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I'm trying to check that certain examples of Young functions in the harmonic analysis literature are actually Young functions, and in doing so need to prove the following convexity-like inequality for $p > 1, \delta > 0$ and $0 \leq a \leq 1 < b$:

\begin{equation}\displaystyle \left( \frac{a+b}{2} \right)^p\left[ 1 + \ln\left( \frac{a+b}{2} \right) \right]^{p-1+\delta} \leq \frac{1}{2}a^p + \frac{1}{2} b^p[1+ \ln b]^{p-1+\delta} \label{ToProve} \end{equation}.

Clearly for fixed $b > 1$ this is true if $a = 0$ and $a = 1$ (the latter by the convexity of $\displaystyle x^p \left[1 + \ln x \right]^{p-1+\delta}$ for $x \geq 1$). Also, this is easy for all $0 \leq a \leq 1 < b$ assuming $p - 2 + \delta \geq 0$:

For fixed $0 \leq a \leq 1$ let $L(b)$ and $R(b)$ denote the left and right hand sides. Then then by convexity and the fact that $1 + \ln\left( \frac{a+1}{2} \right) \in (1-\ln 2, 1]$ we have \begin{align*} L(1) \leq \left( \frac12 a^p + \frac12 \right)\left[ 1 + \ln\left( \frac{a+1}{2} \right) \right]^{p-1+\delta} \leq \frac12 a^p + \frac12 = R(1). \end{align*}using basic calculus and recalling that $a < b$, we see that \begin{align*} L'(b) &= \frac{1}{2} \left(\frac{a+b}{2}\right)^{p-1}\left[p \left( 1+\ln\left(\frac{a+b}{2}\right)\right)^{p-1+\delta} + (p-1+\delta)\left( 1+\ln\left(\frac{a+b}{2}\right)\right)^{p-2+\delta} \right] \\ &\leq \frac12b^{p-1} \left[p \left( 1+\ln b\right)^{p-1+\delta} + (p-1+\delta)\left( 1+\ln b\right)^{p-2+\delta} \right] \\ &= R'(b), \end{align*} as desired. Here, we clearly need $p-2+\delta\geq0$ to use the fact that $\ln$ is increasing.

Any thoughts on how to remove this assumption? Does the inequality I seek follow from anything out in the literature?

P.S. For anyone curious, the Young functions in question are $\Phi(x) = x^p \left[1 + \ln^+ x \right]^{p-1+\delta}$ for $p > 1$ and $\delta > 0$, which induce Orlicz spaces that appear very naturally in modern harmonic analysis (see C. Perez "On sufficient conditions for...", Proc. London Math. Soc., (1995), and other papers that cite this).

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2 Answers 2

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Here is how to remove the assumption that $p-2+\delta\ge0$.

Let \begin{equation*} s:=p-1+\delta. \end{equation*} The conditions $p>1$ and $\delta>0$ imply $s\ge0$. No other conditions on $p$ and $s$ will be used or needed in what follows.

The inequality in question will follow from the inequality \begin{equation*} L\le R(s) \end{equation*} for \begin{equation*} s\in[0,\infty),\ p\in[1,\infty),\ 0<a\le1\le b, \end{equation*} where \begin{equation*} L:=2^{1-p} (a+b)^p,\\ R(s):=R(s,p):=a^p \Big(1+\ln\frac{a+b}2\Big)^{-s}+b^p \Big(\frac{1+\ln b}{1+\ln\frac{a+b}2}\Big)^s. \end{equation*} Clearly, $R(s)$ is convex in $s$.

Lemma 1: $R'(0)\ge0$, and hence $R(s)$ is increasing in $s$.

By Lemma 1 and the power-means inequality, \begin{equation*} R(s)\ge R(0)=a^p + b^p\ge 2^{1-p} (a+b)^p=L, \end{equation*} as desired.

It remains to provide

Proof of Lemma 1: Let \begin{equation*} R_1(p,a,b):=R'(0)/a^p \\ = \Big(\frac{b}{a}\Big)^p \ln\frac{1+\ln b}{1+\ln\frac{a+b}{2}} -\ln\Big(1+\ln\frac{a+b}{2}\Big), \tag{0}\label{0} \end{equation*} which is clearly nondecreasing in $p$. So, \begin{equation*} aR_1(p,a,b)\ge aR_1(1,a,b) \\ =b \ln (1+\ln b) -(a+b) \ln\Big(1+\ln\frac{a+b}{2}\Big), \end{equation*} which latter is clearly decreasing in $a$. So, \begin{equation*} aR_1(p,a,b)\ge aR_1(1,a,b)\ge R_1(1,1,b)=bg(b), \tag{1}\label{1} \end{equation*} where \begin{equation*} g(b):=\ln (1+\ln b) -\frac{1+b}b\,\ln\Big(1+\ln\frac{1+b}2\Big). \end{equation*} So, in view of \eqref{0}, it remains to prove that \begin{equation*} g(b)\overset{\text{(?)}}\ge0 \tag{2}\label{2} \end{equation*} (for real $b>1$).

To begin the proof of \eqref{2}, let \begin{equation*} g_1(b):=g'(b)b^2 \\ = b \Big(\frac{1}{1+\ln b}-\frac{1}{1+\ln\frac{1+b}2}\Big) +\ln\Big(1+\ln\frac{1+b}{2}\Big). \tag{3}\label{3} \end{equation*} Now use the substitution $b=2e^t-1$, so that $t>0$ and $\ln\frac{1+b}{2}=t$. Let \begin{equation*} \begin{aligned} G_2(t)&:=\frac{dg_1(2e^t-1)}{dt}\, \frac{(1+t)^2 (1+\ln (2 e^t-1))^2}{(2 e^t-1)t} \\ & =\frac{2 (t+e^t (1+t^2)) \ln (2e^t-1)}{(2 e^t-1) t}-1-\ln ^2(2 e^t-1). \end{aligned} \tag{4}\label{4} \end{equation*} Next, note that \begin{equation*} h(t):=t^3+t^2+t+2 e^t \left(t^2+1\right)-1>0 \end{equation*} (for $t>0$) and let \begin{equation*} \begin{aligned} G_3(t)&:=G_2'(t)\frac{t^2(2 e^t-1)^2}{2e^t h(t)} \\ & =\frac{2 t (t + e^t (1 + t^2))}{h(t)}- \ln(2e^t-1), \end{aligned} \tag{5}\label{5} \end{equation*} \begin{equation*} G_4(t):=G'_3(t)\frac{(2e^t-1)h(t)^2}{2t} \\ =2 - t + t^3 + 2 e^{2 t} (1 - 7 t - 4 t^2 - 2 t^3 - t^4 + t^5) - e^t (5 - 8 t + 6 t^3 + 3 t^4 + 2 t^5), \tag{5.5}\label{5.5} \end{equation*} \begin{equation*} G_5:=G_4',\ G_6:=G_5',\ G_7:=G_6', \tag{6}\label{6} \end{equation*} \begin{equation*} G_8(t):=G_7'(t)\frac{e^{-t}}{P(t)} =16 e^t \frac{Q(t)}{P(t)}-1, \tag{7}\label{7} \end{equation*} where \begin{equation*} P(t):=189 + 736 t + 768 t^2 + 294 t^3 + 43 t^4 + 2 t^5,\ \\ Q(t):=-65 - 91 t - 8 t^2 + 40 t^3 + 18 t^4 + 2 t^5. \end{equation*} Next, \begin{equation*} G'_8(t)=16 e^t \frac{S(t)}{P(t)^2}>0, \end{equation*} where \begin{equation*} S(t):=18356 + 31777 t + 25602 t^2 + 49850 t^3 + 84244 t^4 + 72903 t^5 + 34758 t^6 + 9548 t^7 + 1492 t^8 + 122 t^9 + 4 t^{10}. \end{equation*}

So, $G_8(t)$ is increasing (in $t>0$). Also, $G_8(0)<0$ and $G_8(\infty-):=\lim_{t\to\infty}G_8(t)=\infty>0$. So, $G_8$ is $-+$ (on $(0,\infty)$) -- that is, for some real $c>0$ we have $G_8<0$ on $(0,c)$ and $G_8>0$ on $(c,\infty)$.

So, by \eqref{7}, $G_7$ is down-up (on $(0,\infty)$) -- that is, for some real $c>0$ we have that $G_7$ is decreasing on $(0,c)$ and increasing on $(c,\infty)$. Also, $G_7(0)<0$ and $G_7(\infty-)=\infty>0$. So, $G_7$ is $-+$.

So, by \eqref{6}, $G_6$ is down-up. Also, $G_6(0)<0$ and $G_6(\infty-)=\infty>0$. So, $G_6$ is $-+$.

So, by \eqref{6}, $G_5$ is down-up. Also, $G_5(0)<0$ and $G_5(\infty-)=\infty>0$. So, $G_5$ is $-+$.

So, by \eqref{6}, $G_4$ is down-up. Also, $G_4(0)<0$ and $G_4(\infty-)=\infty>0$. So, $G_4$ is $-+$.

So, by \eqref{5.5}, $G_3$ is down-up. Also, $G_3(0)=0$ and $G_3(\infty-)=-\ln2<0$. So, $G_3<0$.

So, by \eqref{5}, $G_2$ is decreasing. Also, $G_2(0+):=\lim_{t\downarrow0}G_2(t)=3>0$ and $G_2(\infty-)=-\infty<0$. So, $G_2$ is $+-$ (on $(0,\infty)$) -- that is, for some real $c>0$ we have $G_2>0$ on $(0,c)$ and $G_2<0$ on $(c,\infty)$.

So, by \eqref{4}, $g_1$ is up-down -- that is, for some real $c>1$ we have that $g_1$ is increasing on $(1,c)$ and decreasing on $(c,\infty)$. Also, $g_1(1)=0$ and $g_1(\infty-)=-\infty<0$. So, $g_1$ is $+-$ (on $(1,\infty)$).

So, by \eqref{3}, $g$ is up-down. Also, $g(1+)=0$ and $g(\infty-)=0$. Thus, \eqref{2} follows. $\quad\Box$

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  • $\begingroup$ Wow, thanks for the quick response! I'll try to read this today or tomorrow and then count this as answered once I do. :) $\endgroup$ Nov 3, 2022 at 16:25
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Here is a much simpler proof, actually of the more general fact that $$f(x):=x^p(1+\ln^+ x)^s$$ is convex in $x\ge0$ for any real $p\ge1$ and $s\ge0$, where $\ln^+ x:=\ln\max(1,x)$.

For the left and right derivatives $f_-$ and $f_+$ of $f$ one has $f_+(1)=p+s\ge p=f_-(1)$.

So, it remains to show that $f''\ge0$ on $(0,1)$ and on $(1,\infty)$.

We have $f''(x)=p(p-1)x^{p-2}\ge0$ for $x\in(0,1)$.

Consider finally the case $x\in(1,\infty)$. Then $$f''(x)=x^{p-2}(1+\ln x)^{s-2}f_2(x),$$ where $$f_2(x):=F_2(p,s,\ln x)$$ and $$F_2(p,s,u):=(p-1) p u^2+u \left(2 p^2+2 p (s-1)-s\right)+p^2+p (2 s-1)+(s-2) s. $$

It remains to show that $$F_2(p,s,u)\ge0$$ for $p\ge1$, $s\ge0$, and $u\ge0$. Note that the partial derivative of $F_2(p,s,u)$ in $u$ at $u=0$ is $2 (p-1) p + (2 p-1) s\ge0$.

It remains to note that $$F_2(p,s,0)=(p-1) p + 2 (p-1) s + s^2\ge0.\quad\Box$$

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