Linear Subspace is Convex Set

Theorem

Let $V$ be a vector space over $\R$ or $\C$, and let $L$ be a linear subspace of $V$.

Then $L$ is a convex set.

Proof

Let $x, y \in L, t \in \closedint 0 1$ be arbitrary.

Then, as $L$ is by definition closed under addition and scalar multiplication, it follows immediately that

$t x + \paren {1 - t} y \in L$

Hence $L$ is convex.

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality