Let $ 0<a<b $, $ f\in C^1\left([0,b]\right)$. Assume that $ f $ is concave on $ [0,a] $ and convex on $ [a,b] $ with $ f'(0)>f'(b) $. Please prove that there exist $ n_0\in\mathbb{N} $ which is sufficiently large, such that for any $ n\geq n_0 $ and $ a_1+a_2+...+a_n=b $ with $ a_i\geq 0 $ ($ i=1,2,...,n $), we have $ \sum_{i=1}^nf(a_i)\leq nf\left(\frac{b}{n}\right) $.

Simple observation shows that if $ a_i\in [0,a] $ for all $ i $, the result is easy to prove. However, I cannot deal with the situation that there exists some $ i_0 $ such that $ a_{i_0}\in(0,b] $. I guess that such case is related with the statement that $ f'(0)>f'(b) $, but I do not know how to continue the proof. Could you give me some references or hints?


2 Answers 2


Denote $b/n=s$. We want a pointwise bound $$f(x)\leqslant f(s)+(x-s)f'(s),\label{1}\tag{$\heartsuit$}$$ then summing \eqref{1} up for $x=a_1,\ldots,a_n$ we get the desired inequality. Note that if $s<a$ (that holds for $n>b/a$) we get \eqref{1} on $[0,a]$ by concavity. For proving \eqref{1} on $[a,b]$, by convexity it suffices to verify \eqref{1} for $x=a$ and $x=b$. For $x=a$ this is already done, for $x=b$ it reads as $$ f(b)\leqslant f(s)+(b-s)f'(s). $$ When $n$ is large, RHS converges to $f(0)+bf'(0)$. Thus it suffices to check that $$ f(b)<f(0)+bf'(0). $$ Assume the contrary: $$ f(b)\geqslant f(0)+bf'(0). $$ We have $f(x)\leqslant f(0)+f'(0)x$ for all $x\in [0,a]$ by concavity. Denote by $c$ the endpoint of the maximal segment $[0,c]$ on which we have $f(x)\leqslant f(0)+f'(0)x$. Then $c\in [a,b]$ and we have $f(c)=f(0)+f'(0)c$ (otherwise $c$ is not maximal). This yields $$ f'(c)=\lim_{x\to c-0}\frac{f(c)-f(x)}{c-x}\geqslant f'(0). $$ Since $f'$ increases on $[a,b]$ by convexity we get $f'(b)\geqslant f'(c)\geqslant f'(0)$, a contradiction.


Without loss of generality (wlog), \begin{equation*} 0\le a_1\le\cdots\le a_p\le a\le a_{p+1}\le\cdots\le a_n\le b \tag{1}\label{1} \end{equation*} for some $p\in\{0,\dots,n\}$. Also, by approximation, wlog $f$ is strictly convex on $[a,b]$.

If $p=0$, then $a_i\ge a$ for all $i\in\{0,\dots,n\}$ and hence $b=\sum_1^n a_i\ge na$, which is impossible if \begin{equation*} n>b/a, \tag{2}\label{2} \end{equation*} which will be assumed henceforth. So, $p\ge1$. Letting now \begin{equation*} s:=\sum_1^p a_i \tag{3}\label{3} \end{equation*} and recalling that $f$ is concave on $[0,a]$, we have \begin{equation*} S:=\sum_{i=1}^n f(a_i)\le T:=pf(s/p)+\sum_{i=p+1}^n f(a_i). \tag{4}\label{4} \end{equation*}

It is enough to show that \begin{equation*} T\le nf(b/n) \tag{5}\label{5} \end{equation*} for all large enough $n$. Since $f$ is continuous on $[a,b]$, the sum $\sum_{i=p+1}^n f(a_i)$ attains a maximum over all $(a_{p+1},\dots,a_n)$ such that $a\le a_{p+1}\le\cdots\le a_n\le b$ and $\sum_{i=p+1}^n a_i=b-s$ (with $n,p,s$ fixed). In what follows, let $(a_{p+1},\dots,a_n)$ be a corresponding maximizer.

One of the following two cases takes place:

Case 1: $a_n=b$ or Case 2: $a_n<b$.

Consider Case 1. Then $a_1=\cdots=a_{n-1}=0$ and $p=n-1$, whence \eqref{5} can be rewritten as \begin{equation*} f(b)-f(0)\le n[f(b/n)-f(0)]. \tag{6}\label{6} \end{equation*} We have (i) $f'\le f'(0)$ on $[0,a]$, since $f$ is concave on $[0,a]$ and (ii) $f'\le f'(b)<f'(0)$ on $[a,b]$, since $f$ is convex on $[a,b]$. Therefore and because $a<b$, we have $f(b)-f(0)=\int_0^b f'<bf'(0)$. On the other hand, $n[f(b/n)-f(0)]\to bf'(0)$ as $n\to\infty$. So, for all large enough $n$, \eqref{6} holds, and hence \eqref{5} holds.

Now Consider Case 2. Then, by the strict convexity of $f$ on $[a,b]$, at most one of the coordinates $a_{p+1},\cdots,a_n$ of the maximizer $(a_{p+1},\dots,a_n)$ can be in the interval $(a,b)$, and hence at most one of the numbers $a_{p+1},\dots,a_n$ can be in the interval $(a,b]$. So,
\begin{equation*} T=g(s):=pf(s/p)+(n-p-1)f(a)+f(b-s-(n-p-1)a) \tag{7}\label{7} \end{equation*} and \begin{equation*} 0\le s\le pa,\quad n-p-1\ge0,\quad b-s-(n-p-1)a\ge a. \tag{8}\label{8} \end{equation*} The last inequality in \eqref{8} means \begin{equation} s\le b-(n-p)a. \tag{9}\label{9} \end{equation} For all $t\in[0,b-(n-p)a]$ we have $t/p\in[0,a]$ (by \eqref{2}) and $b-t-(n-p-1)a\in[a,b]$, so that, by the convexity of $f$ on $[0,a]$ and the convexity of $f$ on $[a,b]$,
\begin{equation*} g'(t)=f'(t/p)-f'(b-t-(n-p-1)a)\ge f'(a)-f'(a)=0, \end{equation*}
so that $g$ is nondecreasing on $[0,b-(n-p)a]$. So, by \eqref{7} and \eqref{9},
\begin{equation*} T\le g(b-(n-p)a)=pf\Big(\frac{b-(n-p)a}p\Big)+(n-p)f(a)\le nf\Big(\frac bn\Big), \end{equation*} since $f$ is concave on $[0,a]$ and $\frac{b-(n-p)a}p\in[0,a]$ (by \eqref{8} and \eqref{2}). Thus, \eqref{5} holds in Case 2 as well. $\quad\Box$


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