Parrallelogram is convex

Let $S$ be the parallelogram consisting of all linear combinations of $t_{1}v_{1} + t_{2}v_{2}$ with $0 \leq t_{1} \leq $ and $0 \leq t_{2} \leq $, or equivlently $0 \leq t_{i} \leq $.

We remember that the line segment $PQ$ consists of all points $(1-t)P + tQ$ with $0\leq t \leq 1$, and that $PQ$ exists in vector space $S$ if all points $P, Q$ exist in $S$.

Proof. Let $P=t_{1}v_{1} + t_{2}v_{2}$ and $Q=t_{1}v_{1} + t_{2}v_{2}$ be points in $S$.

Then

$$(1-t)P + tQ = (1-t)(t_{1}v_{1} + t_{2}v_{2}) + t(s_{1}v_{1} + s_{2}v_{2})$$ $$=(1-t)(t_{1}v_{1} + (1-t)(t_{2}v_{2}) + t(s_{1}v_{1}) + t(s_{2}v_{2})$$ $$=r_{1}v_{1} + r_{2}v_{2}$$

where $r_{1} = (1-t)t_{1} + ts_{1}$ and $r_{2} = (1-t)t_{2} + ts_{2}$.

We have from the exposition that $0 \leq (1-t)t_{1} + ts_{1} \leq 1$ and $0 \leq (1-t)t_{2} + ts_{2} \leq 1$, so $(1-t)P + tQ = r_{1}v_{1} + r_{2}v_{2}$ with $0\leq r_{i} \leq 1$.

This proves that $(1-t)P + tQ$ is in the paralleogram, which is therefore convex. $\blacksquare$