I was wondering if anyone could think of a proof that the set of all functions in C[a,b] satisfying the same Lipschitz condition is closed, and its closure is the set of all differentiable functions on [a,b] such that the abs value of the derivative f is less than K (the constant of the condition).
In addition, could someone show that M, the union of all the functions satisfying the condition for any K is not closed, though its closure is the whole space C[a,b]?
Thanks in advance for any help! My apologies for the length of the problem.
Answer to the second part. The set of all functions in $C[a,b]$ satisfying a Lipschitz condition (i.e. the union of all the functions satisfying the condition for any $K$) is not closed since by Stone-Weierstrass theorem every continuous function (even not Lipschitz) on $[a,b]$ can be uniformly approximated by polynomials (which are Lipschitz). For the same reason its closure is $C[a,b]$.
As regards the first part, if a sequence $(f_n)_n$ converges uniformly to $f$ and each $f_n$ is $K$-Lipschitz then $$|f(x)-f(y)|\leq 2||f-f_n||+|f_n(x)-f_n(y)|\leq 2||f-f_n||+K|x-y|.$$ By taking the limit as $n$ goes to infinity we obtain $$|f(x)-f(y)|\leq K|x-y|$$ and the set of $K$-Lipschitz is closed. Note that $f(x)=K|x-c|$ with $c=(a+b)/2$ is
$K$-Lipschitz but it is not differentiable at $c\in (a,b)$.