Let $f:\mathbb{R}_+\to\mathbb{R}_+$ be an Orlicz function, or sometimes referred to as an Young function, i.e. it is a convex, nondecreasing function such that $f(0)=0$. I am trying to study the convexity of the function $\phi:\mathbb{R}^n\to\mathbb{R}$, such that $\phi(\mathbf{x})=f^{1}\left(\frac{1}{s}\sum_{j=1}^s f({x}_j)\right)$. Note that when $f(x)=x^p,\ p\ge 1$, then $\phi(\mathbf{x})=\frac{\\mathbf{x}\_p}{s^{1/p}}$, which is a convex function. Similarly, a bit of calculation shows that the function $\phi$ is also convex when $f(x)=e^{ax}1$ for any $a>0$. So intuitively I thought that this result might hold for all Orlicz functions $f$. However, I have not been able to prove this result, and in fact, I think this does not hold for many functions, for example, $f(x)=e^{x^2}1$. Then, I am left with the investigation of such Orlicz functions $f$, that make the corresponding $\phi$ convex. However, assuming that $f$ is double differentiable, the Hessian of $\phi$ is turning out to be too difficult to analyze. At this point, I am not sure how to proceed to find the properties of $f$ that make $\phi$ convex. Can anyone kindly suggest some ideas, or point me to some relevant references that can help me proceed in this investigation? Thanks in advance.

1$\begingroup$ It is convex if in addition $f’/f’’$ is concave. I think this is also if and only if. There is a theorem of Hardy—Littlewood—Polya which says that if $F’>0, F’’>0$ and $F’/F’’$ is concave then the functional $h \mapsto F^{1}(\int_{\mathbb{R}^{n}} F(h)d\mu)$ is convex on the set of all nonnegative functions h. Here $d\mu$ is an arbitrary probability measure on $\mathbb{R}^{n}$. Now you just need to choose piecewise constant functions h to get the convexity $\endgroup$– Paata IvanishviliJan 20, 2019 at 21:32

$\begingroup$ @PaataIvanishvili, thanks! this is a big help for me. Can you also kindly mention the name of the theorem so that I can get more familiarized with it? $\endgroup$– Samrat MukhopadhyayJan 21, 2019 at 6:13

$\begingroup$ I decided to post the full answer. I am sorry that I did not write all the details in my previous comment. I was in a grocery store, I was typing through my iPhone, and it was almost impossible to write everything. $\endgroup$– Paata IvanishviliJan 21, 2019 at 17:12

$\begingroup$ I am so glad that you wrote this full answer. This is a huge help. Thanks! $\endgroup$– Samrat MukhopadhyayJan 22, 2019 at 6:44
1 Answer
It follows from HardyLittlewoodPolya (see section 3.16, page 86), that if for $x>0$ we have $f, f', f''>0$, and $f'/f''$ is concave, and $d\mu$ is a probability measure on the probability space $\Omega$, then the functional
$$ h \mapsto f^{1}\left(\int_{\Omega} f(h(\omega)) d\mu(\omega) \right) \qquad (*) $$ is convex on the set of nonnegative functions $h :\Omega \to \mathbb{R}_{+}$. In fact these is "if and only if" characterization (see the reference for the details).
Now to obtain the convexity that you are looking for, consider $\Omega = [0,1]$, and let $d\mu = dx$ be a standard Lebesgue measure on $[0,1]$. Next, partition $[0,1]$ into $s$ subintervals $I_{1}, \ldots, I_{s}$ of equal length $\frac{1}{s}$. Given $p=(p_{1}, \ldots, p_{s}) \in \mathbb{R}^{s}$, define $h_{p}(x) =p_{j}$ if $s \in I_{j}$. Next, let $q = (q_{1}, \ldots, q_{s})\in \mathbb{R}^{s}$. Then convexity of the functional (*) means $$ f^{1}\left(\int_{0}^{1} f\left(\frac{h_{p}(x)+h_{q}(x)}{2}\right)dx \right) \leq \frac{1}{2}\left(f^{1}\left(\int_{0}^{1} f\left(h_{p}(x)\right)dx \right)+f^{1}\left(\int_{0}^{1} f\left(h_{q}(x)\right)dx \right) \right). $$ On the other hand we have $$ f^{1}\left(\int_{0}^{1} f\left(h_{p}(x)\right)dx \right) = f^{1}\left(\frac{1}{s}\sum_{1\leq j \leq s} f(p_{j}) \right)\\ f^{1}\left(\int_{0}^{1} f\left(h_{q}(x)\right)dx \right) = f^{1}\left(\frac{1}{s}\sum_{1\leq j \leq s} f(q_{j}) \right)\\ $$ and $$ f^{1}\left(\int_{0}^{1} f\left(\frac{h_{p}(x)+h_{q}(x)}{2}\right)dx \right) = f^{1}\left(\frac{1}{s}\sum_{1\leq j \leq s} f\left(\frac{p_{j}+q_{j}}{2}\right) \right) \geq f^{1}\left(\frac{1}{s}\sum_{1\leq j \leq s} f\left(\frac{p_{j}+q_{j}}{2}\right) \right) $$ where the last inequality follows because $a+b\leq a+b$, and $f, f^{1}$ are strictly increasing.