Proving a set of Lipschitz Continuous functions is closed Let $X:=C[0,1]$ denote the set of all continuous functions $f:[0,1]\to\mathbb{R}$. For $f,g\in C[0,1]$, define
$$d(f,g):=\max_{x\in[0,1]}|f(x)-f(y)|.$$
Consider the subset of X containing all Lipschitz continuous functions with Lipschitz constant $l$:
$$\Omega:=\{f\in C[0,1]\mid |f(x)-f(y)|\leq l|x-y|\}.$$
Prove that $\Omega$ is a closed set.
 A: Suppose $f$ is a limit point of $\Omega$. 
Let $\varepsilon>0$ be given. Since $f$ is a limit point of $\Omega$ there exists a sequence of functions $\{g_n\}_{n=1}^\infty$ in $\Omega$ such that $\max_{x \in [0, 1]} |f(x)-g_n(x)| \to 0$ as $n \to \infty$. In other words the sequence $\{g_n\}_{n=1}^\infty$ converges uniformly to $f$ on $[0,1]$. So there is a positive integer $N$ such that $|f(x)-g_n(x)|<\frac{\varepsilon}{2} \,$ for all $x \in [0, 1]$.
Thus \begin{equation}
\begin{split}
 |f(x)-f(y)|= & |f(x)-g_N(x)+g_N(x)-g_N(y)+g_N(y)-f(y)| \\
& \leq |f(x)-g_N(x)|+|g_N(x)-g_N(y)|+|g_N(y)-f(y)| \\
& < \,l|x-y|+\varepsilon.
\end{split} 
\end{equation}
Since this holds for all $\varepsilon>0$ we have that $ |f(x)-f(y)| \leq \, l|x-y|$, and thus $f \in \Omega$.
A: Hint: If $f_n \to f$ in $C[0,1],$ then for any pair $x,y\in [0,1],$ $f(y) - f(x) = \lim_{n\to \infty} (f_n(y) - f_n(x)).$
A: $\Omega$ is closed iff $\overline{\Omega} \subseteq \Omega$. Let $f \in C[0,1]$ such that every neighborhood of $f$ contains $g \in \Omega$. Let $\epsilon >0$; then $\max_{x \in [0,1]} |g(x) - f(x)| < \epsilon$. Let $x, y \in [0,1]$.
\begin{align*} 
|f(x)-f(y)| &=|f(x) +g(x)-g(x)+g(y)-g(y)-f(y)| \\
&\le |f(x)-g(x)|+|f(y)-g(y)|+|g(x)-g(y)| \\
&\le 2\epsilon +l|x-y|
\end{align*}
and now you conclude since $\epsilon$ is arbitrary.
