Proving Set is Connected and Complete

Question

Problem 1. Let (C([0,1])) denote the space of continuous real-valued functions on ([0,1]) equipped with the distance $$ d(f, g)=\sup _{x \in[0,1]}|f(x)-g(x)| $$ Let (X) be a subset of (C([0,1])) defined via (X={f:[0,1] $\rightarrow \mathbb{R}: f(0)=0 \text { and }|f(x)-f(y)| \leq|x-y|\}$ Show that (X) is connected and complete.

I have been having trouble understanding this question. In my "proof," I essentially said that since connected sets map to connected sets, then the set determined by $f(x)$ is connected. I have tried doing it by contradiction, but I am not getting anywhere.

Any clarification on what the problem actually wants would be great.

Thanks.

Answer 1

Why not proving that $X$ is path-connected?

For $f,g\in X$ and $t\in[0,1]$, we are to show that $tf+(1-t)g\in X$.

Note that $(tf+(1-t)g)(0)=0$ and \begin{align*} |(tf+(1-t)g)(x)-(tf+(1-t)g)(y)|&\leq t|f(x)-f(y)|+(1-t)|g(x)-g(y)|\\ &\leq t|x-y|+(1-t)|x-y|\\ &=|x-y|. \end{align*}

Answer 2

Hints: Show that $X$ is a closed subset of $C[0,1]$. This implies that $X$ is complete.

Show that$f,g \in X$ implies that $tf+(1-t)g \in X$ for $0\leq t \leq 1$. This implies that $t \to tf+(1-t)g $ is a path connecting $f$ and $g$. Hence $X$ is connected.