I am reading a preliminary version of a paper which focuses on some minimization problems connected to a class of integral functionals. Reading the assumptions of one of the theorems I cannot convince myself that a "concrete" example exists (beside the "trivial" function, $x \mapsto \alpha \vert x \vert$ for some $\alpha >0$: the authors want to extend some previous result on this case to a more general framework).

Question. Does there exist a function $f \colon \mathbb R^N \to [0,+\infty)$ such that

  1. $f$ is convex;
  2. $f(\lambda x) = \lambda f(x)$ for any $\lambda >0$ and $\forall x \in \mathbb R^N$;
  3. there are $a>0, b \ge 0$ and $\gamma \in \mathbb R^N$ such that $$ a|x| \le f(x) + \langle \gamma, x \rangle + b $$ for any $x \in \mathbb R^N$?

I have some problems in finding a function satisfying the three points... It does not have to be smooth, still I do not see an example beside $\alpha |x|$, $\alpha >0$.

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    $\begingroup$ Does $f(x) = |x|$ not work? $\endgroup$ – πr8 Jan 15 '19 at 11:58
  • $\begingroup$ @πr8 Sure, I forgot to mention that I am looking for another function than the norm :-) You are perfectly right, thanks! $\endgroup$ – Y.B. Jan 15 '19 at 12:15
  • 1
    $\begingroup$ Then take $k|x|$, $k>0$. $\endgroup$ – Alexandre Eremenko Jan 15 '19 at 13:00
  • $\begingroup$ @AlexandreEremenko I see your point and thanks for the comment, yet I am looking for a completely "different" example. I have not written it in the OP (to avoid being too long and verbose) but this function $f$ plays the role of integrand of a min problem in CoV, i.e. $\min \int_\Omega f(Du) dx$ among suitable competitors $u$. I am not interested into the case $f(\cdot) = \vert \cdot \vert$ (which is well-known) and, as you can now see, multiplicative constants do not play any role. $\endgroup$ – Y.B. Jan 15 '19 at 13:18
  • $\begingroup$ The form of (3) suggests $a|x| - \langle \gamma, x\rangle$ as a possibility. Doesn't that work (for small enough $|\gamma|$)? $\endgroup$ – Yoav Kallus Jan 15 '19 at 14:07

Let $G$ be the set of functions $g\colon\mathbb R\to\mathbb R$ such that for some strictly positive real $a$ and $b$ and all real $x$ we have $g(x)=-ax$ if $x\le0$ and $g(x)=bx$ if $x\ge0$. Let $l_1,\dots,l_N$ be any linearly independent linear functionals on $\mathbb R^N$. Then any function $f$ on $\mathbb R^N$ of the form \begin{equation*} f=\sum_1^N g_i\circ l_i \end{equation*} with the $g_i$'s in $G$ will be nonnegative and satisfy your conditions 1, 2, 3. More generally, we can take \begin{equation*} f=\sum_1^n g_i\circ l_i \tag{0} \end{equation*} with the $g_i$'s in $G$, where $l_1,\dots,l_n$ are any linear functionals on $\mathbb R^N$ spanning $(\mathbb R^N)^*$.

Added in response to a comment by the OP: Here are details on why the so-constructed $f$ will satisfy condition 3. Let \begin{equation} c:=\inf_{x\ne0}\frac{f(x)}{|x|}=\min_{|x|=1}f(x), \tag{1} \end{equation} since $f$ is positive homogeneous and continuous. Suppose that $f(x)=0$ for some $x\in\mathbb R^N$. Since $g_i\ge0$, (0) implies $g_i(l_i(x))=0$ for all $i$. Since $g_i(u)=0\implies u=0$, we have $l_i(x)=0$ for all $i$, and hence $l(x)=0$ for all $l\in(\mathbb R^N)^*$, since the $l_i$'s span $(\mathbb R^N)^*$. So, $f(x)=0$ implies $x=0$. So, (1) implies $c>0$ and $c|x|\le f(x)$ for all $x\in\mathbb R^N$, so that condition 3 indeed holds.

  • $\begingroup$ Marvellous idea, thanks a lot! I have checked without problems the (non-strict) convexity and the positive homogeneity of degree 1. I fail to see why they also satisfy point (3), would you mind expanding a bit your answer about this point? Thank you very much for your interest and for your answer! $\endgroup$ – Y.B. Jan 15 '19 at 14:23
  • $\begingroup$ @Y.B. : I have added details on condition 3. $\endgroup$ – Iosif Pinelis Jan 15 '19 at 16:45
  • $\begingroup$ Thanks a lot for the details. Your post has been very helpful, as usual. $\endgroup$ – Y.B. Jan 16 '19 at 14:10

Another construction that works is the following: Take any convex $A\subset\mathbb{R}^n$ which contains a neighborhood of the origin. Then the associated Minkowski functional $$ \sigma_A(x) = \inf\{\lambda>0\mid x\in\lambda A\} $$ has the desired properties.

  • $\begingroup$ Nice idea, too! I had almost forgotten my old Functional Analysis class :-) Is it possibile to characterize the sets $A$ for which $\sigma_A$ is a strictly convex function?Thank you very much! $\endgroup$ – Y.B. Jan 16 '19 at 14:09
  • $\begingroup$ Yes, that's simple: $\sigma_A$ is never strictly convex (as it is positively homogeneous). $\endgroup$ – Dirk Jan 16 '19 at 14:52
  • $\begingroup$ That's right, thank you very much! $\endgroup$ – Y.B. Jan 16 '19 at 16:59
  • $\begingroup$ I am sorry to bother you again, but I now fail to see why $\sigma_A$ fulfills Property (3). Maybe we need that $A$ is open? I believe in this case probably one can choose $\gamma=0$ and also $b=0$, but I am not sure. Thanks. $\endgroup$ – Y.B. Jan 17 '19 at 15:23
  • $\begingroup$ You need that $A$ contains an open neighborhood of the origin, i.e. a ball $B_\epsilon(0)$. Then $|x|=1$ implies that $\epsilon x\in A$, i.e. $x\in \epsilon^{-1}A$, i.e. $\sigma_A(x)\geq \epsilon^{-1}$. By positive homogeneity this shows that indeed $\gamma=0$, $\beta=0$ and $a = \epsilon^{-1}$ work. $\endgroup$ – Dirk Jan 17 '19 at 15:59

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