Weakest condition on $f\colon \Bbb R^2\to \Bbb R$ so that $f(\|x\|_1,\|x\|_2)$ is a norm. $\newcommand{\norm}[1]{\|#1\|_1}\newcommand{\morm}[1]{\|#1\|_2}\newcommand{\xorm}[1]{\|#1\|_3}$

Let $X$ be a finite dimensional Banach space and $f\colon \Bbb R^2\to \Bbb R$. What is the weakest condition on $f$ that ensures that $\xorm{x}=f\big(\norm{x},\morm{x}\big)$ is a norm on $X$ for any choice of norms $\norm{\cdot},\morm{\cdot}$ on $X$?

I found the following assumptions and wonder if it is possible to do better:  


*

*$f(t,s)\geq 0$ with $\text{"$=$"}\iff s=t=0$.  

*$f$ is positively one homogeneous, $i.e. f(\lambda(s,t))=\lambda f(s,t)$ for every $\lambda >0$.

*$f(s+t,a+b)\leq f(s,a)+f(t,b)$ for every $a,b,s,t>0$.
Somehow this reminds me the notion of sublinearity.
EDIT Actually I realized that if $f$ satisfies these properties, then $f$ is a norm on $\{(s,t)\in\Bbb R^2\mid s,t>0\}\cup\{(0,0)\}$. Now, is this condition necessary?
 A: The conditions can not be weakend. We will show that each one is necessary


*

*Let $\left|\cdot\right|$ be any norm on $X$ and let $x\in X\setminus\left\{ 0\right\} $.
By rescaling, we can achieve $\left|x\right|=1$. For any $\alpha,\beta>0$,
we have that $\left\Vert \cdot\right\Vert _{1}:=\alpha\cdot\left|\cdot\right|$
and $\left\Vert \cdot\right\Vert _{2}:=\beta\cdot\left|\cdot\right|$
are norms on $X$, so that we must have
$$
0<\left\Vert x\right\Vert _{3}=f\left(\left\Vert x\right\Vert _{1},\left\Vert x\right\Vert _{2}\right)=f\left(\alpha\left|x\right|,\beta\left|x\right|\right)=f\left(\alpha,\beta\right).
$$
By choosing $\alpha,\beta$ accordingly, we see $f\left(t,s\right)>0$
for all $t,s>0$. Clearly, we must have $f\left(0,0\right)=0$.

*With the same choices as above, we get
$$
\lambda\cdot f\left(\alpha,\beta\right)=\lambda\cdot f\left(\alpha\left|x\right|,\beta\left|x\right|\right)=\lambda\left\Vert x\right\Vert _{3}=\left\Vert \lambda x\right\Vert _{3}=f\left(\left\Vert \lambda x\right\Vert _{1},\left\Vert \lambda x\right\Vert _{2}\right)=f\left(\lambda\alpha\left|x\right|,\lambda\beta\left|x\right|\right)=f\left(\lambda\left(\alpha,\beta\right)\right).
$$
Again, we can choose $\alpha,\beta>0$ suitably to get $\lambda\cdot f\left(s,t\right)=f\left(\lambda\left(s,t\right)\right)$
for all $s,t>0$ and $\lambda>0$.

*Here, we will assume $\dim\left(X\right)\geq2$ (otherwise this
condition is probably not necessary). Hence, there are linearly independent
$x,y$. Extend $\left(x,y\right)$ to a basis $\left(x,y,z_{1},\dots,z_{k}\right)$
of $X$ and define the norms $\left\Vert \cdot\right\Vert _{1},\left\Vert \cdot\right\Vert _{2}$
by
\begin{eqnarray*}
\left\Vert \left(x,y,z_{1},\dots,z_{k}\right)\left(\begin{matrix}v_{1}\\
v_{2}\\
\vdots\\
v_{k+2}
\end{matrix}\right)\right\Vert _{1} & := & \alpha_{1}\left|v_{1}\right|+\beta_{1}\left|v_{2}\right|+\left|v_{3}\right|+\dots+\left|v_{k+2}\right|,\\
\left\Vert \left(x,y,z_{1},\dots,z_{k}\right)\left(\begin{matrix}v_{1}\\
v_{2}\\
\vdots\\
v_{k+2}
\end{matrix}\right)\right\Vert _{2} & := & \alpha_{2}\left|v_{1}\right|+\beta_{2}\left|v_{2}\right|+\left|v_{3}\right|+\dots+\left|v_{k+2}\right|,
\end{eqnarray*}
for arbitrary $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}>0$. Here,
we get
\begin{eqnarray*}
f\left(\left(\alpha_{1},\alpha_{2}\right)+\left(\beta_{1},\beta_{2}\right)\right) & = & f\left(\alpha_{1}+\beta_{1},\alpha_{2}+\beta_{2}\right)\\
 & = & f\left(\left\Vert x+y\right\Vert _{1},\left\Vert x+y\right\Vert _{2}\right)\\
 & = & \left\Vert x+y\right\Vert _{3}\\
 & \leq & \left\Vert x\right\Vert _{3}+\left\Vert y\right\Vert _{3}\\
 & = & f\left(\left\Vert x\right\Vert _{1},\left\Vert x\right\Vert _{2}\right)+f\left(\left\Vert y\right\Vert _{1},\left\Vert y\right\Vert _{2}\right)\\
 & = & f\left(\alpha_{1},\alpha_{2}\right)+f\left(\beta_{1},\beta_{2}\right),
\end{eqnarray*}
which is the desired estimate.
