# Example of convex functions fulfilling a (strange) lower bound

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 $$\alpha |x|$$, $$\alpha >0$$.

• Does $f(x) = |x|$ not work? – πr8 Jan 15 '19 at 11:58
• @πr8 Sure, I forgot to mention that I am looking for another function than the norm :-) You are perfectly right, thanks! – Y.B. Jan 15 '19 at 12:15
• Then take $k|x|$, $k>0$. – Alexandre Eremenko Jan 15 '19 at 13:00
• @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. – Y.B. Jan 15 '19 at 13:18
• The form of (3) suggests $a|x| - \langle \gamma, x\rangle$ as a possibility. Doesn't that work (for small enough $|\gamma|$)? – 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 $$$$c:=\inf_{x\ne0}\frac{f(x)}{|x|}=\min_{|x|=1}f(x), \tag{1}$$$$ since $$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.

• 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! – Y.B. Jan 15 '19 at 14:23
• @Y.B. : I have added details on condition 3. – Iosif Pinelis Jan 15 '19 at 16:45
• Thanks a lot for the details. Your post has been very helpful, as usual. – 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.

• 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! – Y.B. Jan 16 '19 at 14:09
• Yes, that's simple: $\sigma_A$ is never strictly convex (as it is positively homogeneous). – Dirk Jan 16 '19 at 14:52
• That's right, thank you very much! – Y.B. Jan 16 '19 at 16:59
• 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. – Y.B. Jan 17 '19 at 15:23
• 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. – Dirk Jan 17 '19 at 15:59