Proving $H$ is a subspace of $C$

Question

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.

Answer 1

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.

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

Answer 2

Alternatively, evaluations $\mathrm{ev}_p:C([a,b]) \to \mathbb{R}$; $f \mapsto f(p)$ are obviously linear functionals, so that $l:=\mathrm{ev}_a-\mathrm{ev}_b$ is a linear functional. Since pre-images of subspaces by linear functions are subspaces, we have that $$H=l^{-1}(0)$$ is a subspace.