# Elementary convexity example

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 $$0 \leq a \leq 1 < b$$:

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

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).

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 $$\begin{equation*} s\in[0,\infty),\ p\in[1,\infty),\ 0 where $$\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. $$\quad\Box$$

• Wow, thanks for the quick response! I'll try to read this today or tomorrow and then count this as answered once I do. :) Nov 3, 2022 at 16:25

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 $$F_2(p,s,0)=(p-1) p + 2 (p-1) s + s^2\ge0.\quad\Box$$