Let $V$ be the set of real-valued continuous functions on the interval $[-3, 3]$. $S$ is set of real-valued functions with condition $f(-1) = f(1)$.
Is $S$ a subspace of $V$? Prove, and if not, why?
Let $V$ be the set of real-valued continuous functions on the interval $[-3, 3]$. $S$ is set of real-valued functions with condition $f(-1) = f(1)$.
Is $S$ a subspace of $V$? Prove, and if not, why?
Define $\phi:V \to \mathbb{R}$ by $\phi(f) = f(1)-f(-1)$. Note that $\phi$ is linear, and $S=\ker \phi$.
Suppose that $f$ and $g$ are two real-valued functions with domain $\left[ -3, 3 \right] $ satisfying $f(-1) = f(1)$ and $g(-1) = g(1)$. Then $(f + g)(-1) = f(-1) + g(+1) = f(1) + g(1) =(f+ g)(1)$. Pick $r \in \mathbb{R}$. Then $(rf)(-1) = rf(-1) = rf(1) = (rf)(1)$.