# Which Orlicz functions $f$ make the function $f^{-1}\left(\frac{\sum_{j=1}^s f(x_j)}{s}\right)$ convex?

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, non-decreasing 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.

• 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 – Paata Ivanishvili Jan 20 '19 at 21:32
• @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? – Samrat Mukhopadhyay Jan 21 '19 at 6:13
• 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. – Paata Ivanishvili Jan 21 '19 at 17:12
• I am so glad that you wrote this full answer. This is a huge help. Thanks! – Samrat Mukhopadhyay Jan 22 '19 at 6:44

It follows from Hardy--Littlewood--Polya (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.

Hardy--Littlewood--Polya PDF

• Thanks a lot for the answer! – Samrat Mukhopadhyay Jan 22 '19 at 6:46