Proving $H$ is a subspace of $C$ Let $C([a,b])$ be the vector space of all continuous functions on $[a,b]$ with the usual addition and scalar multiplication. Is $H=\{f\in C([a,b])\,\vert\,f(a)=f(b)\}\,$ a subspace of $C([a,b])$?
Intuitively, the answer would be yes. The following is what I have tried in proving such statement.
A subspace of the vector space $V$ is a subset $H$ that is nonempty, closed under addition and scalar multiplication:
$\bullet\,$ Picking the constant function $f=0\,$ in $C$ so that $f(a)=f(b)=0\,$ shows that the set is nonempty. 
$\bullet\,$ Picking the functions $g,t\in H$ leads to $$+\begin{cases}g(a)=g(b)\\t(a)=t(b)\end{cases}\implies g(a)+t(a)=g(b)+t(b)$$
which leads to $g+t\in H$ showing that $H$ is closed under addition.
$\bullet\,$ Picking the function $cs\in H$ with $c$ being a constant, leads to $cs(a)=cs(b)\Leftrightarrow s(a)=s(b)$
showing that $cs\in H$ and thus that $H$ is closed under scalar multiplication.
I am not sure if this logic is correct. Any help is appreciated.

Yes, this logic is correct. But beware that $cs(a)=cs(b) \iff s(a) = s(b)$ is predicated upon the fact that $c$ is not zero. You shouldn't use $\iff$ there.

In that step, you need to show that $s \in H \implies cs \in H$, there is no need for $\iff$.
