Translation of Convex Set in Vector Space is Convex
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Theorem
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $C \subseteq X$ be convex.
Let $x \in X$.
Then $C + x$, the translation of $C$ by $x$, is convex.
Proof
Let $t \in \closedint 0 1$ and $u, v \in C + x$.
Then there exists $u', v' \in C$ such that:
- $u = u' + x$
and:
- $v = v' + x$
Then, we have:
\(\ds t u + \paren {1 - t} v\) | \(=\) | \(\ds t u' + t x + \paren {1 - t} v' + \paren {1 - t} x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {t u' + \paren {1 - t} v'} + \paren {t + \paren {1 - t} } x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {t u' + \paren {1 - t} v'} + x\) |
Since $C$ is convex, we have:
- $t u' + \paren {1 - t} v' \in C$
so:
- $t u + \paren {1 - t} v \in C + x$
So we have that $C + x$ is convex.
$\blacksquare$