Unit ball of the sum space 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 $\|\cdot\|_2$?

 A: 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).
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.
$\quad\Box$

One may want to note here that the semi-norm  $\|\cdot\|_+$ is the infimal convolution of the norms $\|\cdot\|_1$ and $\|\cdot\|_2$.
A: Iosif Pinelis neatly answered the question. I just want to add another viewpoint. The Minkowski functional (or gauge) $p_B(x)=\inf\{t>0: x\in tB\}$ of an absolutely convex absorbing set is a semi-norm which almost gives back $B$ since $\{p_B<1\}\subseteq B\subseteq \{p_B\le 1\}$.
Since two semi-norms $p_k$ on $V$ with unit balls $A_k=\{p_k\le 1\}$ satisfy $p_1\le p_2$ if and only if $A_2\subseteq A_1$, it is immediate that the Minkowski functional of the convex (=absolutely convex) hull of $B_1\cup B_2$ (where $B_k$ are the unit balls of $\|\cdot\|_k$) is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$ in the partially ordered set of semi-norms on $V$. One has thus to check that $\|\cdot\|_+$ is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$: On the one hand, $\|v\|_+\le\|v\|_k$ since $v=0+v=v+0$ and, on the other hand, if $p$ is a seminorm with $p\le\|\cdot\|_k$ the triangle inequality yields $p\le\|\cdot\|_+$.

I doubt, by the way, that $\|\cdot\|_+$ is always a proper norm
(it is of course a semi-norm): Any two separable Banach spaces have the same algebraic dimension (namely, the cardinality of $\mathbb R$, see, e.g.,
https://math.stackexchange.com/questions/1899365/hamel-dimension-of-infinite-dimensional-separable-banach-space,
however, for a simple example, consider $\ell^2\subseteq c_0$, so that dim$(\ell^2)\le$dim$(c_0)$, but on the other hand, $c_0\to\ell^2$, $(x_n)_{n\in\mathbb N}\mapsto (x_n/n)_{n\in\mathbb N}$ is a linear injection which implies dim$(c_0)\le$dim$(\ell^2)$, talking about Hamel dimension requires the axiom of choice). We can thus find two incomparable complete norms $\|\cdot\|_k$ on a vector space $V$ (e.g., $V=\ell^2$ with the usual norm $\|\cdot\|_2$ and $\|v\|_1=\|T(v)\|_{c_0}$ for a linear bijection $T:\ell_2\to c_0$ which can't be continuous since $\ell_2$ and $c_0$ are not isomorphic as Banach spaces). If $\|\cdot\|_+$ were a proper norm, the identical map $i:(V,\|\cdot\|_1)\to (V,\|\cdot\|_2)$ would have closed graph because it is continuous if both spaces are endowed with the coarser Hausdorff topology generated by $\|\cdot\|_+$ (if one prefers an elementary argument, for $x_n\to x$ in $(V;\|\cdot\|_1)$ with $i(x_n)\to y$ in $(V,\|\cdot\|_2)$ we have $i(x)=y$ because $\|\cdot\|_k$-convergence implies $\|\cdot\|_+$-convergence and $\|\cdot\|_+$-limits are unique). The closed graph theorem implies that $i$ is continuous so that $\|\cdot\|_2\le c\|\cdot\|_1$ for some constant $c$, contradicting the incomparability of the norms.
When talking about closed unit balls one should thus state precisely the topology in which the ball is closed.
