Q: Determine whether the set,

$\displaystyle S=\{f\in\\C[0,1]:\int_{0}^{1}\\f(x)dx=0\}$

is a subspace of $\displaystyle C[0,1]$

A: I said yes, because...

Firstly, the set contains the zero vector, since the set consists of all functions with zero area on the interval $\displaystyle C[0,1]$; thus, $\displaystyle S$ is non-empty.

Now, let $\displaystyle f,g\in{C[0,1]}$, then...

$\displaystyle \int_{0}^{1}\\(f+g)(x)dx\$$\displaystyle =\int_{0}^{1}\\(f(x)+g(x))dx\$$\displaystyle =

\int_{0}^{1}\\f(x)dx+\int_{0}^{1}\\g(x)dx=0\$

$\displaystyle \therefore$ $\displaystyle S$ is closed under addition.

Suppose again that $\displaystyle f\in{C[0,1]}$ and let $\displaystyle k\in\mathbb{R}$, then....

$\displaystyle k\int_{0}^{1}\\f(x)=\int_{0}^{1}\\kf(x)dx=0$ $\displaystyle (\forall{k}\in\mathbb{R})$

$\displaystyle \therefore$ $\displaystyle S$ is closed under scalar multiplication.

In general, is it true that a set made some function equal to zero will automatically contain the zero vector and be nonempty?