le C^q q^{3q}.$$

By a more accurate accounting on the coefficients of the monomials of $P_r(x)$, one can get \begin{equation*} |f^{(q)}(x)|\le C^q q^{2q} \tag{10}\label{10} \end{equation*} and thus $f\in\S_0^2(\R)$.

Indeed, for $r=0,1,\dots$, \begin{equation*} f^{(r+1)}(x)=(P_r(x)f(x))'=(P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}))f(x), \end{equation*} so that \begin{equation*} P_{r+1}(x)=P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}). \end{equation*} Consider any monomial $c(\frac1x)^i(\frac1{1-x})^{p-i}$ of the polynomial $P_r(x)$ in $\frac1x\,,\frac1{1-x}$; this monomial is of total degree $p$ and with coefficient $c$. The derivative of this monomial in $x$ gives is the sum of four monomials of $P_{r+1}(x)$. Two of these four monomials are monomials of $P'_r(x)$ (say they are of the 1st kind) and the other two monomials are monomials of $P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2})$ (say they are of the 2nd kind).

The total degree of each of the two monomials of the 1st kind is $p+1$ and the absolute value of its coefficients is $\le p|c|\le2r|c|$. The total degree of each of the two monomials of the 2nd kind is $p+2$ and the absolute value of its coefficient is $\le |c|$.

So, if a monomial of $P_q(x)$ is obtained by $k$ steps of the 1st kind (and, hence, $q-k$ of the 2nd kind), then it is of the form \begin{equation*} M=c(\tfrac1x)^i(\tfrac1{1-x})^{p-i} \end{equation*} with \begin{equation*} p=k+2(q-k)\quad\text{and}\quad |c|\le (2q)^k\le2^q q^k. \end{equation*} Recalling now the inequality $u^m e^{-u}\le m^m$ for $u>0$ and $m\ge0$ (with $0^0:=1$), we see that \begin{equation*} |Mf(x)|\le2^q q^k i^i (p-i)^{p-i}\le2^q q^k p^p=2^q h(k), \end{equation*} where $h(k):=q^k (k+2(q-k))^{k+2(q-k)}$. Note that $h$ is log convex, $h(0)=(2q)^{2q}$, and $h(q)=q^{2q}$. So, $h(k)\le(2q)^{2q}$ for all $k\in[0,q]$. So, for any monomial $M$ of $P_q(x)$ we have \begin{equation*} |Mf(x)|\le2^q (2q)^{2q}=8^q q^{2q}. \end{equation*} Since $P_q(x)$ contains $\le4^q$ monomials, we conclude that \eqref{10} holds. $\quad\Box$

$\newcommand\S{\mathcal S}\newcommand\R{\mathbb R}$For $x\in(0,1)$, let $g(x):=\frac1x$, and then let $$f(x):=e^{-g(x)-g(1-x)}$$ for $x\in(0,1)$, with $f(x):=0$ for real $x\notin(0,1)$.

Then $f\in\S_0^3(\R)$ -- see the details on this at the end of the answer.

As noted in Willie Wong's comment, you can rescale $f$ vertically and horizontally to get desired $C$ and $A$ (and you said you do not need $B$ to be small).

Details: Everywhere here, $x\in(0,1)$. We have $$f'(x)=P_1(x)f(x),$$ where $P_1(x):=\frac1{x^2}-\frac1{(1-x)^2}$. It is now easy to see by induction that for any $r=0,1,\dots$ $$f^{(r)}(x)=P_r(x)f(x),$$ where $P_r(x)$ is a polynomial in $\frac1x\,,\frac1{1-x}$ of total degree $\le2r$, containing $\le C^r$ monomials with coefficients $\le C^r r!$ in absolute value; here and elsewhere $C$ denotes various universal positive real constants. Also, for any nonnegative integers $m$ and $p$ such that $m+p\le2r\le2q$ we have $$\dfrac1{x^m(1-x)^p}\,e^{-g(x)-g(1-x)}\le e^{-m-p}m^m p^p \le q^{m+p}\le q^{2q}.$$ Thus, $$|f^{(q)}(x)|\le C^q q^{3q}.$$

By a more accurate accounting on the coefficients of the monomials of $P_r(x)$, one can get \begin{equation*} |f^{(q)}(x)|\le C^q q^{2q} \tag{10}\label{10} \end{equation*} and thus $f\in\S_0^2(\R)$.

Indeed, for $r=0,1,\dots$, \begin{equation*} f^{(r+1)}(x)=(P_r(x)f(x))'=(P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}))f(x), \end{equation*} so that \begin{equation*} P_{r+1}(x)=P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}). \end{equation*} Consider any monomial $m(x)=c(\frac1x)^i(\frac1{1-x})^{p-i}$ of the polynomial $P_r(x)$ in $\frac1x\,,\frac1{1-x}$; this monomial is of total degree $p$ and with coefficient $c$. Then $(m(x)f(x))'/f(x)$ is the sum of four monomials of $P_{r+1}(x)$. Two of these four monomials are monomials of $P'_r(x)$ (say they are of the 1st kind) and the other two monomials are monomials of $P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2})$ (say they are of the 2nd kind).

The total degree of each of the two monomials of the 1st kind is $p+1$ and the absolute value of its coefficients is $\le p|c|\le2r|c|$. The total degree of each of the two monomials of the 2nd kind is $p+2$ and the absolute value of its coefficient is $\le |c|$.

So, if a monomial of $P_q(x)$ is obtained by $k$ steps of the 1st kind (and, hence, $q-k$ of the 2nd kind), then it is of the form \begin{equation*} M(x)=c(\tfrac1x)^i(\tfrac1{1-x})^{p-i} \end{equation*} with \begin{equation*} p=k+2(q-k)\quad\text{and}\quad |c|\le (2q)^k\le2^q q^k. \end{equation*} Recalling now the inequality $u^s e^{-u}\le s^s$ for $u>0$ and $s\ge0$ (with $0^0:=1$), we see that \begin{equation*} |M(x)f(x)|\le2^q q^k i^i (p-i)^{p-i}\le2^q q^k p^p=2^q h(k), \end{equation*} where $h(k):=q^k (k+2(q-k))^{k+2(q-k)}$. Note that $h$ is log convex, $h(0)=(2q)^{2q}$, and $h(q)=q^{2q}$. So, $h(k)\le(2q)^{2q}$ for all $k\in[0,q]$. So, for any monomial $M(x)$ of $P_q(x)$ we have \begin{equation*} |M(x)f(x)|\le2^q (2q)^{2q}=8^q q^{2q}. \end{equation*} Since $P_q(x)$ contains $\le4^q$ monomials, we conclude that \eqref{10} holds. $\quad\Box$

$\newcommand\S{\mathcal S}\newcommand\R{\mathbb R}$For $x\in(0,1)$, let $g(x):=\frac1x$, and then let $$f(x):=e^{-g(x)-g(1-x)}$$ for $x\in(0,1)$, with $f(x):=0$ for real $x\notin(0,1)$.

Then $f\in\S_0^3(\R)$ -- see the details on this at the end of the answer.

As noted in Willie Wong's comment, you can rescale $f$ vertically and horizontally to get desired $C$ and $A$ (and you said you do not need $B$ to be small).

Details: Everywhere here, $x\in(0,1)$. We have $$f'(x)=P_1(x)f(x),$$ where $P_1(x):=\frac1{x^2}-\frac1{(1-x)^2}$. It is now easy to see by induction that for any $r=0,1,\dots$ $$f^{(r)}(x)=P_r(x)f(x),$$ where $P_r(x)$ is a polynomial in $\frac1x\,,\frac1{1-x}$ of total degree $\le2r$, containing $\le C^r$ monomials with coefficients $\le C^r r!$ in absolute value; here and elsewhere $C$ denotes various universal positive real constants. Also, for any nonnegative integers $m$ and $p$ such that $m+p\le2r\le2q$ we have $$\dfrac1{x^m(1-x)^p}\,e^{-g(x)-g(1-x)}\le e^{-m-p}m^m p^p \le q^{m+p}\le q^{2q}.$$ Thus, $$|f^{(q)}(x)|\le C^q q^{3q}.$$

By a more accurate accounting on the coefficients of the monomials of $P_r(x)$, one can get \begin{equation*} |f^{(q)}(x)|\le C^q q^{2q} \tag{10}\label{10} \end{equation*} and thus $f\in\S_0^2(\R)$.

Indeed, for $r=0,1,\dots$, \begin{equation*} f^{(r+1)}(x)=(P_r(x)f(x))'=(P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}))f(x), \end{equation*} so that \begin{equation*} P_{r+1}(x)=P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}). \end{equation*} Consider any monomial $c(\frac1x)^i(\frac1{1-x})^{p-i}$ in the polynomial $P_r(x)$ in $\frac1x\,,\frac1{1-x}$; this polynomial is of total degree $p$ and with coefficient $c$. This monomial of $P_r(x)$ gives rise to four monomials of $P_{r+1}(x)$. Two of these four monomials are monomials of $P'_r(x)$ (say they are of the 1st kind) and the other two monomials are monomials of $P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2})$ (say they are of the 2nd kind).

The total degree of each of the monomials of the 1st kind is $p+1$ and the absolute values of their coefficients are $\le p|c|\le2r|c|$. The total degree of each of the monomials of the 2nd kind is $p+2$ and the absolute values of their coefficients are $\le |c|$.

So, if a monomial of $P_q(x)$ is obtained by $k$ steps of the 1st kind (and, hence, $q-k$ of the 2nd kind), then it is of the form \begin{equation*} M=c(\tfrac1x)^i(\tfrac1{1-x})^{p-i} \end{equation*} with \begin{equation*} p=k+2(q-k),\quad |c|\le (2q)^k\le2^q q^k. \end{equation*} Recalling now the inequality $u^m e^{-u}\le m^m$ for $u>0$ and $m\ge0$ (with $0^0:=1$), we see that \begin{equation*} |Mf(x)|\le2^q q^k i^i (p-i)^{p-i}\le2^q q^k p^p=2^q h(k), \end{equation*} where $h(k):=q^k (k+2(q-k))^{k+2(q-k)}$. Note that $h$ is log convex, $h(0)=(2q)^{2q}$ and $h(q)=q^{2q}$. So, $h(k)\le(2q)^{2q}$ for all $k\in[0,q]$. So, for any monomial $M$ of $P_q(x)$ we have \begin{equation*} |Mf(x)|\le2^q (2q)^{2q}=8^q q^{2q}. \end{equation*} Since $P_q(x)$ contains $\le4^q$ monomials, we conclude that \eqref{10} holds. $\quad\Box$

$\newcommand\S{\mathcal S}\newcommand\R{\mathbb R}$For $x\in(0,1)$, let $g(x):=\frac1x$, and then let $$f(x):=e^{-g(x)-g(1-x)}$$ for $x\in(0,1)$, with $f(x):=0$ for real $x\notin(0,1)$.

Then $f\in\S_0^3(\R)$ -- see the details on this at the end of the answer.

As noted in Willie Wong's comment, you can rescale $f$ vertically and horizontally to get desired $C$ and $A$ (and you said you do not need $B$ to be small).

Details: Everywhere here, $x\in(0,1)$. We have $$f'(x)=P_1(x)f(x),$$ where $P_1(x):=\frac1{x^2}-\frac1{(1-x)^2}$. It is now easy to see by induction that for any $r=0,1,\dots$ $$f^{(r)}(x)=P_r(x)f(x),$$ where $P_r(x)$ is a polynomial in $\frac1x\,,\frac1{1-x}$ of total degree $\le2r$, containing $\le C^r$ monomials with coefficients $\le C^r r!$ in absolute value; here and elsewhere $C$ denotes various universal positive real constants. Also, for any nonnegative integers $m$ and $p$ such that $m+p\le2r\le2q$ we have $$\dfrac1{x^m(1-x)^p}\,e^{-g(x)-g(1-x)}\le e^{-m-p}m^m p^p \le q^{m+p}\le q^{2q}.$$ Thus, $$|f^{(q)}(x)|\le C^q q^{3q}.$$

By a more accurate accounting on the coefficients of the monomials of $P_r(x)$, one can get \begin{equation*} |f^{(q)}(x)|\le C^q q^{2q} \tag{10}\label{10} \end{equation*} and thus $f\in\S_0^2(\R)$.

Indeed, for $r=0,1,\dots$, \begin{equation*} f^{(r+1)}(x)=(P_r(x)f(x))'=(P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}))f(x), \end{equation*} so that \begin{equation*} P_{r+1}(x)=P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}). \end{equation*} Consider any monomial $c(\frac1x)^i(\frac1{1-x})^{p-i}$ of the polynomial $P_r(x)$ in $\frac1x\,,\frac1{1-x}$; this monomial is of total degree $p$ and with coefficient $c$. The derivative of this monomial in $x$ gives is the sum of four monomials of $P_{r+1}(x)$. Two of these four monomials are monomials of $P'_r(x)$ (say they are of the 1st kind) and the other two monomials are monomials of $P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2})$ (say they are of the 2nd kind).

The total degree of each of the two monomials of the 1st kind is $p+1$ and the absolute value of its coefficients is $\le p|c|\le2r|c|$. The total degree of each of the two monomials of the 2nd kind is $p+2$ and the absolute value of its coefficient is $\le |c|$.

So, if a monomial of $P_q(x)$ is obtained by $k$ steps of the 1st kind (and, hence, $q-k$ of the 2nd kind), then it is of the form \begin{equation*} M=c(\tfrac1x)^i(\tfrac1{1-x})^{p-i} \end{equation*} with \begin{equation*} p=k+2(q-k)\quad\text{and}\quad |c|\le (2q)^k\le2^q q^k. \end{equation*} Recalling now the inequality $u^m e^{-u}\le m^m$ for $u>0$ and $m\ge0$ (with $0^0:=1$), we see that \begin{equation*} |Mf(x)|\le2^q q^k i^i (p-i)^{p-i}\le2^q q^k p^p=2^q h(k), \end{equation*} where $h(k):=q^k (k+2(q-k))^{k+2(q-k)}$. Note that $h$ is log convex, $h(0)=(2q)^{2q}$, and $h(q)=q^{2q}$. So, $h(k)\le(2q)^{2q}$ for all $k\in[0,q]$. So, for any monomial $M$ of $P_q(x)$ we have \begin{equation*} |Mf(x)|\le2^q (2q)^{2q}=8^q q^{2q}. \end{equation*} Since $P_q(x)$ contains $\le4^q$ monomials, we conclude that \eqref{10} holds. $\quad\Box$

$\newcommand\S{\mathcal S}\newcommand\R{\mathbb R}$For $x\in(0,1)$, let $g(x):=\frac1x$, and then let $$f(x):=e^{-g(x)-g(1-x)}$$ for $x\in(0,1)$, with $f(x):=0$ for real $x\notin(0,1)$.

Then $f\in\S_0^3(\R)$ -- see the details on this at the end of the answer.

As noted in Willie Wong's comment, you can rescale $f$ vertically and horizontally to get desired $C$ and $A$ (and you said you do not need $B$ to be small).

Details: Everywhere here, $x\in(0,1)$. We have $$f'(x)=P_1(x)f(x),$$ where $P_1(x):=\frac1{x^2}-\frac1{(1-x)^2}$. It is now easy to see by induction that for any $r=1,2,\dots$ $$f^{(r)}(x)=P_r(x)f(x),$$ where $P_r(x)$ is a polynomial in $\frac1x\,,\frac1{1-x}$ of total degree $\le2r$, containing $\le C^r$ monomials with coefficients $\le C^r r!$ in absolute value; here and elsewhere $C$ denotes various universal positive real constants. Also, for any nonnegative integers $m$ and $p$ such that $m+p\le2r\le2q$ we have $$\dfrac1{x^m(1-x)^p}\,e^{-g(x)-g(1-x)}\le e^{-m-p}m^m p^p \le q^{m+p}\le q^{2q}.$$ Thus, $$|f^{(q)}(x)|\le C^q q^{3q}.$$

By a more accurate accounting on the coefficients of the monomials of $P_r(x)$, one can get \begin{equation*} |f^{(q)}(x)|\le C^q q^{2q} \tag{10}\label{10} \end{equation*} and thus $f\in\S_0^2(\R)$.

Indeed, \begin{equation*} f^{(r+1)}(x)=(P_r(x)f(x))'=(P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}))f(x), \end{equation*} so that \begin{equation*} P_{r+1}(x)=P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}). \end{equation*} Consider any monomial $c(\frac1x)^i(\frac1{1-x})^{p-i}$ in the polynomial $P_r(x)$ in $\frac1x\,,\frac1{1-x}$; this polynomial is of total degree $p$ and with coefficient $c$. This monomial of $P_r(x)$ gives rise to four monomials of $P_{r+1}(x)$. Two of these four monomials are monomials of $P'_r(x)$ (say they are of the 1st kind) and the other two monomials are monomials of $P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2})$ (say they are of the 2nd kind).

The total degree of each of the monomials of the 1st kind is $p+1$ and the absolute values of their coefficients are $\le p|c|\le2r|c|$. The total degree of each of the monomials of the 2nd kind is $p+2$ and the absolute values of their coefficients are $\le |c|$.

So, if a monomial of $P_q(x)$ is obtained by $k$ steps of the 1st kind (and, hence, $q-k$ of the 2nd kind), then it is of the form \begin{equation*} M=c(\tfrac1x)^i(\tfrac1{1-x})^{p-i} \end{equation*} with \begin{equation*} p=k+2(q-k),\quad |c|\le (2q)^k\le2^q q^k. \end{equation*} Recalling now the inequality $u^m e^{-u}\le m^m$ for $u>0$ and $m\ge0$ (with $0^0:=1$), we see that \begin{equation*} |Mf(x)|\le2^q q^k i^i (p-i)^{p-i}\le2^q q^k p^p=2^q h(k), \end{equation*} where $h(k):=q^k (k+2(q-k))^{k+2(q-k)}$. Note that $h$ is log convex, $h(0)=(2q)^{2q}$ and $h(q)=q^{2q}$. So, $h(k)\le(2q)^{2q}$ for all $k\in[0,q]$. So, for any monomial $M$ of $P_q(x)$ we have \begin{equation*} |Mf(x)|\le2^q (2q)^{2q}=8^q q^{2q}. \end{equation*} Since $P_q(x)$ contains $\le4^q$ monomials, we conclude that \eqref{10} holds. $\quad\Box$

$\newcommand\S{\mathcal S}\newcommand\R{\mathbb R}$For $x\in(0,1)$, let $g(x):=\frac1x$, and then let $$f(x):=e^{-g(x)-g(1-x)}$$ for $x\in(0,1)$, with $f(x):=0$ for real $x\notin(0,1)$.

Then $f\in\S_0^3(\R)$ -- see the details on this at the end of the answer.

As noted in Willie Wong's comment, you can rescale $f$ vertically and horizontally to get desired $C$ and $A$ (and you said you do not need $B$ to be small).

Details: Everywhere here, $x\in(0,1)$. We have $$f'(x)=P_1(x)f(x),$$ where $P_1(x):=\frac1{x^2}-\frac1{(1-x)^2}$. It is now easy to see by induction that for any $r=0,1,\dots$ $$f^{(r)}(x)=P_r(x)f(x),$$ where $P_r(x)$ is a polynomial in $\frac1x\,,\frac1{1-x}$ of total degree $\le2r$, containing $\le C^r$ monomials with coefficients $\le C^r r!$ in absolute value; here and elsewhere $C$ denotes various universal positive real constants. Also, for any nonnegative integers $m$ and $p$ such that $m+p\le2r\le2q$ we have $$\dfrac1{x^m(1-x)^p}\,e^{-g(x)-g(1-x)}\le e^{-m-p}m^m p^p \le q^{m+p}\le q^{2q}.$$ Thus, $$|f^{(q)}(x)|\le C^q q^{3q}.$$

By a more accurate accounting on the coefficients of the monomials of $P_r(x)$, one can get \begin{equation*} |f^{(q)}(x)|\le C^q q^{2q} \tag{10}\label{10} \end{equation*} and thus $f\in\S_0^2(\R)$.

Indeed, for $r=0,1,\dots$, \begin{equation*} f^{(r+1)}(x)=(P_r(x)f(x))'=(P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}))f(x), \end{equation*} so that \begin{equation*} P_{r+1}(x)=P'_r(x)+P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}). \end{equation*} Consider any monomial $c(\frac1x)^i(\frac1{1-x})^{p-i}$ in the polynomial $P_r(x)$ in $\frac1x\,,\frac1{1-x}$; this polynomial is of total degree $p$ and with coefficient $c$. This monomial of $P_r(x)$ gives rise to four monomials of $P_{r+1}(x)$. Two of these four monomials are monomials of $P'_r(x)$ (say they are of the 1st kind) and the other two monomials are monomials of $P_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2})$ (say they are of the 2nd kind).

The total degree of each of the monomials of the 1st kind is $p+1$ and the absolute values of their coefficients are $\le p|c|\le2r|c|$. The total degree of each of the monomials of the 2nd kind is $p+2$ and the absolute values of their coefficients are $\le |c|$.

So, if a monomial of $P_q(x)$ is obtained by $k$ steps of the 1st kind (and, hence, $q-k$ of the 2nd kind), then it is of the form \begin{equation*} M=c(\tfrac1x)^i(\tfrac1{1-x})^{p-i} \end{equation*} with \begin{equation*} p=k+2(q-k),\quad |c|\le (2q)^k\le2^q q^k. \end{equation*} Recalling now the inequality $u^m e^{-u}\le m^m$ for $u>0$ and $m\ge0$ (with $0^0:=1$), we see that \begin{equation*} |Mf(x)|\le2^q q^k i^i (p-i)^{p-i}\le2^q q^k p^p=2^q h(k), \end{equation*} where $h(k):=q^k (k+2(q-k))^{k+2(q-k)}$. Note that $h$ is log convex, $h(0)=(2q)^{2q}$ and $h(q)=q^{2q}$. So, $h(k)\le(2q)^{2q}$ for all $k\in[0,q]$. So, for any monomial $M$ of $P_q(x)$ we have \begin{equation*} |Mf(x)|\le2^q (2q)^{2q}=8^q q^{2q}. \end{equation*} Since $P_q(x)$ contains $\le4^q$ monomials, we conclude that \eqref{10} holds. $\quad\Box$