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Iosif Pinelis
Iosif Pinelis
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$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

In the excellent answer, Willie Wong offered a construction of a measure with a lacunary support set, disproving the conjecture. Let me offer another construction, where the support of the measure is a continuum; however, the construction is still lacunary in the sense that it involves sequences diverging faster than any exponential sequence.

The main idea is this. Let $p>0$ and $s>0$. Let $\|f\|_p:=\|f\|_{L^p(\mu)}$. Let $f(x)=x$ for all real $x$. Let us start with this trivial lemma:

Lemma 1. If $\mu(dx)=\frac1s\,e^{-x/s}I\{x>0\}$, where $I$ denotes the indicator, then \begin{equation*} \|f\|_p^p=\int_0^\infty x^p \,e^{-x/s}\frac{dx}s=s^p\Ga(p+1). \end{equation*}

Then, by Stirling's formula, $\ln\|f\|_p\sim \ln(ps)$ as $s,p\to\infty$. So, if we alternate $s$ between $p$ and $p^2$, $\ln\|f\|_p/\ln p$ will alternate between $2$ and $3$. Now we need to glue pieces of such two alternating sequences of measures, with $s$ alternating between $p$ and $p^2$, to get one measure one. The guiding idea in doing this is that for large $p$ the mass $m_p(dx):=x^p \,e^{-x/s}\frac{dx}s$$m_p(dx):=x^p \,e^{-x/s}I\{x>0\}\frac{dx}s$ is mostly concentrated near the point $ps$.

To proceed, we shall need two more simple lemmas, which will be proved at the end of this answer.

Lemma 2. For every $y\in[0,ps]$ there is some $c\in[1/2,1]$ such that \begin{equation*} \int_y^\infty x^p \,e^{-x/s}\frac{dx}s=cs^p\Ga(p+1). \end{equation*}

Lemma 3. For every $y\ge2ps$, \begin{equation*} \int_y^\infty x^p \,e^{-x/s}\frac{dx}s \le\exp\{p\ln(2ps)-y/(2s)\}. \end{equation*}

Let \begin{equation*} \mu(dx)=\sum_{j=1}^\infty\frac1{s_j}\,e^{-x/s_j}I\{x_j<x<x_{j+1}\}, \end{equation*}\begin{equation*} \mu(dx)=\sum_{j=1}^\infty\frac{dx}{s_j}\,e^{-x/s_j}I\{x_j<x<x_{j+1}\}, \end{equation*} where \begin{equation*} s_j:= \begin{cases} p_j&\text{ if $j$ is odd}, \\ p_j^2&\text{ if $j$ is even}, \end{cases} \qquad p_j:=2^{2^j},\qquad x_j:=p_j s_j, \end{equation*} so that $p_{j+1}=p_j^2$. Then \begin{multline*} \|f\|_p^p=\int_\R x^p\mu(dx)=\sum_{j=1}^\infty I_j(p),\\ I_j(p):=\int_{x_j}^{x_{j+1}} \exp\{g_{j,p}(x)\}\frac{dx}{s_j},\\ g_{j,p}(x):=p\ln x-x/s_j. \end{multline*}

In view of Lemma 1 and Stirling's formula, for large odd $k$ one has \begin{equation*} I_k(p_k) \le s_k^{p_k} p_k^{p_k}=p_k^{2p_k} \end{equation*}\begin{equation*} I_k(p_k) \le s_k^{p_k} p_k^{p_k}=p_k^{2p_k} \tag{1} \end{equation*} and, for $j<k$, \begin{equation*} I_j(p_k)\le s_j^{p_k} p_k^{p_k} \le p_j^{2p_k} p_k^{p_k} \le p_k^{2p_k}, \end{equation*} whence \begin{equation*} \sum_{j<k}I_j(p_k) \le k p_k^{2p_k}=p_k^{(2+o(1))p_k}. \end{equation*}\begin{equation*} \sum_{j<k}I_j(p_k) \le k p_k^{2p_k}=p_k^{(2+o(1))p_k}. \tag{2} \end{equation*}

Next, for large odd $k$ and $j>k$, \begin{equation*} p_k\ln(2p_ks_j)-p_k\ln(p_ks_k) \le p_k\ln s_j\le p_k\ln(p_j^2)\le p_j^{1/2}\ln(p_j^2)\le p_j/4, \end{equation*} whence, by Lemma 3, \begin{multline*} I_j(p_k)\le\exp\{p_k\ln(2p_ks_j)-p_j/2\}\le\exp\{p_k\ln(p_ks_k)-p_j/4\} \\ \le2^{-(j-k)}\exp\{p_k\ln(p_ks_k)\}, \end{multline*} so that \begin{equation*} \sum_{j>k}I_j(p_k) \le p_k^{2p_k}. \end{equation*}\begin{equation*} \sum_{j>k}I_j(p_k) \le p_k^{2p_k}. \tag{3} \end{equation*}

Collecting (1), (2), (3), for large odd $k$ we have \begin{equation*} \ln\|f\|_{p_k}\lesssim 2\ln p_k. \end{equation*}

On the other hand, for large even $k$, by Lemma 2 and Stirling's formula,
\begin{equation*} \ln\|f\|_{p_k}\ge\frac1{p_k}\,\ln I_k(p_k)= \ln(s_k p_k^{(1+o(1))}) = \ln(p_k^{2} p_k^{(1+o(1))}) \sim3\ln p_k. \end{equation*}\begin{equation*} \ln\|f\|_{p_k}\ge\frac1{p_k}\,\ln I_k(p_k)= \ln(s_k p_k^{1+o(1)}) = \ln(p_k^{2} p_k^{1+o(1)}) \sim3\ln p_k. \end{equation*}

So, $\ln\|f\|_p/\ln p$ does not converge as $p\to\infty$.

In conclusion, let us prove the lemmas.

Proof of Lemma 1. This is obvious.

Proof of Lemma 2. Write \begin{equation} x^p \,e^{-x/s}=e^{g(x)},\quad g(x):=p\ln x-x/s. \tag{4} \end{equation} Then $g'(ps)=0$ and $g''$$g''(x)$ is increasing in $x>0$. So, $g(ps-u)<g(ps+u)$ for $u\in(0,ps)$. So, for every $y\in[0,ps]$, \begin{equation*} \int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s \ge\int_y^\infty x^p \,e^{-x/s}\frac{dx}s \ge\frac12\,\int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s. \end{equation*} To complete the proof of Lemma 2, it remains to refer to Lemma 1.

Proof of Lemma 3. For $g$ as in (4) and for for $x>2ps$, we have $g'(x)=\frac px-\frac1s\le-\frac1{2s}$ and hence \begin{equation} g(x)\le g(2ps)-\frac1{2s}(x-2ps)=p\ln(2ps)-\frac x{2s}. \end{equation} Now Lemma 3 easily follows.

$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

In the excellent answer, Willie Wong offered a construction of a measure with a lacunary support set, disproving the conjecture. Let me offer another construction, where the support of the measure is a continuum; however, the construction is still lacunary in the sense that it involves sequences diverging faster than any exponential sequence.

The main idea is this. Let $p>0$ and $s>0$. Let $\|f\|_p:=\|f\|_{L^p(\mu)}$. Let $f(x)=x$ for all real $x$. Let us start with this trivial lemma:

Lemma 1. If $\mu(dx)=\frac1s\,e^{-x/s}I\{x>0\}$, where $I$ denotes the indicator, then \begin{equation*} \|f\|_p^p=\int_0^\infty x^p \,e^{-x/s}\frac{dx}s=s^p\Ga(p+1). \end{equation*}

Then, by Stirling's formula, $\ln\|f\|_p\sim \ln(ps)$ as $s,p\to\infty$. So, if we alternate $s$ between $p$ and $p^2$, $\ln\|f\|_p/\ln p$ will alternate between $2$ and $3$. Now we need glue pieces of such two alternating sequences of measures, with $s$ alternating between $p$ and $p^2$, to get one measure one. The guiding idea in doing this is that for large $p$ the mass $m_p(dx):=x^p \,e^{-x/s}\frac{dx}s$ is mostly concentrated near the point $ps$.

To proceed, we shall need two more simple lemmas, which will be proved at the end of this answer.

Lemma 2. For every $y\in[0,ps]$ there is some $c\in[1/2,1]$ such that \begin{equation*} \int_y^\infty x^p \,e^{-x/s}\frac{dx}s=cs^p\Ga(p+1). \end{equation*}

Lemma 3. For every $y\ge2ps$, \begin{equation*} \int_y^\infty x^p \,e^{-x/s}\frac{dx}s \le\exp\{p\ln(2ps)-y/(2s)\}. \end{equation*}

Let \begin{equation*} \mu(dx)=\sum_{j=1}^\infty\frac1{s_j}\,e^{-x/s_j}I\{x_j<x<x_{j+1}\}, \end{equation*} where \begin{equation*} s_j:= \begin{cases} p_j&\text{ if $j$ is odd}, \\ p_j^2&\text{ if $j$ is even}, \end{cases} \qquad p_j:=2^{2^j},\qquad x_j:=p_j s_j, \end{equation*} so that $p_{j+1}=p_j^2$. Then \begin{multline*} \|f\|_p^p=\int_\R x^p\mu(dx)=\sum_{j=1}^\infty I_j(p),\\ I_j(p):=\int_{x_j}^{x_{j+1}} \exp\{g_{j,p}(x)\}\frac{dx}{s_j},\\ g_{j,p}(x):=p\ln x-x/s_j. \end{multline*}

In view of Lemma 1 and Stirling's formula, for large odd $k$ one has \begin{equation*} I_k(p_k) \le s_k^{p_k} p_k^{p_k}=p_k^{2p_k} \end{equation*} and, for $j<k$, \begin{equation*} I_j(p_k)\le s_j^{p_k} p_k^{p_k} \le p_j^{2p_k} p_k^{p_k} \le p_k^{2p_k}, \end{equation*} whence \begin{equation*} \sum_{j<k}I_j(p_k) \le k p_k^{2p_k}=p_k^{(2+o(1))p_k}. \end{equation*}

Next, for large odd $k$ and $j>k$, \begin{equation*} p_k\ln(2p_ks_j)-p_k\ln(p_ks_k) \le p_k\ln s_j\le p_k\ln(p_j^2)\le p_j^{1/2}\ln(p_j^2)\le p_j/4, \end{equation*} whence, by Lemma 3, \begin{multline*} I_j(p_k)\le\exp\{p_k\ln(2p_ks_j)-p_j/2\}\le\exp\{p_k\ln(p_ks_k)-p_j/4\} \\ \le2^{-(j-k)}\exp\{p_k\ln(p_ks_k)\}, \end{multline*} so that \begin{equation*} \sum_{j>k}I_j(p_k) \le p_k^{2p_k}. \end{equation*}

Collecting (1), (2), (3), for large odd $k$ we have \begin{equation*} \ln\|f\|_{p_k}\lesssim 2\ln p_k. \end{equation*}

On the other hand, for large even $k$, by Lemma 2 and Stirling's formula,
\begin{equation*} \ln\|f\|_{p_k}\ge\frac1{p_k}\,\ln I_k(p_k)= \ln(s_k p_k^{(1+o(1))}) = \ln(p_k^{2} p_k^{(1+o(1))}) \sim3\ln p_k. \end{equation*}

So, $\ln\|f\|_p/\ln p$ does not converge as $p\to\infty$.

In conclusion, let us prove the lemmas.

Proof of Lemma 1. This is obvious.

Proof of Lemma 2. Write \begin{equation} x^p \,e^{-x/s}=e^{g(x)},\quad g(x):=p\ln x-x/s. \tag{4} \end{equation} Then $g'(ps)=0$ and $g''$ is increasing. So, $g(ps-u)<g(ps+u)$ for $u\in(0,ps)$. So, for every $y\in[0,ps]$, \begin{equation*} \int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s \ge\int_y^\infty x^p \,e^{-x/s}\frac{dx}s \ge\frac12\,\int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s. \end{equation*} To complete the proof of Lemma 2, it remains to refer to Lemma 1.

Proof of Lemma 3. For $g$ as in (4) and for for $x>2ps$, we have $g'(x)=\frac px-\frac1s\le-\frac1{2s}$ and hence \begin{equation} g(x)\le g(2ps)-\frac1{2s}(x-2ps)=p\ln(2ps)-\frac x{2s}. \end{equation} Now Lemma 3 easily follows.

$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

In the excellent answer, Willie Wong offered a construction of a measure with a lacunary support set, disproving the conjecture. Let me offer another construction, where the support of the measure is a continuum; however, the construction is still lacunary in the sense that it involves sequences diverging faster than any exponential sequence.

The main idea is this. Let $p>0$ and $s>0$. Let $\|f\|_p:=\|f\|_{L^p(\mu)}$. Let $f(x)=x$ for all real $x$. Let us start with this trivial lemma:

Lemma 1. If $\mu(dx)=\frac1s\,e^{-x/s}I\{x>0\}$, where $I$ denotes the indicator, then \begin{equation*} \|f\|_p^p=\int_0^\infty x^p \,e^{-x/s}\frac{dx}s=s^p\Ga(p+1). \end{equation*}

Then, by Stirling's formula, $\ln\|f\|_p\sim \ln(ps)$ as $s,p\to\infty$. So, if we alternate $s$ between $p$ and $p^2$, $\ln\|f\|_p/\ln p$ will alternate between $2$ and $3$. Now we need to glue pieces of such two alternating sequences of measures, with $s$ alternating between $p$ and $p^2$, to get one measure. The guiding idea in doing this is that for large $p$ the mass $m_p(dx):=x^p \,e^{-x/s}I\{x>0\}\frac{dx}s$ is mostly concentrated near the point $ps$.

To proceed, we shall need two more simple lemmas, which will be proved at the end of this answer.

Lemma 2. For every $y\in[0,ps]$ there is some $c\in[1/2,1]$ such that \begin{equation*} \int_y^\infty x^p \,e^{-x/s}\frac{dx}s=cs^p\Ga(p+1). \end{equation*}

Lemma 3. For every $y\ge2ps$, \begin{equation*} \int_y^\infty x^p \,e^{-x/s}\frac{dx}s \le\exp\{p\ln(2ps)-y/(2s)\}. \end{equation*}

Let \begin{equation*} \mu(dx)=\sum_{j=1}^\infty\frac{dx}{s_j}\,e^{-x/s_j}I\{x_j<x<x_{j+1}\}, \end{equation*} where \begin{equation*} s_j:= \begin{cases} p_j&\text{ if $j$ is odd}, \\ p_j^2&\text{ if $j$ is even}, \end{cases} \qquad p_j:=2^{2^j},\qquad x_j:=p_j s_j, \end{equation*} so that $p_{j+1}=p_j^2$. Then \begin{multline*} \|f\|_p^p=\int_\R x^p\mu(dx)=\sum_{j=1}^\infty I_j(p),\\ I_j(p):=\int_{x_j}^{x_{j+1}} \exp\{g_{j,p}(x)\}\frac{dx}{s_j},\\ g_{j,p}(x):=p\ln x-x/s_j. \end{multline*}

In view of Lemma 1 and Stirling's formula, for large odd $k$ one has \begin{equation*} I_k(p_k) \le s_k^{p_k} p_k^{p_k}=p_k^{2p_k} \tag{1} \end{equation*} and, for $j<k$, \begin{equation*} I_j(p_k)\le s_j^{p_k} p_k^{p_k} \le p_j^{2p_k} p_k^{p_k} \le p_k^{2p_k}, \end{equation*} whence \begin{equation*} \sum_{j<k}I_j(p_k) \le k p_k^{2p_k}=p_k^{(2+o(1))p_k}. \tag{2} \end{equation*}

Next, for large odd $k$ and $j>k$, \begin{equation*} p_k\ln(2p_ks_j)-p_k\ln(p_ks_k) \le p_k\ln s_j\le p_k\ln(p_j^2)\le p_j^{1/2}\ln(p_j^2)\le p_j/4, \end{equation*} whence, by Lemma 3, \begin{multline*} I_j(p_k)\le\exp\{p_k\ln(2p_ks_j)-p_j/2\}\le\exp\{p_k\ln(p_ks_k)-p_j/4\} \\ \le2^{-(j-k)}\exp\{p_k\ln(p_ks_k)\}, \end{multline*} so that \begin{equation*} \sum_{j>k}I_j(p_k) \le p_k^{2p_k}. \tag{3} \end{equation*}

Collecting (1), (2), (3), for large odd $k$ we have \begin{equation*} \ln\|f\|_{p_k}\lesssim 2\ln p_k. \end{equation*}

On the other hand, for large even $k$, by Lemma 2 and Stirling's formula,
\begin{equation*} \ln\|f\|_{p_k}\ge\frac1{p_k}\,\ln I_k(p_k)= \ln(s_k p_k^{1+o(1)}) = \ln(p_k^{2} p_k^{1+o(1)}) \sim3\ln p_k. \end{equation*}

So, $\ln\|f\|_p/\ln p$ does not converge as $p\to\infty$.

In conclusion, let us prove the lemmas.

Proof of Lemma 1. This is obvious.

Proof of Lemma 2. Write \begin{equation} x^p \,e^{-x/s}=e^{g(x)},\quad g(x):=p\ln x-x/s. \tag{4} \end{equation} Then $g'(ps)=0$ and $g''(x)$ is increasing in $x>0$. So, $g(ps-u)<g(ps+u)$ for $u\in(0,ps)$. So, for every $y\in[0,ps]$, \begin{equation*} \int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s \ge\int_y^\infty x^p \,e^{-x/s}\frac{dx}s \ge\frac12\,\int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s. \end{equation*} To complete the proof of Lemma 2, it remains to refer to Lemma 1.

Proof of Lemma 3. For $g$ as in (4) and for $x>2ps$, we have $g'(x)=\frac px-\frac1s\le-\frac1{2s}$ and hence \begin{equation} g(x)\le g(2ps)-\frac1{2s}(x-2ps)=p\ln(2ps)-\frac x{2s}. \end{equation} Now Lemma 3 easily follows.

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Iosif Pinelis
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

In the excellent answer, Willie Wong offered a construction of a measure with a lacunary support set, disproving the conjecture. Let me offer another construction, where the support of the measure is a continuum; however, the construction is still lacunary in the sense that it involves sequences diverging faster than any exponential sequence.

The main idea is this. Let $p>0$ and $s>0$. Let $\|f\|_p:=\|f\|_{L^p(\mu)}$. Let $f(x)=x$ for all real $x$. Let us start with this trivial lemma:

Lemma 1. If $\mu(dx)=\frac1s\,e^{-x/s}I\{x>0\}$, where $I$ denotes the indicator, then \begin{equation*} \|f\|_p^p=\int_0^\infty x^p \,e^{-x/s}\frac{dx}s=s^p\Ga(p+1). \end{equation*}

Then, by Stirling's formula, $\ln\|f\|_p\sim \ln(ps)$ as $s,p\to\infty$. So, if we alternate $s$ between $p$ and $p^2$, $\ln\|f\|_p/\ln p$ will alternate between $2$ and $3$. Now we need glue pieces of such two alternating sequences of measures, with $s$ alternating between $p$ and $p^2$, to get one measure one. The guiding idea in doing this is that for large $p$ the mass $m_p(dx):=x^p \,e^{-x/s}\frac{dx}s$ is mostly concentrated near the point $ps$.

To proceed, we shall need two more simple lemmas, which will be proved at the end of this answer.

Lemma 2. For every $y\in[0,ps]$ there is some $c\in[1/2,1]$ such that \begin{equation*} \int_y^\infty x^p \,e^{-x/s}\frac{dx}s=cs^p\Ga(p+1). \end{equation*}

Lemma 3. For every $y\ge2ps$, \begin{equation*} \int_y^\infty x^p \,e^{-x/s}\frac{dx}s \le\exp\{p\ln(2ps)-y/(2s)\}. \end{equation*}

Let \begin{equation*} \mu(dx)=\sum_{j=1}^\infty\frac1{s_j}\,e^{-x/s_j}I\{x_j<x<x_{j+1}\}, \end{equation*} where \begin{equation*} s_j:= \begin{cases} p_j&\text{ if $j$ is odd}, \\ p_j^2&\text{ if $j$ is even}, \end{cases} \qquad p_j:=2^{2^j},\qquad x_j:=p_j s_j, \end{equation*} so that $p_{j+1}=p_j^2$. Then \begin{multline*} \|f\|_p^p=\int_\R x^p\mu(dx)=\sum_{j=1}^\infty I_j(p),\\ I_j(p):=\int_{x_j}^{x_{j+1}} \exp\{g_{j,p}(x)\}\frac{dx}{s_j},\\ g_{j,p}(x):=p\ln x-x/s_j. \end{multline*}

In view of Lemma 1 and Stirling's formula, for large odd $k$ one has \begin{equation*} I_k(p_k) \le s_k^{p_k} p_k^{p_k}=p_k^{2p_k} \end{equation*} and, for $j<k$, \begin{equation*} I_j(p_k)\le s_j^{p_k} p_k^{p_k} \le p_j^{2p_k} p_k^{p_k} \le p_k^{2p_k}, \end{equation*} whence \begin{equation*} \sum_{j<k}I_j(p_k) \le k p_k^{2p_k}=p_k^{(2+o(1))p_k}. \end{equation*}

Next, for large odd $k$ and $j>k$, \begin{equation*} p_k\ln(2p_ks_j)-p_k\ln(p_ks_k) \le p_k\ln s_j\le p_k\ln(p_j^2)\le p_j^{1/2}\ln(p_j^2)\le p_j/4, \end{equation*} whence, by Lemma 3, \begin{multline*} I_j(p_k)\le\exp\{p_k\ln(2p_ks_j)-p_j/2\}\le\exp\{p_k\ln(p_ks_k)-p_j/4\} \\ \le2^{-(j-k)}\exp\{p_k\ln(p_ks_k)\}, \end{multline*} so that \begin{equation*} \sum_{j>k}I_j(p_k) \le p_k^{2p_k}. \end{equation*}

Collecting (1), (2), (3), for large odd $k$ we have \begin{equation*} \ln\|f\|_{p_k}\lesssim 2\ln p_k. \end{equation*}

On the other hand, for large even $k$, by Lemma 2 and Stirling's formula,
\begin{equation*} \ln\|f\|_{p_k}\ge\frac1{p_k}\,\ln I_k(p_k)= \ln(s_k p_k^{(1+o(1))}) = \ln(p_k^{2} p_k^{(1+o(1))}) \sim3\ln p_k. \end{equation*}

So, $\ln\|f\|_p/\ln p$ does not converge as $p\to\infty$.

In conclusion, let us prove the lemmas.

Proof of Lemma 1. This is obvious.

Proof of Lemma 2. Write \begin{equation} x^p \,e^{-x/s}=e^{g(x)},\quad g(x):=p\ln x-x/s. \tag{4} \end{equation} Then $g'(ps)=0$ and $g''$ is increasing. So, $g(ps-u)<g(ps+u)$ for $u\in(0,ps)$. So, for every $y\in[0,ps]$, \begin{equation*} \int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s \ge\int_y^\infty x^p \,e^{-x/s}\frac{dx}s \ge\frac12\,\int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s. \end{equation*} To complete the proof of Lemma 2, it remains to refer to Lemma 1.

Proof of Lemma 3. For $g$ as in (4) and for for $x>2ps$, we have $g'(x)=\frac px-\frac1s\le-\frac1{2s}$ and hence \begin{equation} g(x)\le g(2ps)-\frac1{2s}(x-2ps)=p\ln(2ps)-\frac x{2s}. \end{equation} Now Lemma 3 easily follows.