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Let $B_+,B_1,B_2$ denote the closed unit balls w.r. to the norms $\|\cdot\|_+,\|\cdot\|_1,\|\cdot\|_2$, 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 $\|\cdot\|_+$).

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 norm $\|\cdot\|_+$ is the infimal convolution of the norms $\|\cdot\|_1$ and $\|\cdot\|_2$.

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 $\|\cdot\|_2$.

Let $B_+,B_1,B_2$ denote the closed unit balls w.r. to the norms $\|\cdot\|_+,\|\cdot\|_1,\|\cdot\|_2$, 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 $\|\cdot\|_+$).

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 norm $\|\cdot\|_+$ is infimal convolution of the norms $\|\cdot\|_1$ and $\|\cdot\|_2$.

Let $B_+,B_1,B_2$ denote the closed unit balls w.r. to the norms $\|\cdot\|_+,\|\cdot\|_1,\|\cdot\|_2$, 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 $\|\cdot\|_+$).

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 norm $\|\cdot\|_+$ is the infimal convolution of the norms $\|\cdot\|_1$ and $\|\cdot\|_2$.

Let $B_+,B_1,B_2$ denote the closed unit balls w.r. to the norms $\|\cdot\|_+,\|\cdot\|_1,\|\cdot\|_2$, 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 $\|\cdot\|_+$).

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. $\quad\Box$

Let $B_+,B_1,B_2$ denote the closed unit balls w.r. to the norms $\|\cdot\|_+,\|\cdot\|_1,\|\cdot\|_2$, 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 $\|\cdot\|_+$).

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 norm $\|\cdot\|_+$ is infimal convolution of the norms $\|\cdot\|_1$ and $\|\cdot\|_2$.