# 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.

• What did you try? – user99914 Nov 2 '15 at 23:22
• Not much, I'm stuck. Obviously I need to show that $\Omega^C$ is open, so I need to show that any open ball at an arbitrary point in the complement is a proper subset of the complement, but at that point it's so many layers of confusion... – Aidan Nov 2 '15 at 23:24
• Notice also that there is a (small) incongruence in the question you ask: you ask for closedness of the set of Lipschitz functions with Lipschitz constant $l$, but the $\Omega$ you define is the set of Lipschitz function with Lipschitz constant that does not exceed $l$. – Giovanni Nov 2 '15 at 23:25
• @Aidan: For future reference (as per the rules of thus site) you should outline an attempt at a solution or some context behind the question. – Shalop Nov 2 '15 at 23:26
• @Shalop I'll keep that in mind, thanks. I would have for this question but I'm with my study group and this question has seriously stumped all of us. Our exam prep (where this question originates) is significantly harder than the homework so far... – Aidan Nov 2 '15 at 23:32

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 $f \in \Omega$.
Hint: If $f_n \to f$ in $C[0,1],$ then for any pair $x,y\in [0,1],$ $f(y) - f(x) = \lim_{n\to \infty} (f_n(y) - f_n(x)).$