If A is a subset of R^n, and v is an element of R^n, then we define the translate of a set to be: v + A = (v + x : x is an element of A)
Is it necessarily true that v + conv(A) = conv(v + A)?
Since v is a single element of A, then that implies v = conv(v).
Based on this, we have conv(v) + conv(A) = conv(v + A), is this true, and is there a formula or mechanism that can be used to prove this statement?
Any help would be appreciated!
If $\lambda_k \ge 0$ and $\sum_k \lambda_k = 1$, then $v + \sum_k \lambda_k a_k = \sum_k \lambda_k (v + a_k)$ from which the result follows.
Here is another way that relies on the following facts: (i) $\operatorname{co} B$ is the smallest convex set containing $B$. (ii) If $L$ is linear and $C$ is convex, then $L(C)$ is convex. (iii) If $A,B$ are convex, then so is $A \times B$.
Then we see that for any $p$, $\{p\} + \operatorname{co} A $ is convex since $\{p\} \times \operatorname{co} A$ is convex and the map $(x,y) \to x+y$ is linear. In particular, the translates of a convex set by any $p$ are convex.
Since $\{v\} + A \subset \{v\} + \operatorname{co} A $ and the right hand side is convex, we see that $\operatorname{co} ( \{v\} + A ) \subset \{v\} + \operatorname{co} A $.
Now note that $A \subset (\{v\} +A ) + \{-v\} \subset \operatorname{co} ( \{v\} + A ) + \{-v\}$, and since the right hand side is convex, we have $ \operatorname{co} A \subset \operatorname{co} ( \{v\} + A ) + \{-v\}$.
Since $\{v \}$ is a singleton, we can add to both sides to get $\{v \} + \operatorname{co} A \subset \operatorname{co} ( \{v\} + A )$.