Complete space of functions with $f(a)=f(b)$?

Question

Consider M is space of continuous functions (on $[a,b]$) with condition: $f(a)=f(b)$ . Is it complete metric space with $\mu(f,g) =\max\underset{x\in [a,b]}{|(f(x)-g(x))|} $?

In my opinion it's true. But I'm not understanding how I can prove that if $\forall n f_n(a)=f_n(b) \Rightarrow f(a)=f(b)$?

Thanks.

Answer 1

If $\mu(f_n,f) \to 0$ then $f_n(x) \to f(x)$ for every $x \in [a,b]$. In particular this is true for $x=a$ and $x=b$. Hence $f(a)=f(b)$. [In the equation $f_n(a)=f_n(b)$ take limit as $ n \to \infty$].

Answer 2

The map$$\begin{array}{rccc}\eta\colon&C\bigl([a,b]\bigr)&\longrightarrow&\mathbb R\\&f&\mapsto&f(b)-f(a)\end{array}$$is continuous. Therefore $M$ is a closed subset of $C\bigl([a,b]\bigr)$ ($M=\eta^{-1}(0)$). So, since $C\bigl([a,b]\bigr)$ is complete, $M$ is complete too.

And, since $\eta$ is continuous, $\lim_{n\to\infty}f_n=f\implies\lim_{n\to\infty}\eta(f_n)=\eta(f)$.