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Is $F=\left\lbrace f: f\in C_{\left[0, 1\right]}, f\left(0\right)=f\left(1\right)\right\rbrace$ complete metric space?
I know how to prove that $C_{\left[0, 1\right]}$ is complete. But I don't know how to use condition that $f\left(0\right)=f\left(1\right)$.
$F$ is a closed subspace of the complete space $\mathcal C_{\left[0, 1\right]}$. Hence it is complete.
Why is $F$ closed? Because $F$ is the inverse image of (the closed subset) $\{0\}$ under the continuous map $f \mapsto f(1)-f(0)$.
You only need to show that $F$ is a closed subspace of the complete space $C[0,1]$.
Consider a sequence $(f_n)_{n\in\mathbb{N}} \subset F$ with $f_n \stackrel{n\rightarrow \infty}{\Longrightarrow} f$ in $C[0,1]$.
Then $f_n(0) \stackrel{n\rightarrow \infty}{\longrightarrow} f(0)$ and $f_n(1) \stackrel{n\rightarrow \infty}{\longrightarrow} f(1)$.
So, $f \in F$ as well, because $0 = f_n(0)-f_n(1) \stackrel{n\rightarrow \infty}{\longrightarrow} f(0)-f(1) = 0$.
$F$ is closed in the pointwise topology on $C([0,1])$ and hence also in the sup-metric topology. A closed subspace of a complete metric space is complete in the inherited metric.
You have to take a Cauchy sequence in $F$ and show that it converges to a limit in $F$. Since $C([0,1])$ is complete, then taking such a sequence converges to a function $f$ in $C([0,1])$.
Now using the condition for the endpoints for each entry of the sequence you get that $f(0)=f(1)$, hence is in $F$. Thus your subspace is complete.
First try to prove that closed subspace of a metric space is complete.
Then try to show it is a closed subspace.
