Let $V$ be a vector space and $\|\cdot \|\_1$ and $\|\cdot\|\_2$ two norms on $V$.


Let $\|\cdot\|\_+$ be given by
$$ \|v\|\_+ := \inf\_{v = v\_1 + v\_2} \|v\_1\|\_1 + \|v\_2\|\_2 $$
It is well-known that $\|\cdot\|\_+$ is a norm on $V$. ${}{}{}{}$


Is it true that the (closed) unit ball of $\|\cdot\|\_+$ is the closed convex hull of the union of those of $\|\cdot\|\_1$ and $\|\cdot\|\_2$?


* If so: where can I find a proof?
* If not: is there an easy characterization of the unit ball of $\|\cdot\|\_+$ in terms of that of $\|\cdot\|\_1$ and $\|\cdot\|\_2$?



Unit ball of the sum space
Let $B\_+,B\_1,B\_2$ denote the closed unit balls w.r. to $\|\cdot\|\_+,\|\cdot\|\_1,\|\cdot\|\_2$, respectively. Let $C$ be the convex hull of $B\_1\cup B\_2$. Let $\bar C$ be the closure of $C$ (w.r. to the "norm" $\|\cdot\|\_+$ -- which is actually only a semi-norm in general, as pointed out by [Jochen Wengenroth](https://mathoverflow.net/a/429306/36721)).


Take any $v\_1\in B\_1$. Then $v\_1=v\_1+0$ and $\|v\_1\|\_+\le\|v\_1\|\_1 + \|0\|\_2=\|v\_1\|\_1\le1$. So, $B\_1\subseteq B\_+$. Similarly, $B\_2\subseteq B\_+$. So, $C\subseteq B\_+$ and $\bar C\subseteq B\_+$.


Vice versa, take any $v\in B\_+^\circ$, the interior of $B\_+$, so that $\|v\|\_+<1$. Then there are some $v\_1$ and $v\_2$ in $V$ such that $v=v\_1+v\_2$ and $\|v\_1\|\_1 + \|v\_2\|\_2<1$. Let $t:=\|v\_1\|\_1$, so that $t\in[0,1)$ and $\|v\_2\|\_2<1-t\in(0,1]$.


If $t=0$, then $v=v\_2$, $\|v\|\_2=\|v\_2\|\_2<1$, and hence $v\in B\_2\subseteq\bar C$.


It remains to consider the case when $t\in(0,1)$. Then $v=tu\_1+(1-t)u\_2$, where $u\_1:=\frac{v\_1}t\in B\_1$ and $u\_2:=\frac{v\_2}{1-t}\in B\_2$.


So, in either case, $v\in C\subseteq\bar C$, for each $v\in B\_+^\circ$.


So, $B\_+^\circ\subseteq\bar C$ and hence $B\_+\subseteq\bar C$.


Thus, $B\_+=\bar C$, as desired.
$\quad\Box$




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One may want to note here that the semi-norm $\|\cdot\|\_+$ is the [infimal convolution](https://en.wikipedia.org/wiki/Convex_conjugate#Infimal_convolution) of the norms $\|\cdot\|\_1$ and $\|\cdot\|\_2$.


