Return to Answer

Let $G$ be the set of functions $g\colon\mathbb R\to\mathbb R$ such that for some real $a$ and $b$ such that $0<a<b$ 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.

(To simplify the verification of condition 3, I have now replaced the condition $0\le a<b$ in the previous definition of the set $G$ by $0<a<b$. The result will of course hold with the original condition $0\le a<b$ as well, but then we'll need to invoke an appropriate convex separation theorem.)

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.

Let $G$ be the set of functions $g\colon\mathbb R\to\mathbb R$ such that for some real $a$ and $b$ such that $0<a<b$ 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} a:=\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 $a>0$ and $a|x|\le f(x)$ for all $x\in\mathbb R^N$, so that condition 3 indeed holds.

(To simplify the verification of condition 3, I have now replaced the condition $0\le a<b$ in the previous definition of the set $G$ by $0<a<b$. The result will of course hold with the original condition $0\le a<b$ as well, but then we'll need to invoke an appropriate convex separation theorem.)

Let $G$ be the set of functions $g\colon\mathbb R\to\mathbb R$ such that for some real $a$ and $b$ such that $0<a<b$ 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.

(To simplify the verification of condition 3, I have now replaced the condition $0\le a<b$ in the previous definition of the set $G$ by $0<a<b$. The result will of course hold with the original condition $0\le a<b$ as well, but then we'll need to invoke an appropriate convex separation theorem.)

Let $G$ be the set of functions $g\colon\mathbb R\to\mathbb R$ such that for some real $a$ and $b$ such that $0\le a<b$ 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.

Let $G$ be the set of functions $g\colon\mathbb R\to\mathbb R$ such that for some real $a$ and $b$ such that $0<a<b$ 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} a:=\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 $a>0$ and $a|x|\le f(x)$ for all $x\in\mathbb R^N$, so that condition 3 indeed holds.

(To simplify the verification of condition 3, I have now replaced the condition $0\le a<b$ in the previous definition of the set $G$ by $0<a<b$. The result will of course hold with the original condition $0\le a<b$ as well, but then we'll need to invoke an appropriate convex separation theorem.)